problem solving about algebraic expressions

The expand command is used mainly to rewrite polynomials with all brackets and whole number powers multiplied out and all like terms collected together. In the advanced section, you also have the option of expanding trigonometric functions, expanding modulo any integer and leaving certain parts of the expression untouched whilst expanding the rest.

Go to the Expand page

The factor command will try to rewrite an expression as a product of smaller expressions. It takes care of such things as taking out common factors, factoring by pairs, quadratic trinomials, differences of two squares, sums and differences of two cubes, and a whole lot more. The advanced section includes options for factoring trigonometric functions, factoring modulo any integer, factoring over the field of Gaussian integers (just the thing for those tricky sums of squares), and even extending the field over which factoring occurs with your own custom extensions.

Go to the Factor page

Simplifying is perhaps the most difficult of all the commands to describe. The way simplification is performed in QuickMath involves looking at many different combinations of transformations of an expression and choosing the one which has the smallest number of parts. Amongst other things, the Simplify command will take care of canceling common factors from the top and bottom of a fraction and collecting like terms. The advanced options allow you to simplify trigonometric functions or to instruct QuickMath to try harder to find a simplified expression.

Go to the Simplify page

The cancel command allows you to cancel out common factors in the denominator and numerator of any fraction appearing in an expression. This command works by canceling the greatest common divisor of the denominator and numerator.

Go to the Cancel page

Partial Fractions

The partial fractions command allows you to split a rational function into a sum or difference of fractions. A rational function is simply a quotient of two polynomials. Any rational function can be written as a sum of fractions, where the denominators of the fractions are powers of the factors of the denominator of the original expression. This command is especially useful if you need to integrate a rational function. By splitting it into partial fractions first, the integration can often be made much simpler.

Go to the Partial Fractions page

Join Fractions

The join fractions command essentially does the reverse of the partial fractions command. It will rewrite a number of fractions which are added or subtracted as a single fraction. The denominator of this single fraction will usually be the lowest common multiple of the denominators of all the fractions being added or subtracted. Any common factors in the numerator and denominator of the answer will automatically be cancelled out.

Go to the Join Fractions page

Introduction to Algebraic Functions

The notion of correspondence is encountered frequently in everyday life. For example, to each book in a library there corresponds the number of pages in the book. As another example, to each human being there corresponds a birth date. To cite a third example, if the temperature of the air is recorded throughout a day, then at each instant of time there is a corresponding temperature.

The examples of correspondences we have given involve two sets X and Y. In our first example, X denotes the set of books in a library and Y the set of positive integers. For each book x in X there corresponds a positive integer y, namely the number of pages in the book. In the second example, if we let X denote the set of all human beings and Y the set of all possible dates, then to each person x in X there corresponds a birth date y.

We sometimes represent correspondences by diagrams of the type shown in Figure 1.17, where the sets X and Y are represented by points within regions in a plane. The curved arrow indicates that the element y of Y corresponds to the element x of X. We have pictured X and Y as different sets. However, X and Y may have elements in common. As a matter of fact, we often have X = Y.

A function f from a set X to a set Y is a correspondence that assigns to each element x of X a unique element y of Y. The element y is called the image of x under f and is denoted by f(x). The set X is called the domain of the function. The range of the function consists of all images of elements of X.

Earlier, we introduced the notation f(x) for the element of Y which corresponds to x. This is usually read "f of x." We also call f(x) the value of f at x. In terms of the pictorial representation given earlier, we may now sketch a diagram as in Figure 1.18. The curved arrows indicate that the elements f(x), f(w), f(z), and f(a) of Y correspond to the elements x, y, z and a of X. Let us repeat the important fact that to each x in X there is assigned precisely one image f(x) in Y; however, different elements of X such as w and z in Figure 1.18 may have the same image in Y.

Solution As in Example 1, finding images under f is simply a matter of substituting the appropriate number for x in the expression for f(x). Thus:

Many formulas which occur in mathematics and the sciences determine functions. As an illustration, the formula A = pi*r 2 for the area A of a circle of radius r assigns to each positive real number r a unique value of A. This determines a function f, where f(r) = pi*r 2 , and we may write A= f(r). The letter r, which represents an arbitrary number from the domain off, is often called an independent variable. The letter A, which represents a number from the range off, is called a dependent variable, since its value depends on the number assigned tor. When two variables r and A are related in this manner, it is customary to use the phrase A is a function of r. To cite another example, if an automobile travels at a uniform rate of 50 miles per hour, then the distance d (miles) traveled in time t (hours) is given by d = 50t and hence the distance d is a function of time t.

We have seen that different elements in the domain of a function may have the same image. If images are always different, then, as in the next definition, the function is called one-to-one.  

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Algebraic Sentences Problems

Algebraic sentences word problems.

An algebraic sentence when written in equation form involves algebraic expressions (which contain variables such as letters in the alphabet), constants, and an equal symbol. Each algebraic sentence may contain a combination of algebraic expressions and constants, or with just two or more algebraic expressions. Almost always, the word “is” in an algebraic sentence denotes the symbol of equality.

In our example above, the algebraic sentence, “ Five more than twice a number is forty-three “, is translated and written into its equation form: [latex]2x + 5 = 43[/latex].

But before we delve into solving word problems that involve algebraic sentences, it’s crucial that we become familiar with how to translate and write algebraic expressions.

Algebraic Expressions

Learning how to write and translate algebraic expressions is the foundation in writing algebraic sentences. Eventually, we will use this knowledge to write algebraic equations where we solve for the value of the unknown variable.

What is an algebraic expression?

Think of an algebraic expression as the equivalent of a phrase in the English language. It does not convey a complete thought but is an important building block in constructing a sentence. Algebraic expressions consist of numbers, variables, and arithmetic operations.

Let’s take a quick look at some math phrases that were translated and written into algebraic expressions.

• the sum of a number and [latex]8[/latex] [latex]\large{\,\,\, \to \,\,\,\, {\color{red}{c + 8}}}[/latex]
• [latex]3[/latex] less than a number [latex]\large{\,\,\, \to \,\,\,\, {\color{red}{y – 3}}}[/latex]
• the product of [latex]45[/latex] and [latex]m[/latex] [latex]\large{\,\,\, \to \,\,\,\,{\color{red}{45m}}}[/latex]
• a number divided by [latex]2[/latex] [latex]\large{\,\,\, \to \,\,\,\,\Large {\color{red}{x \over 2}}}[/latex]

Translating and Writing Algebraic Sentences

Once you know how to translate math phrases into algebraic expressions, it’ll be easy for you to translate and write algebraic sentences in an equation form as well.

For instance, how do you write “ the difference of 16 and k is seven ” into an algebraic equation?

Let’s deconstruct this algebraic sentence.

You may notice that we translated “the difference of [latex]16[/latex] and [latex]k[/latex]” into an algebraic expression, [latex]16-k[/latex], then used the equal symbol ([latex]=[/latex]) in place of the word “is”.

As I mentioned at the beginning of this lesson, the word “is” in an algebraic sentence, most of the time, signifies the equality symbol which is the case in this example.

Therefore, we can write this algebraic sentence in equation form as [latex]16 – k = 7[/latex].

Now that we are confident in translating algebraic sentences and writing them in an equation format, it’s time for us to take one step further. We will not only translate and write algebraic sentences into algebraic equations but also proceed to solve them.

Examples of Algebraic Sentences Word Problems

The main key when solving word problems with algebraic sentences is to accurately translate the algebraic expressions then set up and write each algebraic equation correctly. In doing so, we can ensure that we are solving the right equation and as a result, will get the correct answer for each word problem.

Example 1: Six more than seven times a number is thirty-four. Find the number.

First, let’s deconstruct the algebraic sentence. It’s important that we identify and separate the algebraic expressions from the constants as well as determine if the problem suggests equality between the terms.

For the unknown value, we’ll use [latex]\large{x}[/latex] as our variable.

Note: Since addition is commutative, changing the order of the addends on the left side of the equation does not change the sum. Therefore, we may also write the algebraic expression as [latex]7x+6[/latex] or the algebraic equation as [latex]7x + 6 = 34[/latex] instead.

However, for our discussion, we will use [latex]6 + 7x = 34[/latex] as our equation.

The original problem asks us to find the number which in this case is [latex]\large{x}[/latex]. So our next step is to solve for [latex]\large{x}[/latex] in our equation, [latex]6 + 7x = 34[/latex].

Now, we’ll evaluate the value of [latex]\large{x = 4}[/latex] to verify if the number we found satisfies the original algebraic sentence.

Remember that we are asked to find the identity of “the number” in our algebraic sentence and NOT the value of the variable [latex]\large{x}[/latex]. Therefore, it will be incorrect to say that [latex]\large{x=4}[/latex] is the answer. This is a common mistake that we always need to pay attention to.

We’ll replace the variable [latex]\large{x}[/latex] with the number [latex]4[/latex] to see if the algebraic expression on the left side of the equation also results to [latex]34[/latex].

Yes, it does! This means that number 4 is the answer .

Example 2: The difference between three times a number and five is sixteen. Find the number.

Right away, the word “difference” in our algebraic sentence gives us a clue that we will be using the subtraction operation. But, unlike, in addition, the order of the terms within an expression matters in subtraction. So we have to make sure that the terms in our algebraic expression are set up in the correct order.

Another keyword is “times”, which suggests that [latex]3[/latex] is being multiplied to a number whose value is currently unknown.

This time, let’s use [latex]\large{g}[/latex] as our variable. Start by translating the algebraic expression correctly then continue to write the equation.

Now that we have our equation, let’s find out what the unknown value is by solving for the variable, [latex]\large{g}[/latex].

As discussed in our first example, it is important that we verify if the result of our solution makes the algebraic equation true. For this one, I will leave it up to you to evaluate the value of [latex]\large{g = 7}[/latex]. Remember to replace the unknown value of the variable [latex]\large{g}[/latex] with the number [latex]7[/latex] to see if the left side of the equation is also equal to [latex]16[/latex].

Since the original algebraic sentence requires us to find the number, then the answer is number 7 .

Example 3: A number decreased by half of the number is four. Find the number.

This algebraic sentence is quite interesting because we don’t just have one unknown value, but two! First, we have the “number” (the unknown) then the “half of the number”, which means one-half of the unknown.

The keyword “half of” also indicates multiplying something by one-half while the “decreased by” tells us that we will be subtracting one term from the other.

To organize our thoughts, let’s deconstruct this algebraic sentence into its meat and potatoes.

Being able to break apart our algebraic sentence into its basic parts allows us to read the sentence thoroughly and understand the relationship between its quantities.

After setting up our algebraic equation, our next step is to solve it for [latex]\large{m}[/latex].

We need to evaluate the value of [latex]\large{m = 8}[/latex] to check if the number we got makes our algebraic equation true.

Great! The values on both sides of the equation are equal to each other. Therefore, the number 8 is the correct answer to our original algebraic sentence.

Example 4: Four times the sum of twice a number and six is thirty-two. Find the number.

Looking closely, we see a few keywords that would help us in translating this algebraic sentence into an equation.

• times – means that we’ll be multiplying [latex]4[/latex] by the quantity, “sum of twice a number and six”
• sum – means that we need to add the terms, “ twice a number ” and six
• twice – means that a number (the unknown value) is multiplied by [latex]2[/latex]

With these in mind, let’s write our algebraic equation.

Notice that we used the variable [latex]\large{d}[/latex] in our equation to stand for our unknown value. Let’s now proceed and solve for [latex]\large{d}[/latex] and afterward, check if the value we get indeed makes the equation true.

Evaluate the value of [latex]\large{d = 1}[/latex]:

Looks like everything checks out, so the answer to our algebraic sentence is the number 1 .

Example 5: Two-thirds of the sum of three times a number and six is ten. What is the number?

This problem involves a fraction and an algebraic expression as you can tell. The keyword “of” indicates multiplication, so it tells us that we need to multiply the fraction [latex]\large{2 \over 3}[/latex] by the sum of the two given quantities (“three times a number” and six).

Let’s move on and write our algebraic equation. We’ll use the variable [latex]\large{y}[/latex] as the placeholder for our “number” whose value is currently unknown.

Awesome! We are able to translate our algebraic sentence into an equation. It’s time for us to solve for [latex]\large{y}[/latex] then I’ll leave it up to you to do a check if the number we find as the value of [latex]\large{y}[/latex] makes the algebraic equation true.

Once you evaluate [latex]\large{y = 3}[/latex], you’ll see that the answer to our word problem is indeed the number 3 .

Example 6: Twice the difference of a number and three is four more than the number. Find the number.

We have an example here that is a little different from the word problems we’ve covered so far. In this algebraic sentence, both sides of the equation contain an algebraic expression. However, the keywords present will help us translate it correctly into an equation so there’s nothing to worry about.

• twice – means that we will multiply the quantity (difference of a number and three) by [latex]2[/latex]
• difference – tells us to use the subtraction operation between our unknown value (“number”) and three
• more than – means that we need to add four to our unknown value

Let’s solve for our unknown value, [latex]\large{a}[/latex], then verify that our result makes the algebraic equation true.

Evaluate the value of [latex]\large{a = 10}[/latex]:

Perfect! The value we got for [latex]\large{a}[/latex] which is [latex]10[/latex], made our equation equal on both sides. Thus, the answer is the number 10 .

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Algebra Practice Problems – Master Your Skills with Exercises and Solutions!

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Understanding Algebra Fundamentals

Working with equations, practicing algebra with worksheets, mastering fractions and percentages, utilizing algebra in geometry, solving real-world problems.

To master algebra , practicing problems and going through their solutions is crucial. It’s a bit like learning to play an instrument – practice is key to improving.

For many students, the leap from arithmetic to algebra can be challenging. It requires a shift from working with concrete numbers to thinking abstractly about numbers and variables. But I’ve found that with a set repertoire of practice problems, from basic equations to more complex ones involving exponents and polynomials, it becomes much clearer.

What’s more, each problem has a set of steps to reach the solution, revealing patterns and strategies that can be applied to future challenges. Having a diverse array of practice problems is beneficial not only for students but also for educators and parents seeking to support their learner’s educational journey.

Hands-on experience with algebraic expressions and equations solidifies understanding and boosts confidence. So let’s get our hands on some algebra problems and start tackling them – who knows what new connections we’ll uncover in the intricacies of algebraic thinking.

Algebra is like a puzzle where I use mathematical symbols and letters (known as variables) to find unknown quantities. Variables represent values that can change; for example, in the equation $x + 2 = 5$, the variable $x$ takes the value $3$.

I need to understand that variables allow me to describe general math truths or perform operations without knowing the exact values.

Here’s what I often start with when exploring fundamental algebraic concepts :

Expressions and Equations : An expression is a combination of numbers and variables, like $3x + 4$. When an expression equals a number, like $3x + 4 = 19$, it becomes an equation. Equations set the stage for finding the value of the variable.

Operations with Variables : I treat variables the same way as numbers in operations. Addition, subtraction, multiplication, and division acts on variables just like on numbers, except that I don’t combine unlike terms (e.g., $x + y$ remains as it is unless I know the relationship between $x$ and $y$).

Below is a simple table detailing how basic arithmetic operations work with variables:

• Solving the Equations : To solve an equation, I perform operations that isolate the variable on one side. For example, if $x – 4 = 10$, I add $4$ to both sides to get $x = 14$.

Remembering mathematical order of operations is vital. In algebra, I follow this hierarchy: operations inside parentheses first, exponents second, then multiplication and division, and finally, addition and subtraction. This rule ensures I simplify expressions and solve equations correctly.

When I solve equations, my goal is to isolate the variable I’m interested in. This often involves moving terms from one side of the equation to the other.

For example, in a simple linear equation like ( ax + b = c ), I would solve for ( x ) by subtracting ( b ) from both sides and then dividing by ( a ).

When dealing with absolute values , like ( |x + a| = b ), I remember that the solution considers both the positive and negative counterparts because the absolute value represents the distance from zero without considering direction.

I always check my solutions by substituting them back into the original equation to ensure they satisfy the equation. This verification step is crucial, especially when initial equations involve absolute values or higher degree terms.

For practice problems, Khan Academy and Paul’s Online Math Notes are great resources that offer a variety of algebra problems, including linear equations and absolute value equations. I appreciate how Khan Academy’s platform provides instant feedback.

Working through these problems helps reinforce my understanding. Solving equations can be like a puzzle, and I find it satisfying when I find the correct solution. Each equation brings a new challenge, which keeps my algebra skills sharp and ready.

When I dive into algebra, I find that worksheets are a vital tool for mastering concepts. They offer a structured approach, enabling me to tackle one problem at a time.

While practice problems may vary in complexity, they often share a common focus on fundamental algebraic structures, such as linear equations and variables. 

For linear equations, I often start with simple formats like ( y = mx + b ) and gradually work up to more complex problems. This incremental approach helps me understand the relationship between variables and coefficients.

Sample Linear Equation Worksheet

Italicized tip: Always isolate the variable you’re solving for by performing inverse operations.

I make sure worksheets are varied, combining problems that involve solving for one variable with those requiring multiple steps. It’s the repetition and gradual escalation in difficulty that really solidify my understanding.

Variables Practice

• Identify variables and constants
• Evaluate expressions like ( 3$x^2$ – 2x + 7 ) for given values of ( x )
• Translate word problems into algebraic equations

Algebra worksheets offer distinct advantages, as they can be tailored to focus on just the areas I’m looking to improve, providing immediate feedback when answer keys are included. Through disciplined practice , I find myself not just solving problems but also gaining a deeper appreciation of algebra’s expressive power.

I find that a solid grasp of fractions and percentages is crucial in algebra. These concepts are not only foundational in mathematics but also widely applicable in real-world scenarios. Let’s take a closer look at how we can tackle problems involving these topics.

The Fraction Fundamentals For me, working with fractions always starts with understanding the roles of the numerator (top number) and denominator (bottom number). When performing addition or subtraction with fractions, I remember that the denominator must be the same.

For example, to add $$ \frac{3}{4} $$ and $$ \frac{5}{8} $$, I first find a common denominator. Multiplying both the numerator and denominator of $$ \frac{3}{4} $$ by 2 gives me $$ \frac{6}{8} $$. Now, I can easily add the two fractions:

$$ \frac{6}{8} + \frac{5}{8} = \frac{11}{8} $$.

Multiplication and Division Multiplication of fractions is more straightforward. I simply multiply the numerators together and the denominators together. For instance:

$$ \frac{3}{5} \times \frac{2}{3} = \frac{6}{15} $$.

But when it comes to division , I flip the second fraction (find the reciprocal) and then multiply.

$$ \frac{4}{7} \div \frac{2}{5} = \frac{4}{7} \times \frac{5}{2} = \frac{20}{14} $$.

Converting to Percentages When I need to express a fraction as a percent , I multiply the fraction by 100%. For example:

$$ \frac{3}{4} = \frac{3}{4} \times 100% = 75% $$.

To solve for a percentage of a number , I convert the percent to a decimal and then multiply. If I have to find 60% of 50, I do:

$$ 60% = 0.60; \quad 0.60 \times 50 = 30 $$.

By keeping these strategies in mind, I can confidently solve a variety of problems involving fractions and percentages. Experimenting with different problems is an enjoyable way to get better at these concepts!

In exploring the relationship between algebra and geometry, I find it essential to recognize how algebra serves as a robust tool for solving geometric problems. Take the circle, for example; its properties can be unraveled using algebraic methods.

The equation of a circle in a coordinate plane, which is ( $(x – h)^2$ + $(y – k)^2$ = $r^2$ ), allows me to calculate the radius, or find points on its circumference by substituting values for ( x ) and ( y ).

When it comes to finding areas or lengths, algebra is my go-to. For a rectangle or a square, if I know one side and the area, I can easily find the missing side by setting up an algebraic equation. Let’s say the area (( A )) of a rectangle is given and also one side (( l )). The other side (( w )) can be found using ( A = l \times w ).

In geometry, I also use systems of equations, which is an algebraic approach, to find the intersection between lines, a crucial aspect in defining points in polygons or linear graphs. Here’s a scenario: I have two lines with equations ( y = 2x + 3 ) and ( y = -x + 5 ). Solving this system will give me the exact point where both lines cross.

My journey through geometric challenges often involves these algebraic techniques, showing that algebra isn’t just numbers and letters, it’s a gateway to unlocking the mysteries laid out in shapes and spaces.

When I tackle real-world problems using algebra, I begin by defining the variables that represent the unknowns I’m trying to find. Let’s say I’m trying to figure out how many apples and oranges I can buy with a certain amount of money. I’d let ( x ) represent the number of apples and ( y ) the number of oranges.

I then translate the situation into algebraic equations, usually involving addition, subtraction, multiplication, or division. For instance, if apples cost $2 each and oranges $1.50, and I have $10, the equations would look like this:

[ 2x + 1.5y = 10 ]

Next, I would use function notation to express relationships. If I want to determine the relationship between the cost of fruit and the number I buy, I might write a function:

[ f(x) = 2x ] [ g(y) = 1.5y ]

These equations and functions help me visualize the problem and perform calculations to solve for ( x ) and ( y ).

• Define variables : Represent unknown quantities with variables.
• Translate into equations : Use the context to form equations.
• Function notation : Express relationships with functions.
• Solve the equations : Find the value of the variables.

Algebra acts as a bridge between basic arithmetic and the more complex real-world scenarios. It allows me to take practical questions and find concrete answers, such as how many items I can purchase or how long a trip might take given a constant speed.

As I practice, I become adept at forming and solving these algebraic equations, making me better equipped to handle similar problems in the future.

In my experience with algebra practice problems and solutions, I’ve observed that consistency is key. Complete a variety of problems from different domains regularly to sharpen skills. It’s essential to tackle problems that challenge various difficulty levels. I’ve appreciated websites like Paul’s Online Math Notes, which provide a range of problems and detailed solutions.

I often use web filters to ensure that my search results come from educational resources, especially those from trusted domains like kastatic.org and kasandbox.org , which are known for their reliable content. Interactive platforms, rather than static worksheets, offer immediate feedback and keep practice engaging.

Khan Academy, for example, offers an interactive approach that I find particularly effective. The immediate feedback helps me adjust my learning path in real-time.

When approaching algebraic word problems , I make sure to define my variables clearly, as this can often simplify complex equations. It’s also crucial to understand concepts like Venn diagrams, which visually represent logical relationships, as they frequently appear in algebra problems.

Consistency, a variety of sources, and clear problem-solving strategies have greatly contributed to my understanding of algebra . Remember, practice doesn’t just make perfect—it makes permanent. So find resources that work for you and stick to them!

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Algebra Problems

Algebra problems are not only based on algebraic expressions but also on various types of equations in Maths where a quantity or variable is unknown to us. Many of us are familiar with the word problem, but are we aware of the fact and problems related to variables and constants? When we say 5 it means a number but what if we say x=5 or 5y or something like that?

This is where algebra came into existence algebra is that branch of mathematics which not only deals with numbers but also variable and alphabets. The versatility of Algebra is very deep and very conceptual, all the non-numeric character represents variable and numeric as constants. Let us solve some problems based algebra with solutions which will cover the syllabus for class 6, 7, 8. Below are some of the examples of algebraic expressions .

For example.

Algebra Word Problems deal with real-time situations and solutions which can be solved using algebra.

Basic Algebra Identities

• (a + b) 2  = a 2 + b 2 + 2ab
• (a – b) 2 = a 2  + b 2 – 2ab
• a 2 – b 2 = (a + b)(a – b)
• a 2 + b 2 = (a + b) 2 – 2ab = (a – b) 2 + 2ab
• a 3 + b 3 = (a + b)(a 2 – ab + b 2 )
• a 3 – b 3 = (a – b)(a 2 + ab + b 2 )
• (a + b) 3 = a 3 + 3ab(a + b) + b 3
• (a – b) 3 = a 3 – 3ab(a – b) – b 3

Algebra problems With Solutions

Example 1: Solve, (x-1) 2 = [4√(x-4)] 2 Solution: x 2 -2x+1 = 16(x-4)

x 2 -2x+1 = 16x-64

x 2 -18x+65 = 0

(x-13) (x-5) = 0

Hence, x = 13 and x = 5.

Algebra Problems for Class 6

In class 6, students will be introduced with an algebra concept. Here, you will learn how the unknown values are represented in terms of variables.  The given expression can be solved only if we know the value of unknown variable. Let us see some examples.

Example: Solve, 4x + 5 when, x = 3.

Solution: Given, 4x + 5

Now putting the value of x=3, we get;

4 (3) + 5 = 12 + 5 = 17.

Example: Give expressions for the following cases:

(i) 12 added to 2x

(ii) 6 multiplied by y

(iii) 25 subtracted from z

(iv) 17 times of m 

(i) 12 + 2x

Algebra Problems for Class 7

In class 7, students will deal with algebraic expressions like x+y, xy, 32x 2 -12y 2 , etc. There are different kinds of the terminology used in case algebraic equations such as;

• Coefficient

Let us understand these terms with an example. Suppose 4x + 5y is an algebraic expression, then 4x and 5y are the terms. Since here the variables used are x and y, therefore, x and y are the factors of 4x + 5y. And the numerical factor attached to the variables are the coefficient such as 4 and 5 are the coefficient of x and y in the given expression.

Any expression with one or more terms is called a polynomial. Specifically, a one-term expression is called a monomial; a two-term expression is called a binomial, and a three-term expression is called a trinomial.

Terms which have the same algebraic factors are like terms . Terms which have different algebraic factors are unlike terms . Thus, terms 4xy and – 3xy are like terms; but terms 4xy and – 3x are not like terms.

Example: Add 3x + 5x

Solution: Since 3x and 5x have the same algebraic factors, hence, they are like terms and can be added by their coefficient.

3x + 5x = 8x

Example: Collect like terms and simplify the expression: 12x 2 – 9x + 5x – 4x 2 – 7x + 10.

Solution: 12x 2 – 9x + 5x – 4x 2 – 7x + 10

= (12 – 4)x 2 – 9x + 5x – 7x + 10

=  8x 2 – 11x + 10

Algebra Problems for Class 8

Here, students will deal with algebraic identities. See the examples.

Example: Solve (2x+y) 2

Solution: Using the identity: (a+b) 2  = a 2 + b 2 + 2 ab, we get;

(2x+y) = (2x) 2 + y 2 + 2.2x.y = 4x 2 + y 2 + 4xy

Example: Solve (99) 2 using algebraic identity.

Solution: We can write, 99 = 100 -1

Therefore, (100 – 1 ) 2

= 100 2 + 1 2 – 2 x 100 x 1  [By identity: (a -b) 2 = a 2 + b 2 – 2ab

= 10000 + 1 – 200

Algebra Word Problems

Question 1: There are 47 boys in the class. This is three more than four times the number of girls. How many girls are there in the class?

Solution: Let the number of girls be x

As per the given statement,

4 x + 3 = 47

4x = 47 – 3

Question 2: The sum of two consecutive numbers is 41. What are the numbers?

Solution: Let one of the numbers be x.

Then the other number will x+1

Now, as per the given questions,

x + x + 1 = 41

2x + 1 = 41

So, the first number is 20 and second number is 20+1 = 21

Linear Algebra Problems

There are various methods For Solving the Linear Equations

• Cross multiplication method
• Replacement method or Substitution method
• Hit and trial method

There are Variety of different Algebra problem present and are solved depending upon their functionality and state. For example, a linear equation problem can’t be solved using a quadratic equation formula and vice verse for, e.g., x+x/2=7 then solve for x is an equation in one variable for x which can be satisfied by only one value of x. Whereas x 2 +5x+6 is a quadratic equation which is satisfied for two values of x the domain of algebra is huge and vast so for more information. Visit BYJU’S. where different techniques are explained different algebra problem.

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Simplifying Algebraic Expressions

Simplifying algebraic expressions is a fundamental skill in algebra that helps in solving equations, graphing functions and understanding mathematical relationships. This process involves reducing the expressions to their simplest form by combining like terms, applying mathematical operations and following the algebraic rules. This article will guide us through the essential steps, rules and techniques for simplifying algebraic expressions, along with examples and practice problems to enhance our understanding.

Table of Content

• Understanding Algebraic Expressions

Combining Like Terms

Applying the distributive property, combining distributive property and like terms, using the foil method.

• Examples with Solutions
• Practical Questions

What are Algebraic Expressions?

An algebraic expression is a combination of numbers, variables and mathematical operators (such as +, -, × and /). For example:

3x + 5 − 2x

In this expression, 3x and −2x are like terms, while 5 is a constant term.

Terms are terms that have the same variable raised to the same power. To simplify an expression combine these terms by performing the arithmetic operations.

Example: Simplify the expression 4x + 7 − 3x + 2.

• Combine 4x and −3x: 4x − 3x = x
• Combine 7 and 2: 7 + 2 = 9
• The simplified expression is x + 9.

The distributive property states that a(b + c) = ab + ac. Use this property to the eliminate parentheses in an expression.

Example: Simplify the expression 3(x + 4).

Apply the distributive property: 3(x + 4) = 3⋅x + 3⋅4 Perform the multiplication: 3x + 12 The simplified expression is 3x+12.

In more complex expressions, use both the distributive property and combining like terms to the simplify.

Example: Simplify the expression 2(x − 3) + 4x.

Apply the distributive property: 2(x − 3) = 2x − 6 Combine with 4x: 2x − 6 + 4x Combine like terms: 2x + 4x = 6x Simplified expression: 6x − 6

The FOIL method is used to the simplify the product of the two binomials. The FOIL stands for First, Outer, Inner, Last referring to the terms to be multiplied.

Example: Simplify (x + 2)(x + 3) using the FOIL method.

First: x ⋅ x = x 2 Outer: x⋅3 = 3x Inner: 2⋅x = 2x Last: 2⋅3 = 6 Combine the results: x 2 + 3x + 2x + 6 Combine like terms: x 2 + 5x + 6 The simplified expression is x 2 + 5x + 6.

Solved Examples on Simplifying Algebraic Expressions

Example 1: Simplify 4x + 3 − 2x + 5.

Combine like terms: 4x − 2x + 3 + 5. Perform the addition/subtraction: 2x + 8. The simplified expression is 2x + 8.

Example 2: Simplify 3(a+4)−2(a−1).

Apply the distributive property: 3(a + 4) = 3a + 12 −2(a − 1) = −2a + 2 Combine the results: 3a + 12 − 2a + 2 Combine like terms: 3a − 2a + 12 + 2. Perform the addition/subtraction: a + 14. The simplified expression is a + 14.

Example 3: Simplify [Tex]\frac{6x^2 – 3x + 2x^2 + 5}{2}[/Tex] .

Combine like terms in the numerator: [Tex]6x^2 + 2x^2 – 3x + 5 = 8x^2 – 3x + 5 [/Tex] Divide each term by 2: [Tex]\frac{8x^2}{2} – \frac{3x}{2} + \frac{5}{2} = 4x^2 – \frac{3x}{2} + \frac{5}{2} [/Tex] The simplified expression is [Tex]4x^2 – \frac{3x}{2} + \frac{5}{2}[/Tex] .

Example 4: Simplify (x + 2)(x – 3).

Use the FOIL method: First: [Tex]x \cdot x = x^2[/Tex] Outer: [Tex]x \cdot (-3) = -3x[/Tex] Inner: [Tex]2 \cdot x = 2x[/Tex] Last: [Tex]2 \cdot (-3) = -6[/Tex] Combine the results: [Tex]x^2 – 3x + 2x – 6 = x^2 – x – 6 [/Tex] The simplified expression is [Tex]x^2 – x – 6[/Tex] .

Example 5: Simplify [Tex]\frac{2(x + 4) – 3(x – 2)}{x}[/Tex] .

Apply the distributive property in the numerator: [Tex]2(x + 4) = 2x + 8 [/Tex] [Tex]-3(x – 2) = -3x + 6 [/Tex] Combine the results: [Tex]2x + 8 – 3x + 6 = -x + 14[/Tex] Divide by x: [Tex]\frac{-x + 14}{x} = -1 + \frac{14}{x}[/Tex] The simplified expression is [Tex]-1 + \frac{14}{x}[/Tex] .

Practice Questions

Q1: Simplify: [Tex]7m – 4n + 2m – 3n[/Tex] .

Q2: Simplify: [Tex]\frac{5x^2 – 2x + 3x^2 – 4}{3}[/Tex] .

Q3: Simplify: (2x + 1)(x – 2) + 3(x + 1).

Q4: Simplify: [Tex]\frac{4y – 3(2y – 1)}{y}[/Tex] .

Q5: Simplify: 5(a + 3b) – 2(a – b).

Q6: Simplify: [Tex](3x – 4)^2 – (x – 2)^2[/Tex] .

Q7: Simplify: 6 – 2(3 – x) + 4x.

Q8: Simplify: [Tex]\frac{3(x^2 – 4) + 2(x^2 + 1)}{x}[/Tex] .

Q9: Simplify: (x + 2)(x + 3) – (x – 1)(x – 2).

Q10: Simplify: 2(3x – 5) + 4x – 7x.

What is the difference between combining like terms and applying the distributive property?

The Combining like terms involves adding or subtracting terms with the same variables. The distributive property involves the multiplying a term by the sum or difference inside parentheses.

How do you handle negative signs when simplifying expressions?

Be careful with the negative signs by the distributing them correctly across the terms in the parentheses and combining them properly with the other terms.

Can algebraic expressions have more than one variable?

Yes, expressions can have multiple variables. Simplify them by the combining like terms that share the same variables and powers.

What is a binomial?

A binomial is a polynomial with the exactly two terms such as the x+3 or 2x 2 -y.

Why is it important to simplify algebraic expressions?

The Simplifying expressions helps in the solving the equations making calculations easier and understanding the mathematical relationships more clearly.

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Unit 1 – sat math – algebra.

Solving linear equations and inequalities

Linear equation word problems

Linear relationship word problems

Graphs of linear equations and functions

Solving systems of linear equations

Systems of linear equations word problems

Linear inequality word problems

Graphs of linear systems and inequalities

Unit 2 – SAT Math – Problem Solving and Data Analysis

Ratios, rates, and proportions

Unit Conversion

Percentages

Center, spread, and shape of distributions

Data representations

Scatterplots

Linear and exponential growth

Probability and relative frequency

Data inferences

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Unit 3 – SAT Math – Advanced Math

Factoring quadratic and polynomial expressions

Radicals and rational exponents

Operations with polynomials

Operations with rational expressions

Nonlinear functions

Isolating quantities

Solving quadratic equations

Linear and quadratic systems

Radical, rational, and absolute value equations

Quadratic and exponential word problems

Quadratic graphs

Exponential graphs

Polynomial and other nonlinear graphs

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Area and volume

Congruence, similarity, and angle relationships

Right triangle trigonometry

Circle theorems

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Circle equations

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Command of Evidence: Textual

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Unit 6 – SAT Reading & Writing – Craft and Structure

Words in Context

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Unit 7 – SAT Reading & Writing – Expression of Ideas + Standard English Conventions

Transitions

Rhetorical Synthesis

Form, Structure, and Sense

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Subject-verb agreement

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Plurals and possessives

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