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Simple Algebra Problems – Easy Exercises with Solutions for Beginners

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Understanding Algebraic Expressions

Breaking down algebra problems, solving algebraic equations, tackling algebra word problems, types of algebraic equations, algebra for different grades.

Simple Algebra Problems Easy Exercises with Solutions for Beginners

For instance, solving the equation (3x = 7) for (x) helps us understand how to isolate the variable to find its value.

I always find it fascinating how algebra serves as the foundation for more advanced topics in mathematics and science. Starting with basic problems such as ( $(x-1)^2 = [4\sqrt{(x-4)}]^2$ ) allows us to grasp key concepts and build the skills necessary for tackling more complex challenges.

So whether you’re refreshing your algebra skills or just beginning to explore this mathematical language, let’s dive into some examples and solutions to demystify the subject. Trust me, with a bit of practice, you’ll see algebra not just as a series of problems, but as a powerful tool that helps us solve everyday puzzles.

Simple Algebra Problems and Strategies

When I approach simple algebra problems, one of the first things I do is identify the variable.

The variable is like a placeholder for a number that I’m trying to find—a mystery I’m keen to solve. Typically represented by letters like ( x ) or ( y ), variables allow me to translate real-world situations into algebraic expressions and equations.

An algebraic expression is a mathematical phrase that can contain ordinary numbers, variables (like ( x ) or ( y )), and operators (like add, subtract, multiply, and divide). For example, ( 4x + 7 ) is an algebraic expression where ( x ) is the variable and the numbers ( 4 ) and ( 7 ) are terms. It’s important to manipulate these properly to maintain the equation’s balance.

Solving algebra problems often starts with simplifying expressions. Here’s a simple method to follow:

Combine like terms : Terms that have the same variable can be combined. For instance, ( 3x + 4x = 7x ).
Isolate the variable : Move the variable to one side of the equation. If the equation is ( 2x + 5 = 13 ), my job is to get ( x ) by itself by subtracting ( 5 ) from both sides, giving me ( 2x = 8 ).

With algebraic equations, the goal is to solve for the variable by performing the same operation on both sides. Here’s a table with an example:

Equation	Strategy	Solution
( x + 3 = 10 )	Subtract 3 from both sides	( x = 7 )

Algebra word problems require translating sentences into equations. If a word problem says “I have six less than twice the number of apples than Bob,” and Bob has ( b ) apples, then I’d write the expression as ( 2b – 6 ).

Understanding these strategies helps me tackle basic algebra problems efficiently. Remember, practice makes perfect, and each problem is an opportunity to improve.

In algebra, we encounter a variety of equation types and each serves a unique role in problem-solving. Here, I’ll brief you about some typical forms.

Linear Equations : These are the simplest form, where the highest power of the variable is one. They take the general form ( ax + b = 0 ), where ( a ) and ( b ) are constants, and ( x ) is the variable. For example, ( 2x + 3 = 0 ) is a linear equation.

Polynomial Equations : Unlike for linear equations, polynomial equations can have variables raised to higher powers. The general form of a polynomial equation is ( $a_nx^n + a_{n-1}x^{n-1} + … + a_2x^2 + a_1x + a_0 = 0$ ). In this equation, ( n ) is the highest power, and ( $a_n$ ), ( $a_{n-1} $), …, ( $a_0$ ) represent the coefficients which can be any real number.

Binomial Equations : They are a specific type of polynomial where there are exactly two terms. Like ($ x^2 – 4 $), which is also the difference of squares, a common format encountered in factoring.

To understand how equations can be solved by factoring, consider the quadratic equation ( $x^2$ – 5x + 6 = 0 ). I can factor this into ( (x-2)(x-3) = 0 ), which allows me to find the roots of the equation.

Here’s how some equations look when classified by degree:


1	Linear	( ax + b = 0 )
2	Quadratic	( a$x^2$ + bx + c = 0 )
3	Cubic	( a$x^3$ + b$x^2$ + cx + d = 0 )
n	Polynomial	( $a_nx^n$ + … +$ a_1x $+ a_0 = 0 )

Remember, identification and proper handling of these equations are essential in algebra as they form the basis for complex problem-solving.

In my experience with algebra, I’ve found that the journey begins as early as the 6th grade, where students get their first taste of this fascinating subject with the introduction of variables representing an unknown quantity.

I’ve created worksheets and activities aimed specifically at making this early transition engaging and educational.

6th Grade :

Concept	Description
Variables	Students learn to use letters to represent numbers.
Basic Equations	Solving for an unknown, such as ( x + 5 = 9 ), where ( x = 4 ).
Negative Numbers	Introduction to numbers less than zero is important for understanding a range of quantities.

Moving forward, the complexity of algebraic problems increases:

7th and 8th Grades :

Mastery of negative numbers: students practice operations like ( -3 – 4 ) or ( -5 $\times$ 2 ).
Exploring the rules of basic arithmetic operations with negative numbers.
Worksheets often contain numeric and literal expressions that help solidify their concepts.

Advanced topics like linear algebra are typically reserved for higher education. However, the solid foundation set in these early grades is crucial. I’ve developed materials to encourage students to understand and enjoy algebra’s logic and structure.

Remember, algebra is a tool that helps us quantify and solve problems, both numerical and abstract. My goal is to make learning these concepts, from numbers to numeric operations, as accessible as possible, while always maintaining a friendly approach to education.

I’ve walked through various simple algebra problems to help establish a foundational understanding of algebraic concepts. Through practice, you’ll find that these problems become more intuitive, allowing you to tackle more complex equations with confidence.

Remember, the key steps in solving any algebra problem include:

Identifying variables and what they represent.
Setting up the equation that reflects the problem statement.
Applying algebraic rules such as the distributive property ($a(b + c) = ab + ac$), combining like terms, and inverse operations.
Checking your solutions by substituting them back into the original equations to ensure they work.

As you continue to engage with algebra, consistently revisiting these steps will deepen your understanding and increase your proficiency. Don’t get discouraged by mistakes; they’re an important part of the learning process.

I hope that the straightforward problems I’ve presented have made algebra feel more manageable and a little less daunting. Happy solving!

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Solving Word Questions

With LOTS of examples!

In Algebra we often have word questions like:

Example: Sam and Alex play tennis.

On the weekend Sam played 4 more games than Alex did, and together they played 12 games.

How many games did Alex play?

How do we solve them?

The trick is to break the solution into two parts:

Turn the English into Algebra.

Then use Algebra to solve.

Turning English into Algebra

To turn the English into Algebra it helps to:

Read the whole thing first
Do a sketch if possible
Assign letters for the values
Find or work out formulas

You should also write down what is actually being asked for , so you know where you are going and when you have arrived!

Also look for key words:

When you see		Think
add, total, sum, increase, more, combined, together, plus, more than		+
minus, less, difference, fewer, decreased, reduced		−
multiplied, times, of, product, factor		×
divided, quotient, per, out of, ratio, percent, rate		÷
maximize or minimize		geometry formulas
rate, speed		distance formulas
how long, days, hours, minutes, seconds		time

Thinking Clearly

Some wording can be tricky, making it hard to think "the right way around", such as:

Example: Sam has 2 dollars less than Alex. How do we write this as an equation?

Let S = dollars Sam has
Let A = dollars Alex has

Now ... is that: S − 2 = A

or should it be: S = A − 2

or should it be: S = 2 − A

The correct answer is S = A − 2

( S − 2 = A is a common mistake, as the question is written "Sam ... 2 less ... Alex")

Example: on our street there are twice as many dogs as cats. How do we write this as an equation?

Let D = number of dogs
Let C = number of cats

Now ... is that: 2D = C

or should it be: D = 2C

Think carefully now!

The correct answer is D = 2C

( 2D = C is a common mistake, as the question is written "twice ... dogs ... cats")

Let's start with a really simple example so we see how it's done:

Example: A rectangular garden is 12m by 5m, what is its area ?

Turn the English into Algebra:

Use w for width of rectangle: w = 12m
Use h for height of rectangle: h = 5m

Formula for Area of a Rectangle : A = w × h

We are being asked for the Area.

A = w × h = 12 × 5 = 60 m 2

The area is 60 square meters .

Now let's try the example from the top of the page:

Example: Sam and Alex play Tennis. On the weekend Sam played 4 more games than Alex did, and together they played 12 games. How many games did Alex play?

Use S for how many games Sam played
Use A for how many games Alex played

We know that Sam played 4 more games than Alex, so: S = A + 4

And we know that together they played 12 games: S + A = 12

We are being asked for how many games Alex played: A

Which means that Alex played 4 games of tennis.

Check: Sam played 4 more games than Alex, so Sam played 8 games. Together they played 8 + 4 = 12 games. Yes!

A slightly harder example:

Example: Alex and Sam also build tables. Together they make 10 tables in 12 days. Alex working alone can make 10 in 30 days. How long would it take Sam working alone to make 10 tables?

Use a for Alex's work rate
Use s for Sam's work rate

12 days of Alex and Sam is 10 tables, so: 12a + 12s = 10

30 days of Alex alone is also 10 tables: 30a = 10

We are being asked how long it would take Sam to make 10 tables.

30a = 10 , so Alex's rate (tables per day) is: a = 10/30 = 1/3

Which means that Sam's rate is half a table a day (faster than Alex!)

So 10 tables would take Sam just 20 days.

Should Sam be paid more I wonder?

And another "substitution" example:

Example: Jenna is training hard to qualify for the National Games. She has a regular weekly routine, training for five hours a day on some days and 3 hours a day on the other days. She trains altogether 27 hours in a seven day week. On how many days does she train for five hours?

The number of "5 hour" days: d
The number of "3 hour" days: e

We know there are seven days in the week, so: d + e = 7

And she trains 27 hours in a week, with d 5 hour days and e 3 hour days: 5d + 3e = 27

We are being asked for how many days she trains for 5 hours: d

The number of "5 hour" days is 3

Check : She trains for 5 hours on 3 days a week, so she must train for 3 hours a day on the other 4 days of the week.

3 × 5 hours = 15 hours, plus 4 × 3 hours = 12 hours gives a total of 27 hours

Some examples from Geometry:

Example: A circle has an area of 12 mm 2 , what is its radius?

Use A for Area: A = 12 mm 2
Use r for radius

And the formula for Area is: A = π r 2

We are being asked for the radius.

We need to rearrange the formula to find the area

Example: A cube has a volume of 125 mm 3 , what is its surface area?

Make a quick sketch:

Use V for Volume
Use A for Area
Use s for side length of cube
Volume of a cube: V = s 3
Surface area of a cube: A = 6s 2

We are being asked for the surface area.

First work out s using the volume formula:

Now we can calculate surface area:

An example about Money:

Example: Joel works at the local pizza parlor. When he works overtime he earns 1¼ times the normal rate. One week Joel worked for 40 hours at the normal rate of pay and also worked 12 hours overtime. If Joel earned $660 altogether in that week, what is his normal rate of pay?

Joel's normal rate of pay: $N per hour
Joel works for 40 hours at $N per hour = $40N
When Joel does overtime he earns 1¼ times the normal rate = $1.25N per hour
Joel works for 12 hours at $1.25N per hour = $(12 × 1¼N) = $15N
And together he earned $660, so:

$40N + $(12 × 1¼N) = $660

We are being asked for Joel's normal rate of pay $N.

So Joel’s normal rate of pay is $12 per hour

Joel’s normal rate of pay is $12 per hour, so his overtime rate is 1¼ × $12 per hour = $15 per hour. So his normal pay of 40 × $12 = $480, plus his overtime pay of 12 × $15 = $180 gives us a total of $660

More about Money, with these two examples involving Compound Interest

Example: Alex puts $2000 in the bank at an annual compound interest of 11%. How much will it be worth in 3 years?

This is the compound interest formula:

So we will use these letters:

Present Value PV = $2,000
Interest Rate (as a decimal): r = 0.11
Number of Periods: n = 3
Future Value (the value we want): FV

We are being asked for the Future Value: FV

Example: Roger deposited $1,000 into a savings account. The money earned interest compounded annually at the same rate. After nine years Roger's deposit has grown to $1,551.33 What was the annual rate of interest for the savings account?

The compound interest formula:

Present Value PV = $1,000
Interest Rate (the value we want): r
Number of Periods: n = 9
Future Value: FV = $1,551.33

We are being asked for the Interest Rate: r

So the annual rate of interest is 5%

Check : $1,000 × (1.05) 9 = $1,000 × 1.55133 = $1,551.33

And an example of a Ratio question:

Example: At the start of the year the ratio of boys to girls in a class is 2 : 1 But now, half a year later, four boys have left the class and there are two new girls. The ratio of boys to girls is now 4 : 3 How many students are there altogether now?

Number of boys now: b
Number of girls now: g

The current ratio is 4 : 3

Which can be rearranged to 3b = 4g

At the start of the year there was (b + 4) boys and (g − 2) girls, and the ratio was 2 : 1

b + 4 g − 2 = 2 1

Which can be rearranged to b + 4 = 2(g − 2)

We are being asked for how many students there are altogether now: b + g

There are 12 girls !

And 3b = 4g , so b = 4g/3 = 4 × 12 / 3 = 16 , so there are 16 boys

So there are now 12 girls and 16 boys in the class, making 28 students altogether .

There are now 16 boys and 12 girls, so the ratio of boys to girls is 16 : 12 = 4 : 3 At the start of the year there were 20 boys and 10 girls, so the ratio was 20 : 10 = 2 : 1

And now for some Quadratic Equations :

Example: The product of two consecutive even integers is 168. What are the integers?

Consecutive means one after the other. And they are even , so they could be 2 and 4, or 4 and 6, etc.

We will call the smaller integer n , and so the larger integer must be n+2

And we are told the product (what we get after multiplying) is 168, so we know:

n(n + 2) = 168

We are being asked for the integers

That is a Quadratic Equation , and there are many ways to solve it. Using the Quadratic Equation Solver we get −14 and 12.

Check −14: −14(−14 + 2) = (−14)×(−12) = 168 YES

Check 12: 12(12 + 2) = 12×14 = 168 YES

So there are two solutions: −14 and −12 is one, 12 and 14 is the other.

Note: we could have also tried "guess and check":

We could try, say, n=10: 10(12) = 120 NO (too small)
Next we could try n=12: 12(14) = 168 YES

But unless we remember that multiplying two negatives make a positive we might overlook the other solution of (−14)×(−12).

Example: You are an Architect. Your client wants a room twice as long as it is wide. They also want a 3m wide veranda along the long side. Your client has 56 square meters of beautiful marble tiles to cover the whole area. What should the length of the room be?

Let's first make a sketch so we get things right!:

the length of the room: L
the width of the room: W
the total Area including veranda: A
the width of the room is half its length: W = ½L
the total area is the (room width + 3) times the length: A = (W+3) × L = 56

We are being asked for the length of the room: L

This is a quadratic equation , there are many ways to solve it, this time let's use factoring :

And so L = 8 or −14

There are two solutions to the quadratic equation, but only one of them is possible since the length of the room cannot be negative!

So the length of the room is 8 m

L = 8, so W = ½L = 4

So the area of the rectangle = (W+3) × L = 7 × 8 = 56

There we are ...

... I hope these examples will help you get the idea of how to handle word questions. Now how about some practice?

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Algebra Topics - Introduction to Word Problems

Algebra topics -, introduction to word problems, algebra topics introduction to word problems.

Algebra Topics: Introduction to Word Problems

Lesson 9: introduction to word problems.

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What are word problems?

A word problem is a math problem written out as a short story or scenario. Basically, it describes a realistic problem and asks you to imagine how you would solve it using math. If you've ever taken a math class, you've probably solved a word problem. For instance, does this sound familiar?

Johnny has 12 apples. If he gives four to Susie, how many will he have left?

You could solve this problem by looking at the numbers and figuring out what the problem is asking you to do. In this case, you're supposed to find out how many apples Johnny has left at the end of the problem. By reading the problem, you know Johnny starts out with 12 apples. By the end, he has 4 less because he gave them away. You could write this as:

12 - 4 = 8 , so you know Johnny has 8 apples left.

Word problems in algebra

If you were able to solve this problem, you should also be able to solve algebra word problems. Yes, they involve more complicated math, but they use the same basic problem-solving skills as simpler word problems.

You can tackle any word problem by following these five steps:

Read through the problem carefully, and figure out what it's about.
Represent unknown numbers with variables.
Translate the rest of the problem into a mathematical expression.
Solve the problem.
Check your work.

We'll work through an algebra word problem using these steps. Here's a typical problem:

The rate to rent a small moving van is $30 per day, plus $0.50 per mile. Jada rented a van to drive to her new home. It took two days, and the van cost $360. How many miles did she drive?

It might seem complicated at first glance, but we already have all of the information we need to solve it. Let's go through it step by step.

Step 1: Read through the problem carefully.

With any problem, start by reading through the problem. As you're reading, consider:

What question is the problem asking?
What information do you already have?

Let's take a look at our problem again. What question is the problem asking? In other words, what are you trying to find out?

The rate to rent a small moving van is $30 per day, plus $0.50 per mile. Jada rented a van to drive to her new home. It took 2 days, and the van cost $360. How many miles did she drive?

There's only one question here. We're trying to find out how many miles Jada drove . Now we need to locate any information that will help us answer this question.

There are a few important things we know that will help us figure out the total mileage Jada drove:

The van cost $30 per day.
In addition to paying a daily charge, Jada paid $0.50 per mile.
Jada had the van for 2 days.
The total cost was $360 .

Step 2: Represent unknown numbers with variables.

In algebra, you represent unknown numbers with letters called variables . (To learn more about variables, see our lesson on reading algebraic expressions .) You can use a variable in the place of any amount you don't know. Looking at our problem, do you see a quantity we should represent with a variable? It's often the number we're trying to find out.

Since we're trying to find the total number of miles Jada drove, we'll represent that amount with a variable—at least until we know it. We'll use the variable m for miles . Of course, we could use any variable, but m should be easy to remember.

Step 3: Translate the rest of the problem.

Let's take another look at the problem, with the facts we'll use to solve it highlighted.

The rate to rent a small moving van is $30 per day , plus $0.50 per mile . Jada rented a van to drive to her new home. It took 2 days , and the van cost $360 . How many miles did she drive?

We know the total cost of the van, and we know that it includes a fee for the number of days, plus another fee for the number of miles. It's $30 per day, and $0.50 per mile. A simpler way to say this would be:

$30 per day plus $0.50 per mile is $360.

If you look at this sentence and the original problem, you can see that they basically say the same thing: It cost Jada $30 per day and $0.50 per mile, and her total cost was $360 . The shorter version will be easier to translate into a mathematical expression.

Let's start by translating $30 per day . To calculate the cost of something that costs a certain amount per day, you'd multiply the per-day cost by the number of days—in other words, 30 per day could be written as 30 ⋅ days, or 30 times the number of days . (Not sure why you'd translate it this way? Check out our lesson on writing algebraic expressions .)

$30 per day and $.50 per mile is $360

$30 ⋅ day + $.50 ⋅ mile = $360

As you can see, there were a few other words we could translate into operators, so and $.50 became + $.50 , $.50 per mile became $.50 ⋅ mile , and is became = .

Next, we'll add in the numbers and variables we already know. We already know the number of days Jada drove, 2 , so we can replace that. We've also already said we'll use m to represent the number of miles, so we can replace that too. We should also take the dollar signs off of the money amounts to make them consistent with the other numbers.

30 ⋅ 2 + .5 ⋅ m = 360

Now we have our expression. All that's left to do is solve it.

Step 4: Solve the problem.

This problem will take a few steps to solve. (If you're not sure how to do the math in this section, you might want to review our lesson on simplifying expressions .) First, let's simplify the expression as much as possible. We can multiply 30 and 2, so let's go ahead and do that. We can also write .5 ⋅ m as 0.5 m .

60 + .5m = 360

Next, we need to do what we can to get the m alone on the left side of the equals sign. Once we do that, we'll know what m is equal to—in other words, it will let us know the number of miles in our word problem.

We can start by getting rid of the 60 on the left side by subtracting it from both sides .

60	+ .5m =	360
-60		-60

The only thing left to get rid of is .5 . Since it's being multiplied with m , we'll do the reverse and divide both sides of the equation with it.

.5m	=	300
.5		.5

.5 m / .5 is m and 300 / 0.50 is 600 , so m = 600 . In other words, the answer to our problem is 600 —we now know Jada drove 600 miles.

Step 5: Check the problem.

To make sure we solved the problem correctly, we should check our work. To do this, we can use the answer we just got— 600 —and calculate backward to find another of the quantities in our problem. In other words, if our answer for Jada's distance is correct, we should be able to use it to work backward and find another value, like the total cost. Let's take another look at the problem.

According to the problem, the van costs $30 per day and $0.50 per mile. If Jada really did drive 600 miles in 2 days, she could calculate the cost like this:

$30 per day and $0.50 per mile

30 ⋅ day + .5 ⋅ mile

30 ⋅ 2 + .5 ⋅ 600

According to our math, the van would cost $360, which is exactly what the problem says. This means our solution was correct. We're done!

While some word problems will be more complicated than others, you can use these basic steps to approach any word problem. On the next page, you can try it for yourself.

Let's practice with a couple more problems. You can solve these problems the same way we solved the first one—just follow the problem-solving steps we covered earlier. For your reference, these steps are:

If you get stuck, you might want to review the problem on page 1. You can also take a look at our lesson on writing algebraic expressions for some tips on translating written words into math.

Try completing this problem on your own. When you're done, move on to the next page to check your answer and see an explanation of the steps.

A single ticket to the fair costs $8. A family pass costs $25 more than half of that. How much does a family pass cost?

Here's another problem to do on your own. As with the last problem, you can find the answer and explanation to this one on the next page.

Flor and Mo both donated money to the same charity. Flor gave three times as much as Mo. Between the two of them, they donated $280. How much money did Mo give?

Problem 1 Answer

Here's Problem 1:

A single ticket to the fair costs $8. A family pass costs $25 more than half that. How much does a family pass cost?

Answer: $29

Let's solve this problem step by step. We'll solve it the same way we solved the problem on page 1.

Step 1: Read through the problem carefully

The first in solving any word problem is to find out what question the problem is asking you to solve and identify the information that will help you solve it . Let's look at the problem again. The question is right there in plain sight:

So is the information we'll need to answer the question:

A single ticket costs $8 .
The family pass costs $25 more than half the price of the single ticket.

Step 2: Represent the unknown numbers with variables

The unknown number in this problem is the cost of the family pass . We'll represent it with the variable f .

Step 3: Translate the rest of the problem

Let's look at the problem again. This time, the important facts are highlighted.

A single ticket to the fair costs $8 . A family pass costs $25 more than half that . How much does a family pass cost?

In other words, we could say that the cost of a family pass equals half of $8, plus $25 . To turn this into a problem we can solve, we'll have to translate it into math. Here's how:

First, replace the cost of a family pass with our variable f .

f equals half of $8 plus $25

Next, take out the dollar signs and replace words like plus and equals with operators.

f = half of 8 + 25

Finally, translate the rest of the problem. Half of can be written as 1/2 times , or 1/2 ⋅ :

f = 1/2 ⋅ 8 + 25

Step 4: Solve the problem

Now all we have to do is solve our problem. Like with any problem, we can solve this one by following the order of operations.

f is already alone on the left side of the equation, so all we have to do is calculate the right side.
First, multiply 1/2 by 8 . 1/2 ⋅ 8 is 4 .
Next, add 4 and 25. 4 + 25 equals 29 .

That's it! f is equal to 29. In other words, the cost of a family pass is $29 .

Step 5: Check your work

Finally, let's check our work by working backward from our answer. In this case, we should be able to correctly calculate the cost of a single ticket by using the cost we calculated for the family pass. Let's look at the original problem again.

We calculated that a family pass costs $29. Our problem says the pass costs $25 more than half the cost of a single ticket. In other words, half the cost of a single ticket will be $25 less than $29.

We could translate this into this equation, with s standing for the cost of a single ticket.

1/2s = 29 - 25

Let's work on the right side first. 29 - 25 is 4 .
To find the value of s , we have to get it alone on the left side of the equation. This means getting rid of 1/2 . To do this, we'll multiply each side by the inverse of 1/2: 2 .

According to our math, s = 8 . In other words, if the family pass costs $29, the single ticket will cost $8. Looking at our original problem, that's correct!

So now we're sure about the answer to our problem: The cost of a family pass is $29 .

Problem 2 Answer

Here's Problem 2:

Answer: $70

Let's go through this problem one step at a time.

Start by asking what question the problem is asking you to solve and identifying the information that will help you solve it . What's the question here?

To solve the problem, you'll have to find out how much money Mo gave to charity. All the important information you need is in the problem:

The amount Flor donated is three times as much the amount Mo donated
Flor and Mo's donations add up to $280 total

The unknown number we're trying to identify in this problem is Mo's donation . We'll represent it with the variable m .

Here's the problem again. This time, the important facts are highlighted.

Flor and Mo both donated money to the same charity. Flor gave three times as much as Mo . Between the two of them, they donated $280 . How much money did Mo give?

The important facts of the problem could also be expressed this way:

Mo's donation plus Flor's donation equals $280

Because we know that Flor's donation is three times as much as Mo's donation, we could go even further and say:

Mo's donation plus three times Mo's donation equals $280

We can translate this into a math problem in only a few steps. Here's how:

Because we've already said we'll represent the amount of Mo's donation with the variable m , let's start by replacing Mo's donation with m .

m plus three times m equals $280

Next, we can put in mathematical operators in place of certain words. We'll also take out the dollar sign.

m + three times m = 280

Finally, let's write three times mathematically. Three times m can also be written as 3 ⋅ m , or just 3 m .

m + 3m = 280

It will only take a few steps to solve this problem.

To get the correct answer, we'll have to get m alone on one side of the equation.
To start, let's add m and 3 m . That's 4 m .
We can get rid of the 4 next to the m by dividing both sides by 4. 4 m / 4 is m , and 280 / 4 is 70 .

We've got our answer: m = 70 . In other words, Mo donated $70 .

The answer to our problem is $70 , but we should check just to be sure. Let's look at our problem again.

If our answer is correct, $70 and three times $70 should add up to $280 .

We can write our new equation like this:

70 + 3 ⋅ 70 = 280

The order of operations calls for us to multiply first. 3 ⋅ 70 is 210.

70 + 210 = 280

The last step is to add 70 and 210. 70 plus 210 equals 280 .

280 is the combined cost of the tickets in our original problem. Our answer is correct : Mo gave $70 to charity.

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Education and Communications
Mathematics

How to Solve Word Problems in Algebra

Last Updated: December 19, 2022 Fact Checked

This article was co-authored by Daron Cam . Daron Cam is an Academic Tutor and the Founder of Bay Area Tutors, Inc., a San Francisco Bay Area-based tutoring service that provides tutoring in mathematics, science, and overall academic confidence building. Daron has over eight years of teaching math in classrooms and over nine years of one-on-one tutoring experience. He teaches all levels of math including calculus, pre-algebra, algebra I, geometry, and SAT/ACT math prep. Daron holds a BA from the University of California, Berkeley and a math teaching credential from St. Mary's College. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 75,195 times.

You can solve many real world problems with the help of math. In order to familiarize students with these kinds of problems, teachers include word problems in their math curriculum. However, word problems can present a real challenge if you don't know how to break them down and find the numbers underneath the story. Solving word problems is an art of transforming the words and sentences into mathematical expressions and then applying conventional algebraic techniques to solve the problem.

Assessing the Problem

For example, you might have the following problem: Jane went to a book shop and bought a book. While at the store Jane found a second interesting book and bought it for $80. The price of the second book was $10 less than three times the price of he first book. What was the price of the first book?
In this problem, you are asked to find the price of the first book Jane purchased.

Step 3 Summarize what you know, and what you need to know.

For example, you know that Jane bought two books. You know that the second book was $80. You also know that the second book cost $10 less than 3 times the price of the first book. You don't know the price of the first book.

Step 4 Assign variables to the unknown quantities.

Multiplication keywords include times, of, and f actor. [9] X Research source
Division keywords include per, out of, and percent. [10] X Research source
Addition keywords include some, more, and together. [11] X Research source
Subtraction keywords include difference, fewer, and decreased. [12] X Research source

Finding the Solution

Completing a Sample Problem

Robyn and Billy run a lemonade stand. They are giving all the money that they make to a cat shelter. They will combine their profits from selling lemonade with their tips. They sell cups of lemonade for 75 cents. Their mom and dad have agreed to double whatever amount they receive in tips. Write an equation that describes the amount of money Robyn and Billy will give to the shelter.

Step 2 Read the problem carefully and determine what you are asked to find.

Since you are combining their profits and tips, you will be adding two terms. So, x = __ + __.

$.75c$

Expert Q&A

Word problems can have more than one unknown and more the one variable. Thanks Helpful 2 Not Helpful 1
The number of variables is always equal to the number of unknowns. Thanks Helpful 1 Not Helpful 0
While solving word problems you should always read every sentence carefully and try to extract all the numerical information. Thanks Helpful 1 Not Helpful 0

Find the Maximum or Minimum Value of a Quadratic Function Easily

↑ Daron Cam. Academic Tutor. Expert Interview. 29 May 2020.
↑ http://www.purplemath.com/modules/translat.htm
↑ https://www.mathsisfun.com/algebra/word-questions-solving.html
↑ https://www.wtamu.edu/academic/anns/mps/math/mathlab/int_algebra/int_alg_tut8_probsol.htm
↑ http://www.virtualnerd.com/algebra-1/algebra-foundations/word-problem-equation-writing.php
↑ https://www.khanacademy.org/test-prep/praxis-math/praxis-math-lessons/praxis-math-algebra/a/gtp--praxis-math--article--algebraic-word-problems--lesson

About This Article

To solve word problems in algebra, start by reading the problem carefully and determining what you’re being asked to find. Next, summarize what information you know and what you need to know. Then, assign variables to the unknown quantities. For example, if you know that Jane bought 2 books, and the second book cost $80, which was $10 less than 3 times the price of the first book, assign x to the price of the 1st book. Use this information to write your equation, which is 80 = 3x - 10. To learn how to solve an equation with multiple variables, keep reading! Did this summary help you? Yes No

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Problem of the Week has wrapped up for the 2023-2024 school year. Below you can find all problems and solutions that were sent. Problem of the Week will resume in September 2024.

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Visit the archive page to access links to booklets containing all the problems and solutions from the last two years. The problems are organized into themes, grouping problems into various areas of the curriculum.

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Problems are organized into five themes: Algebra (A), Computational Thinking (C), Data Management (D), Geometry and Measurement (G), Number Sense (N). A problem may have more than one theme.


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factor\:x^{2}-5x+6
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The mystic and the mathematician: What the towering 20th-century thinkers Simone and André Weil can teach today’s math educators

Professor of Mathematics, Colby College

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Scott Taylor receives funding from the National Science Foundation and the John and Mary Neff Foundation.

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On the left is a photo of André Weil as an adult. On the right is a photo of Simone Weil as a teenager.

Like most mathematicians , I hear confessions from complete strangers: the inevitable “I was always bad at math.” I suppress the response, “You are forgiven, my child.”

Why does it feel like a sin to struggle in math? Why are so many traumatized by their mathematics education? Is learning math worthwhile?

Sometimes agreeing and sometimes disagreeing, André and Simone Weil were the sort of siblings who would argue about such questions. André achieved renown as a mathematician; Simone was a formidable philosopher and mystic. André focused on applying algebra and geometry to deep questions about the structures of whole numbers, while Simone was concerned with how the world can be soul-crushing.

Both wrestled with the best way to teach math. Their insights and contradictions point to the fundamental role that mathematics and mathematics education play in human life and culture.

André Weil’s rigorous mathematics

André Weil, aged about 50, stands on a sunny path. A child is standing next to him shielding their face.

Unlike the prominent French mathematicians of previous generations, André, who was born in 1906 and died in 1998, spent little time philosophizing. For him, mathematics was a living subject endowed with a long and substantial history, but as he remarked, he saw “ no need to defend (it).”

In his interactions with people, André was an unsparing critic. Although admired by some colleagues, he was feared by and at times disdainful of his students. He co-founded the Bourbaki mathematics collective that used abstraction and logical rigor to restructure mathematics from the ground up.

Nicolas Bourbaki’s commitment to proceeding from first principles, however, did not completely encapsulate his conception of what constituted worthwhile mathematics. André was attuned to how math should be taught differently to different audiences.

Tempering the Bourbaki spirit, he defined rigor as “(not) proving everything, but … endeavoring to assume as little as possible at every stage.”

A black and white photo of one woman and six men standing in front of a doorway.

In other words, absolute rigor has its place, but teachers must be willing to take their audience into account. He believed that teachers must motivate students by providing them meaningful problems and provocative examples. Excitement for advanced students comes in encountering the unknown; for beginning students, it emerges from solving questions of, as he put it, “theoretical or practical importance.” He insisted that math “must be a source of intellectual excitement.”

André’s own sense of intellectual excitement came from applying insights from one part of mathematics to other parts. In a letter to his sister , André described his work as seeking a metaphorical “ Rosetta stone ” of analogies between advanced versions of three basic mathematical objects: numbers, polynomials and geometric spaces.

André described mathematics in romantic terms . Initially, the relationship between the different parts of mathematics is that of passionate lovers, exchanging “furtive caresses” and having “inexplicable quarrels.” But as the analogies eventually give way to a single unified theory, the affair grows cold: “Gone is the analogy: gone are the two theories, their conflicts and their delicious reciprocal reflections … alas, all is just one theory, whose majestic beauty can no longer excite us.”

Despite being passionless, this theory that unifies numbers, polynomials and geometry gets to the heart of mathematics; André pursued it intensely. In the words of a colleague, André sought the “ real meaning of every basic mathematical phenomenon.” For him, unlike his sister, this real meaning was found in the careful definitions, precisely articulated theorems and rigorous proofs of the most advanced mathematics of his time. Romantic language simply described the emotions of the mathematician encountering the mathematics; it did not point to any deeper significance.

Simone Weil and the philosophy of mathematics

On the other hand, Simone, who was born three years after André and died 55 years before him, used philosophy and religion to investigate the value of mathematics for nongeniuses, in addition to her work on politics, war, science and suffering.

All of her writing – indeed, her life – has a maddening quality to it. In her polished essays, as well as her private letters and journals, she will often make an extreme assertion or enigmatic comment. Such assertions might concern the motivations of scientists, the psychological state of a sufferer, the nature of labor, an analysis of labor unions or an interpretation of Greek philosophers and mathematicians. She is not a systematic thinker but rather circles around and around clusters of ideas and themes. When I read her writing, I am often taken aback. I start to argue with her, bringing up counterexamples and qualifications, but I eventually end up granting the essence of her point. Simone was known for the single-minded pursuit of her ideals.

Despite the discomfort her viewpoints provoke, they are worth engaging. Although her childhood was largely happy , her whole life she felt stupid in comparison with her brother. She channeled her feelings of inadequacy into an exploration of how to experience a meaningful existence in the face of oppression and affliction. Over her life, she developed an interpretation of beauty and suffering intertwined with geometry.

Along with her lifelong mathematical discussions with André, her views were influenced by one of her first jobs as a teacher . In a letter to a colleague, she described her pupils as struggling because they “regarded the various sciences as compilations of cut-and-dried knowledge.” Like André, Simone saw the ability to motivate students as the key to good teaching. She taught mathematics as a subject embedded in culture, emphasizing overarching historical themes. Even those students who were “most ignorant in science” followed her lectures with “passionate interest.”

For Simone, however, the primary purpose of mathematics education was to develop the virtue of attention. Mathematics confronts us with our mistakes, and the contemplation of these inadequacies brings the ability to concentrate on one thing, at the exclusion of all else, to the fore. As a math teacher, I frequently see students grit their teeth and furrow their brow, developing only a headache and resentment. According to Simone, however, true attention arises from joy and desire. We hold our knowledge lightly and wait with detached thought for light to arrive.

For Simone, the “first duty” of teachers is to help students develop, through their studies, the ability to apprehend God, which she conceptualized as a blending of Plato’s description of the ultimate Good with Christian conceptions of the self-abnegating God. A true understanding of God results in love for the afflicted.

Simone might even locate the lingering anxiety and frustration of many former math students in the absence of attention paid to them by their teachers.

Authors grapple with the Weil legacy

Recently, others have wrestled with the Weil legacy.

Sylvie Weil, André’s daughter, was born shortly before Simone’s death. Her family experience was that of being mistaken for her aunt, ignored or demeaned by her father and not being acknowledged and appreciated by those in her orbit.

Similarly, author Karen Olsson uses Simone and André to explore her own conflicted relationship with mathematics. Her forlorn quest to understand André’s mathematics eerily reflects Simone’s desire to understand André’s work and Sylvie’s desire to be seen as her own person, to not be in Simone’s shadow. Olsson studied with exceptional math teachers and students, all the while feeling out of place, overwhelmed and intimidated by her fellow students. Most painfully, in the process of writing her book on the Weil siblings, Olsson asks a mathematician, who had been a student with her, for help in understanding some aspect of André’s mathematics. She was ignored. Both Sylvie Weil and Karen Olsson are living witnesses to Simone’s observation that each of us cries out to be seen.

Christopher Jackson, on the other hand, gives testimony to how mathematics can live up to Simone’s vision. Jackson is incarcerated in a federal prison but found a new life through mathematics. His correspondence with mathematician Francis Su is the backbone of Su’s 2020 book “ Mathematics for Human Flourishing ,” which uses Simone’s observation that “every being cries out silently to be read differently” as a leitmotif. Su identifies aspects of mathematics that promote human flourishing, such as beauty, truth, freedom and love. In their own ways, both Simone and André would likely agree.

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What is JUMP Math, and why are some teachers raving about it? Try 13 of its brain-teasing problems to find out

Michelle Jones says her students have been far more engaged since she started using the JUMP Math method at Jarvis Traditional Elementary School in Delta, B.C. Try the quiz at the end of this article to see whether you could solve the problems that Ms. Jones's Grade 6 students are expected to handle. Jennifer Gauthier/The Globe and Mail

In her more than two decades in front of a classroom, Michelle Jones has used five different math textbooks and, until recently, had grown increasingly frustrated in her inability to reach many of her students.

The story-based math problems that filled those textbooks left most of the children either checked out or confused. She’d draw on videos and other materials to supplement her lessons, but it didn’t feel like that was enough to help her students build their confidence in the subject.

Then, three years ago, her board – the Delta School District in Delta, B.C. – piloted a program that incorporated JUMP Math, a resource originally developed by John Mighton, an accomplished playwright and entrepreneur in Toronto. The program, run by a charity established in 2002, emphasizes students rehearsing basic arithmetic operations so they can see patterns and break problems down into smaller parts, gradually raising the level of difficulty. “We were just ready for a shift, to try something different,” said Ms. Jones, who teaches Grades 6 and 7 at Delta’s Jarvis Traditional Elementary School.

John Mighton helps Grade 4 pupils practice JUMP techniques at a Toronto school in 2007, five years after he developed the method. Deborah Baic/The Globe and Mail

For Ms. Jones, using JUMP Math – JUMP stands for Junior Undiscovered Math Prodigies – represented a sea change.

She is now armed with new strategies to teach the subject and has been able to reintroduce some rote learning so that students can engage with the material more quickly. The students use whiteboards, check in with their partners and practise on their own. As a result, she’s noticed they are less anxious and take more risks in class.

“In my experience, I have never seen students so engaged, relaxed and enjoying a math lesson,” she said.

The pilot at Delta has since expanded: In 14 of the district’s 24 elementary schools, most of the teachers are now using JUMP.

Focusing on math fact fluency may seem like an obvious recipe for success, but the way math is taught in schools has been the subject of a long-standing and divisive debate, much like reading.

On one side, some experts and educators believe rote learning creates anxiety and dread, and that children should approach the subject with playfulness and curiosity by learning through problem solving, pattern discovery and open-ended exploration.

Others have advocated for a so-called back-to-basics approach and pushed governments to initiate curriculum changes so students have the ability to quickly recall addition, subtraction, multiplication and division through repetition and memorization. Rote learning shouldn’t be considered a dirty phrase, they argue.

The debate comes at a critical time: Although Canada performs well compared with other countries globally, Canadian students’ scores on an international test administered by the Organization for Economic Co-operation and Development have been slipping for almost two decades – and the latest results from late last year show that slide continuing.

Neil Stephenson, director of learning services at Delta, brought in JUMP Math because he felt something needed to change in his district.

Educators were doing a “hodgepodge of things” to help students meet curriculum expectations, he said, which put an incredible strain on them to find and build lesson plans.

After doing some research and finding JUMP, he approached an elementary school that hadn’t been scoring well on provincial tests to see if any teachers there would try the program. Around three-quarters of them raised their hands – and assessments at the end of that school year showed that several students had progressed multiple grade-levels, and teacher confidence in how they approach the subject rose, he said.

“Absolutely we want kids to be doing creative work and solving interesting questions and synthesizing their knowledge. But there has to be some building up of that knowledge somewhere else first,” Mr. Stephenson said.

In Jarvis Elementary's district, more than half the schools now use JUMP for math education. Jennifer Gauthier/The Globe and Mail

That is heartening to JUMP’s founder.

Mr. Mighton didn’t fare well in math in school and nearly failed first-year calculus in university. But he slowly overcame his own math anxiety and, as a playwright trying to make a living, started tutoring the subject later in life. Teaching children encouraged him to break down difficult concepts into smaller parts, and, in turn, grasp the subject better. He relearned concepts he had missed along the way, and then returned to school in his early 30s to earn a PhD in math at the University of Toronto.

“Math is actually accessible, very accessible,” he said.

He explained that the current method – investigating ideas through problem solving, pattern discovery and open-ended exploration – rushes children past learning math facts in the hopes of making the subject more engaging. It has the opposite effect, he said, because children actually just become confused and disengaged.

His program provides lesson plans for teachers that allows for an incremental approach to problem solving. There’s a workbook for students, but Mr. Mighton said that should only be used after the lessons. “You want to get to those problems, but that’s not where you start. That’s the mistake we’re making,” he said. “We always think kids are experts. And we give them problems that are designed for experts when they’re novice learners.”

Math professor Anna Stokke feels that methods of teaching introduced in the 1980s have done a 'disservice to children' in the decades since. John Woods/the Globe and Mail

Anna Stokke, a mathematics professor at the University of Winnipeg and a vocal proponent for schools to once again focus on fundamentals, said the change in how math was taught began in the late 1980s under the school of thought called constructivism. The theory suggests students should not passively acquire knowledge through direct instruction but rather learn through experiences and interactions. At the time, the National Council of Teachers of Mathematics in the U.S. released a set of standards where problem solving became the focus of instruction, she said. The movement then spread to Canada.

Prof. Stokke said the change in instruction has been a “disservice to children” because students should be practising math procedures and memorizing facts before they can grasp more complex problems. “I’m a mathematician and, believe me, I know how to solve complex problems. And you can’t do complex problems without having a web of knowledge in your brain.”

The result of this change has been a widening equity gap, she said, where families who have the means provide tutoring for their children, while others continue to struggle in the subject.

However, Jason To, a math co-ordinator at the Toronto District School Board, said the argument that schools are teaching one way over another is misplaced. He worries that some experts are latching onto international test scores and insinuating that inquiry-based instruction is dominating the education space. But teachers, he said, are doing both: instructing their students on math fluency and immersing them in complex problems.

“This debate to me is you got to do one versus the other, and it’s not productive. It’s more like, how do these co-exist?”

Math fluency, and the way it is measured in standardized test, can be polarizing subjects in the world of education. Justin Tang/The Globe and Mail

Janelle Feenan, a teacher and peer support co-ordinator at the Delta school division, echoed the sentiment. For years, she and her Grade 3 teacher colleague would spend an evening a week researching and pulling resources to help their students with math fluency and to develop a more comprehensive understanding for concepts.

“We were struggling a little bit to make sure our students were understanding what we were doing. We’re going through the motions, but we just didn’t feel that they were where they needed to be,” she said.

They raised their hands to participate in the pilot that introduced JUMP Math to students.

Having worked with the program, Ms. Feenan found that there’s a place for both the structural approach that JUMP provides as well as allowing for problem solving and conceptual understanding. She uses JUMP as her main lesson plan, and then supports that with games and visual aids to deepen understanding.

“Neither of those approaches alone would be adequate to prepare kids for success in math,” she said. “I think you have to supplement lessons with activities and resources that are fun and engaging to build their understanding and enrich their learning experience.”

Pop quiz: Test your math skills, JUMP-style

These are Grade 6-level problems from JUMP Math assessment and practice books. Get out your calculator app and give them a try!

c. If the pitcher pitches on the first game (or on the second, or on the third), she will pitch a total of 10 games, ending on the 46th game (or 47th, or 48th, respectively).

Photo: Jon Blacker/The Canadian Press

d. The lake with the longest shoreline is Huron, at 6,164 km. The shortest is Lake Ontario, 1,146 km. The difference is 6,164 – 1,146 = 5,018 km.

a. Avril’s grade sold 10 + 15 + 25 + 10 = 60 tickets in total. Of those 60 tickets, 30 (half) are adult tickets and sell for $5 each, and the other 30 sell for $3 each. So Avril’s grade raises (30 × $5) + (30 × $3) = $150 + $90 = $240. Since the bus costs $320, there is still $320 – $240 = $80 needed.

c. Round 3,128 to 3,000, and 4,956 to 5,000. So 3,128 × 4,956 is approximately equal to 3,000 × 5,000 = 15,000,000, i.e., 15 million.

b. 821 × 4 = 3,284. To calculate mentally, multiply the digits separately.

c. The perimeter of the field is 921 × 5 = 4,605 m. The farmer needs 4,605 – 4,500 = 105 more metres of fence to surround the field.

Photo illustration (source: Ina Fassbender/AFP/Getty Images, JUMP Math

b. Add the digits and check if the sum makes a multiple of nine.

c. 40 per cent of 20 = 8, and 25 per cent of 20 = 5. Since 8 + 5 = 13, there are 20 – 13 = 7 green fish.

a. $7.21 × 3 = $21.63. To multiply mentally, multiply the digits separately.

Photo: Kham/Reuters

d. 84.8 mm ÷ 4 = 21.2 mm. To divide mentally, divide the digits separately.

b. 220 kg ÷ 4 = 55 kg.

c. 15 × 8 = 120, 120 ÷ 100 = 1.20. They will pay $1.20 in taxes.

d. Three quarters of 12 is nine, so nine green balloons have writing on them. Sixty per cent of 15 is nine, so nine blue balloons have writing on them. So, 9 + 9 = 18 balloons in total have writing on them.

Photo: Vadim Ghirda/AP

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Solvo - Math Homework Helper 4+

Problem solver & essay writer.

#96 in Education
4.7 • 167 Ratings
Offers In-App Purchases

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Description.

Solvo is your new superpower in education and beyond Maximize your academic potential with your own personal AI homework helper! Meet Solvo—an AI-powered math, chemistry, biology, physics solver & essay writer that revolutionizes the way you manage study assignments. Simply scan, type, or upload the task in front of you and let the app work its magic! Check out what Solvo can help you with: Scan & Solve • Scan and solve math problems, equations, and more Faced with a boss-level problem (e.g., you need science answers) and don’t know where to start? Snap a picture of it—you’ll see the result and in-depth solution steps. This way, you gain more insights into how to tackle certain tasks and become more confident solving them yourself next time! Math, science answers, and more—you name it, our AI homework helper helps with it in a flash. • Ace any test and quiz Our AI homework helper can answer all sorts of questions typically used in tests and quizzes, including true or false, multiple-choice, and open questions. Biology solver? Chemistry solver? It’s already in your pocket! Simply tap Text-Based Problems, snap a picture of the question, and get your answer in seconds. This feature can also help you test your knowledge and prepare for exams. Streamline Reading & Writing • Write killer essays in a breeze Have excellent ideas for your essay but find it hard to articulate them clearly? No problem—Solvo is an experienced essay writer! Simply tap Create Essay and type your subject. You can go ahead and use the output directly or to get your creative juices flowing. • Improve and reword your writing Solvo isn’t just an essay writer—it’s a great editor! Already prepared a draft of your text and need help with polishing it into something truly A grade-worthy? Just upload your writing to our AI homework helper, and the app will offer suggestions to reword and improve it. This can be a game-changer if you feel stuck with a writing assignment. • Read smarter, not harder Our AI homework helper can be a lifesaver if you need a quick overview of a book. Type the name of the book or its author, or upload the book if you've got a file, and no matter how long or complex, tap Generate Summary. Get the essentials in a breeze! Math solver, physics homework solver, essay writer, biology solver, chemistry solver—Solvo wears many hats! Yes, studies can be challenging, but with our AI homework helper, you're well-equipped to handle them! Get answers to all your problems—including tricky science answers—with prompt assistance for your tasks whenever and wherever you need it and enjoy studying with less anxiety. Be unstoppable in class with Premium! A subscription allows you to: • Remove usage limits • Get more detailed answers • Use text recognition (OCR) • Get instant responses Subscriptions are auto-billed based on the chosen plan. Privacy Policy - https://aiby.mobi/ai_study_ios/privacy Terms of Use - https://aiby.mobi/ai_study_ios/terms

Version 1.30

Get ready for the back-to-school season with Solvo! In this update: — Improved scanning and solving of visual tasks (including graphs, geometry, tables, and more) — Leave feedback after any solution, so we can continue improving task-solving — Copy and share any solution Don’t forget to send feedback to [email protected] and leave your review on the App Store! It helps us make the app even better.

Ratings and Reviews

167 Ratings

A wonderful app

Very helpful app I was needing something to help me with my daughter’s homework my daughter is horrible on taking notes in school. And if I have notes to see the task I’m good at figuring it out how to do the rest so I tried this help because I couldn’t find anything online to help me. So when this can’t up I was like why not. And I was glad I did it tells u how to solve it. And I could figure out the rest by their help. The only thing I would like if they make another app or add on this this one for younger kids. I know a couple of parents that also need help with there kid’s homework (how to help there child I don’t do it for my kids do there homework but I have to explain it to them sometimes and for that I need to refresh my mind as will) and this is a great app for that just hope they for something god younger students grades 2nd to 5th graders would help parents a lot.

Great App / One Major Issue

I love this app. Its saved me multiple times on upcoming tests, and the great thing about it is it thoroughly goes through the topic step-by-step making sure you understand how the AI got to the solution. All that to be said, I really wish there was a feature to edit the text that was scanned in the picture. I think its already an intended design because theres text displayed saying if you’d made typos heres the time to fix it, but it doesn't work. Tapping on the screen doesn't do anything. You can copy and paste the text but theres no way to edit it where the users keyboard opens. This is a 10/10 if I could edit the prompt.

Great academic support

Solvo has truly been a remarkable discovery for me as a busy working mom. My son has been facing difficulties with certain subjects in school, and finding the time and energy to assist him with homework has been a challenge for me. Since we found Solvo, everything has changed for the better. My son doesn’t give up on his assignments when they’re difficult. Solvo gives me peace of mind. I know my son receives the help he needs. The app has empowered my son to become more independent in tackling his academic challenges without unnecessary stress. I highly recommend Solvo to all working parents who want to actively support their children’s education. It’s an invention that has made a significant difference in our lives.

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Solving algebra problems often starts with simplifying expressions. Here's a simple method to follow: Combine like terms: Terms that have the same variable can be combined. For instance, ( 3x + 4x = 7x ). Isolate the variable: Move the variable to one side of the equation. If the equation is ( 2x + 5 = 13 ), my job is to get ( x ) by itself ...
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Free math problem solver answers your algebra homework questions with step-by-step explanations. Mathway. Visit Mathway on the web. Start 7-day free trial on the app. Start 7-day free trial on the app. Download free on Amazon. Download free in Windows Store. Take a photo of your math problem on the app. get Go. Algebra. Basic Math.
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Intermediate Algebra Online Math Course
A comprehensive textbook covering Algebra 2 and topics in Precalculus. This book is the follow-up to the acclaimed Introduction to Algebra textbook. In addition to offering standard Algebra 2 and Precalculus curriculum, the text includes advanced topics such as those problem solving strategies required for success on the AMC and AIME competitions.
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What is JUMP Math, and why are some teachers raving about it? Try 13 of
At the time, the National Council of Teachers of Mathematics in the U.S. released a set of standards where problem solving became the focus of instruction, she said. The movement then spread to ...
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‎Solvo
Meet Solvo—an AI-powered math, chemistry, biology, physics solver & essay writer that revolutionizes the way you manage study assignments. Simply scan, type, or upload the task in front of you and let the app work its magic! Check out what Solvo can help you with: Scan & Solve • Scan and solve math problems, equations, and more

- \twostack{▭}{▭}	\lt	7	8	9	\div	AC
+ \twostack{▭}{▭}	\gt	4	5	6	\times	\square\frac{\square}{\square}
\times \twostack{▭}{▭}	\left(	1	2	3	-	x
▭\:\longdivision{▭}	\right)	.	0	=	+	y