- Mayfield City Schools
- Mayfield High School
- Mayfield Middle School
- Center Elementary
- Gates Mills Elementary
- Lander Elementary
- Millridge Elementary
- Mayfield Preschool Program
- Board Policies
- Information and Times
- Meeting videos on YouTube
- Math 1 - Unit 2 Downloads
- Math 1 - Unit 1 Downloads
- Math 1 - Unit 3 Downloads
- Math 1 - Midterm Review Downloads
- Math 1 - Unit 4 Downloads
- Math 1 - Unit 5 Downloads
- Math 1 - Unit 6 Downloads
- Math 1 - Final Exam Review Downloads
- Math 1 AIR Review
- Math 1 - Unit 7 Downloads
Unit 2 Downloads
- $ 0.00 0 items
Unit 2 – Linear Expressions, Equations, and Inequalities
This unit is all about linear topics, which is a major focus of Common Core Algebra I. We develop general methods for solving linear equations using properties of equality and inverse operations. Thorough review is given to review of equation solving from Common Core 8th Grade Math. Solutions to equations and inequalities are defined in terms of making statements true. This theme is emphasized throughout the unit. Modeling with both linear equations and inequalities is stressed.
Equations and Their Solutions
LESSON/HOMEWORK
LECCIÓN/TAREA
LESSON VIDEO
EDITABLE LESSON
EDITABLE KEY
Seeing Structure to Solve Equations
A Linear Equation Solving Review
Justifying Steps in Solving an Equation
Linear Word Problems
More Linear Equations and Consecutive Integer Games
Solving Linear Equations with Unspecified Constants
Inequalities
Solving Linear Inequalities
Compound Inequalities
More Work with Compound Inequalities
Interval Notation
Modeling with Inequalities
Unit Review
Unit #2 Review – Linear Equations and Inequalities
UNIT REVIEW
REPASO DE LA UNIDAD
EDITABLE REVIEW
Unit #2 Assessment Form A
EDITABLE ASSESSMENT
Unit #2 Assessment Form B
Unit #2 Assessment Form C
Unit #2 Assessment Form D
Unit #2 Exit Tickets
Unit #2 Mid-Unit Quiz (Through Lesson #6) – Form A
Unit #2 Mid-Unit Quiz (Through Lesson #6) – Form B
Unit #2 Mid-Unit Quiz (Through Lesson #6) – Form C
U02.AO.01 – Basic Equation Solving
EDITABLE RESOURCE
U02.AO.02 – More Work with Linear Word Problems (After Lesson #7)
U02.AO.03 – Inequality Warm-Up Set (Prior to Lesson #8)
U02.AO.04 – Equation and Inequality Practice
U02.AO.05 – Using Structure to Understand Solutions to Equations (Enrichment)
U02.AO.06 – Solving Linear Equations – Extra Practice
Thank you for using eMATHinstruction materials. In order to continue to provide high quality mathematics resources to you and your students we respectfully request that you do not post this or any of our files on any website. Doing so is a violation of copyright. Using these materials implies you agree to our terms and conditions and single user license agreement .
The content you are trying to access requires a membership . If you already have a plan, please login. If you need to purchase a membership we offer yearly memberships for tutors and teachers and special bulk discounts for schools.
Sorry, the content you are trying to access requires verification that you are a mathematics teacher. Please click the link below to submit your verification request.
Linear Equation Worksheets. Free worksheets with answer keys
Enjoy these free printable worksheets . Each one has model problems worked out step by step, practice problems, as well as challenge questions at the sheets end. Plus each one comes with an answer key.
- Mixed Problems on Writing Equations of Lines
- Slope Intercept Form Worksheet
- Standard Form Worksheet
- Point Slope Worksheet
- Write Equation of Line From the Slope and 1 Point
- Write Equation of Line From Two Points
- Equation of Line Parallel to Another Line and Through a Point
- Equation of Line Perpendicular to Another Line and Through a Point
- Slope of a Line
- Perpendicular Bisector of Segment
- Write Equation of Line Mixed Review
- Word Problems
Ultimate Math Solver (Free) Free Algebra Solver ... type anything in there!
Popular pages @ mathwarehouse.com.
2.2 Linear Equations in One Variable
Learning objectives.
In this section, you will:
- Solve equations in one variable algebraically.
- Solve a rational equation.
- Find a linear equation.
- Given the equations of two lines, determine whether their graphs are parallel or perpendicular.
- Write the equation of a line parallel or perpendicular to a given line.
Caroline is a full-time college student planning a spring break vacation. To earn enough money for the trip, she has taken a part-time job at the local bank that pays $15.00/hr, and she opened a savings account with an initial deposit of $400 on January 15. She arranged for direct deposit of her payroll checks. If spring break begins March 20 and the trip will cost approximately $2,500, how many hours will she have to work to earn enough to pay for her vacation? If she can only work 4 hours per day, how many days per week will she have to work? How many weeks will it take? In this section, we will investigate problems like this and others, which generate graphs like the line in Figure 1 .
Solving Linear Equations in One Variable
A linear equation is an equation of a straight line, written in one variable. The only power of the variable is 1. Linear equations in one variable may take the form a x + b = 0 a x + b = 0 and are solved using basic algebraic operations.
We begin by classifying linear equations in one variable as one of three types: identity, conditional, or inconsistent. An identity equation is true for all values of the variable. Here is an example of an identity equation.
The solution set consists of all values that make the equation true. For this equation, the solution set is all real numbers because any real number substituted for x x will make the equation true.
A conditional equation is true for only some values of the variable. For example, if we are to solve the equation 5 x + 2 = 3 x − 6 , 5 x + 2 = 3 x − 6 , we have the following:
The solution set consists of one number: { − 4 } . { − 4 } . It is the only solution and, therefore, we have solved a conditional equation.
An inconsistent equation results in a false statement. For example, if we are to solve 5 x − 15 = 5 ( x − 4 ) , 5 x − 15 = 5 ( x − 4 ) , we have the following:
Indeed, −15 ≠ −20. −15 ≠ −20. There is no solution because this is an inconsistent equation.
Solving linear equations in one variable involves the fundamental properties of equality and basic algebraic operations. A brief review of those operations follows.
Linear Equation in One Variable
A linear equation in one variable can be written in the form
where a and b are real numbers, a ≠ 0. a ≠ 0.
Given a linear equation in one variable, use algebra to solve it.
The following steps are used to manipulate an equation and isolate the unknown variable, so that the last line reads x = _________, x = _________, if x is the unknown. There is no set order, as the steps used depend on what is given:
- We may add, subtract, multiply, or divide an equation by a number or an expression as long as we do the same thing to both sides of the equal sign. Note that we cannot divide by zero.
- Apply the distributive property as needed: a ( b + c ) = a b + a c . a ( b + c ) = a b + a c .
- Isolate the variable on one side of the equation.
- When the variable is multiplied by a coefficient in the final stage, multiply both sides of the equation by the reciprocal of the coefficient.
Solving an Equation in One Variable
Solve the following equation: 2 x + 7 = 19. 2 x + 7 = 19.
This equation can be written in the form a x + b = 0 a x + b = 0 by subtracting 19 19 from both sides. However, we may proceed to solve the equation in its original form by performing algebraic operations.
The solution is 6.
Solve the linear equation in one variable: 2 x + 1 = −9. 2 x + 1 = −9.
Solving an Equation Algebraically When the Variable Appears on Both Sides
Solve the following equation: 4 ( x −3 ) + 12 = 15 −5 ( x + 6 ) . 4 ( x −3 ) + 12 = 15 −5 ( x + 6 ) .
Apply standard algebraic properties.
This problem requires the distributive property to be applied twice, and then the properties of algebra are used to reach the final line, x = − 5 3 . x = − 5 3 .
Solve the equation in one variable: −2 ( 3 x − 1 ) + x = 14 − x . −2 ( 3 x − 1 ) + x = 14 − x .
Solving a Rational Equation
In this section, we look at rational equations that, after some manipulation, result in a linear equation. If an equation contains at least one rational expression, it is a considered a rational equation .
Recall that a rational number is the ratio of two numbers, such as 2 3 2 3 or 7 2 . 7 2 . A rational expression is the ratio, or quotient, of two polynomials. Here are three examples.
Rational equations have a variable in the denominator in at least one of the terms. Our goal is to perform algebraic operations so that the variables appear in the numerator. In fact, we will eliminate all denominators by multiplying both sides of the equation by the least common denominator (LCD).
Finding the LCD is identifying an expression that contains the highest power of all of the factors in all of the denominators. We do this because when the equation is multiplied by the LCD, the common factors in the LCD and in each denominator will equal one and will cancel out.
Solve the rational equation: 7 2 x − 5 3 x = 22 3 . 7 2 x − 5 3 x = 22 3 .
We have three denominators; 2 x , 3 x , 2 x , 3 x , and 3. The LCD must contain 2 x , 3 x , 2 x , 3 x , and 3. An LCD of 6 x 6 x contains all three denominators. In other words, each denominator can be divided evenly into the LCD. Next, multiply both sides of the equation by the LCD 6 x . 6 x .
A common mistake made when solving rational equations involves finding the LCD when one of the denominators is a binomial—two terms added or subtracted—such as ( x + 1 ) . ( x + 1 ) . Always consider a binomial as an individual factor—the terms cannot be separated. For example, suppose a problem has three terms and the denominators are x , x , x − 1 , x − 1 , and 3 x − 3. 3 x − 3. First, factor all denominators. We then have x , x , ( x − 1 ) , ( x − 1 ) , and 3 ( x − 1 ) 3 ( x − 1 ) as the denominators. (Note the parentheses placed around the second denominator.) Only the last two denominators have a common factor of ( x − 1 ) . ( x − 1 ) . The x x in the first denominator is separate from the x x in the ( x − 1 ) ( x − 1 ) denominators. An effective way to remember this is to write factored and binomial denominators in parentheses, and consider each parentheses as a separate unit or a separate factor. The LCD in this instance is found by multiplying together the x , x , one factor of ( x − 1 ) , ( x − 1 ) , and the 3. Thus, the LCD is the following:
So, both sides of the equation would be multiplied by 3 x ( x − 1 ) . 3 x ( x − 1 ) . Leave the LCD in factored form, as this makes it easier to see how each denominator in the problem cancels out.
Another example is a problem with two denominators, such as x x and x 2 + 2 x . x 2 + 2 x . Once the second denominator is factored as x 2 + 2 x = x ( x + 2 ) , x 2 + 2 x = x ( x + 2 ) , there is a common factor of x in both denominators and the LCD is x ( x + 2 ) . x ( x + 2 ) .
Sometimes we have a rational equation in the form of a proportion; that is, when one fraction equals another fraction and there are no other terms in the equation.
We can use another method of solving the equation without finding the LCD: cross-multiplication. We multiply terms by crossing over the equal sign.
Multiply a ( d ) a ( d ) and b ( c ) , b ( c ) , which results in a d = b c . a d = b c .
Any solution that makes a denominator in the original expression equal zero must be excluded from the possibilities.
Rational Equations
A rational equation contains at least one rational expression where the variable appears in at least one of the denominators.
Given a rational equation, solve it.
- Factor all denominators in the equation.
- Find and exclude values that set each denominator equal to zero.
- Find the LCD.
- Multiply the whole equation by the LCD. If the LCD is correct, there will be no denominators left.
- Solve the remaining equation.
- Make sure to check solutions back in the original equations to avoid a solution producing zero in a denominator.
Solving a Rational Equation without Factoring
Solve the following rational equation:
We have three denominators: x , x , 2 , 2 , and 2 x . 2 x . No factoring is required. The product of the first two denominators is equal to the third denominator, so, the LCD is 2 x . 2 x . Only one value is excluded from a solution set, 0. Next, multiply the whole equation (both sides of the equal sign) by 2 x . 2 x .
The proposed solution is −1, which is not an excluded value, so the solution set contains one number, −1 , −1 , or { −1 } { −1 } written in set notation.
Solve the rational equation: 2 3 x = 1 4 − 1 6 x . 2 3 x = 1 4 − 1 6 x .
Solving a Rational Equation by Factoring the Denominator
Solve the following rational equation: 1 x = 1 10 − 3 4 x . 1 x = 1 10 − 3 4 x .
First find the common denominator. The three denominators in factored form are x , 10 = 2 ⋅ 5 , x , 10 = 2 ⋅ 5 , and 4 x = 2 ⋅ 2 ⋅ x . 4 x = 2 ⋅ 2 ⋅ x . The smallest expression that is divisible by each one of the denominators is 20 x . 20 x . Only x = 0 x = 0 is an excluded value. Multiply the whole equation by 20 x . 20 x .
The solution is 35 2 . 35 2 .
Solve the rational equation: − 5 2 x + 3 4 x = − 7 4 . − 5 2 x + 3 4 x = − 7 4 .
Solving Rational Equations with a Binomial in the Denominator
Solve the following rational equations and state the excluded values:
- ⓐ 3 x − 6 = 5 x 3 x − 6 = 5 x
- ⓑ x x − 3 = 5 x − 3 − 1 2 x x − 3 = 5 x − 3 − 1 2
- ⓒ x x − 2 = 5 x − 2 − 1 2 x x − 2 = 5 x − 2 − 1 2
The denominators x x and x − 6 x − 6 have nothing in common. Therefore, the LCD is the product x ( x − 6 ) . x ( x − 6 ) . However, for this problem, we can cross-multiply.
The solution is 15. The excluded values are 6 6 and 0. 0.
The LCD is 2 ( x − 3 ) . 2 ( x − 3 ) . Multiply both sides of the equation by 2 ( x − 3 ) . 2 ( x − 3 ) .
The solution is 13 3 . 13 3 . The excluded value is 3. 3.
The least common denominator is 2 ( x − 2 ) . 2 ( x − 2 ) . Multiply both sides of the equation by x ( x − 2 ) . x ( x − 2 ) .
The solution is 4. The excluded value is 2. 2.
Solve − 3 2 x + 1 = 4 3 x + 1 . − 3 2 x + 1 = 4 3 x + 1 . State the excluded values.
Solving a Rational Equation with Factored Denominators and Stating Excluded Values
Solve the rational equation after factoring the denominators: 2 x + 1 − 1 x − 1 = 2 x x 2 − 1 . 2 x + 1 − 1 x − 1 = 2 x x 2 − 1 . State the excluded values.
We must factor the denominator x 2 −1. x 2 −1. We recognize this as the difference of squares, and factor it as ( x − 1 ) ( x + 1 ) . ( x − 1 ) ( x + 1 ) . Thus, the LCD that contains each denominator is ( x − 1 ) ( x + 1 ) . ( x − 1 ) ( x + 1 ) . Multiply the whole equation by the LCD, cancel out the denominators, and solve the remaining equation.
The solution is −3. −3. The excluded values are 1 1 and −1. −1.
Solve the rational equation: 2 x − 2 + 1 x + 1 = 1 x 2 − x − 2 . 2 x − 2 + 1 x + 1 = 1 x 2 − x − 2 .
Finding a Linear Equation
Perhaps the most familiar form of a linear equation is the slope-intercept form, written as y = m x + b , y = m x + b , where m = slope m = slope and b = y -intercept . b = y -intercept . Let us begin with the slope.
The Slope of a Line
The slope of a line refers to the ratio of the vertical change in y over the horizontal change in x between any two points on a line. It indicates the direction in which a line slants as well as its steepness. Slope is sometimes described as rise over run.
If the slope is positive, the line slants to the right. If the slope is negative, the line slants to the left. As the slope increases, the line becomes steeper. Some examples are shown in Figure 2 . The lines indicate the following slopes: m = −3 , m = −3 , m = 2 , m = 2 , and m = 1 3 . m = 1 3 .
The slope of a line, m , represents the change in y over the change in x. Given two points, ( x 1 , y 1 ) ( x 1 , y 1 ) and ( x 2 , y 2 ) , ( x 2 , y 2 ) , the following formula determines the slope of a line containing these points:
Finding the Slope of a Line Given Two Points
Find the slope of a line that passes through the points ( 2 , −1 ) ( 2 , −1 ) and ( −5 , 3 ) . ( −5 , 3 ) .
We substitute the y- values and the x- values into the formula.
The slope is − 4 7 . − 4 7 .
It does not matter which point is called ( x 1 , y 1 ) ( x 1 , y 1 ) or ( x 2 , y 2 ) . ( x 2 , y 2 ) . As long as we are consistent with the order of the y terms and the order of the x terms in the numerator and denominator, the calculation will yield the same result.
Find the slope of the line that passes through the points ( −2 , 6 ) ( −2 , 6 ) and ( 1 , 4 ) . ( 1 , 4 ) .
Identifying the Slope and y- intercept of a Line Given an Equation
Identify the slope and y- intercept, given the equation y = − 3 4 x − 4. y = − 3 4 x − 4.
As the line is in y = m x + b y = m x + b form, the given line has a slope of m = − 3 4 . m = − 3 4 . The y- intercept is b = −4. b = −4.
The y -intercept is the point at which the line crosses the y- axis. On the y- axis, x = 0. x = 0. We can always identify the y- intercept when the line is in slope-intercept form, as it will always equal b. Or, just substitute x = 0 x = 0 and solve for y.
The Point-Slope Formula
Given the slope and one point on a line, we can find the equation of the line using the point-slope formula.
This is an important formula, as it will be used in other areas of college algebra and often in calculus to find the equation of a tangent line. We need only one point and the slope of the line to use the formula. After substituting the slope and the coordinates of one point into the formula, we simplify it and write it in slope-intercept form.
Given one point and the slope, the point-slope formula will lead to the equation of a line:
Finding the Equation of a Line Given the Slope and One Point
Write the equation of the line with slope m = −3 m = −3 and passing through the point ( 4 , 8 ) . ( 4 , 8 ) . Write the final equation in slope-intercept form.
Using the point-slope formula, substitute −3 −3 for m and the point ( 4 , 8 ) ( 4 , 8 ) for ( x 1 , y 1 ) . ( x 1 , y 1 ) .
Note that any point on the line can be used to find the equation. If done correctly, the same final equation will be obtained.
Given m = 4 , m = 4 , find the equation of the line in slope-intercept form passing through the point ( 2 , 5 ) . ( 2 , 5 ) .
Finding the Equation of a Line Passing Through Two Given Points
Find the equation of the line passing through the points ( 3 , 4 ) ( 3 , 4 ) and ( 0 , −3 ) . ( 0 , −3 ) . Write the final equation in slope-intercept form.
First, we calculate the slope using the slope formula and two points.
Next, we use the point-slope formula with the slope of 7 3 , 7 3 , and either point. Let’s pick the point ( 3 , 4 ) ( 3 , 4 ) for ( x 1 , y 1 ) . ( x 1 , y 1 ) .
In slope-intercept form, the equation is written as y = 7 3 x − 3. y = 7 3 x − 3.
To prove that either point can be used, let us use the second point ( 0 , −3 ) ( 0 , −3 ) and see if we get the same equation.
We see that the same line will be obtained using either point. This makes sense because we used both points to calculate the slope.
Standard Form of a Line
Another way that we can represent the equation of a line is in standard form . Standard form is given as
where A , A , B , B , and C C are integers. The x- and y- terms are on one side of the equal sign and the constant term is on the other side.
Finding the Equation of a Line and Writing It in Standard Form
Find the equation of the line with m = −6 m = −6 and passing through the point ( 1 4 , −2 ) . ( 1 4 , −2 ) . Write the equation in standard form.
We begin using the point-slope formula.
From here, we multiply through by 2, as no fractions are permitted in standard form, and then move both variables to the left aside of the equal sign and move the constants to the right.
This equation is now written in standard form.
Find the equation of the line in standard form with slope m = − 1 3 m = − 1 3 and passing through the point ( 1 , 1 3 ) . ( 1 , 1 3 ) .
Vertical and Horizontal Lines
The equations of vertical and horizontal lines do not require any of the preceding formulas, although we can use the formulas to prove that the equations are correct. The equation of a vertical line is given as
where c is a constant. The slope of a vertical line is undefined, and regardless of the y- value of any point on the line, the x- coordinate of the point will be c .
Suppose that we want to find the equation of a line containing the following points: ( −3 , −5 ) , ( −3 , 1 ) , ( −3 , 3 ) , ( −3 , −5 ) , ( −3 , 1 ) , ( −3 , 3 ) , and ( −3 , 5 ) . ( −3 , 5 ) . First, we will find the slope.
Zero in the denominator means that the slope is undefined and, therefore, we cannot use the point-slope formula. However, we can plot the points. Notice that all of the x- coordinates are the same and we find a vertical line through x = −3. x = −3. See Figure 3 .
The equation of a horizontal line is given as
where c is a constant. The slope of a horizontal line is zero, and for any x- value of a point on the line, the y- coordinate will be c .
Suppose we want to find the equation of a line that contains the following set of points: ( −2 , −2 ) , ( 0 , −2 ) , ( 3 , −2 ) , ( −2 , −2 ) , ( 0 , −2 ) , ( 3 , −2 ) , and ( 5 , −2 ) . ( 5 , −2 ) . We can use the point-slope formula. First, we find the slope using any two points on the line.
Use any point for ( x 1 , y 1 ) ( x 1 , y 1 ) in the formula, or use the y -intercept.
The graph is a horizontal line through y = −2. y = −2. Notice that all of the y- coordinates are the same. See Figure 3 .
Finding the Equation of a Line Passing Through the Given Points
Find the equation of the line passing through the given points: ( 1 , −3 ) ( 1 , −3 ) and ( 1 , 4 ) . ( 1 , 4 ) .
The x- coordinate of both points is 1. Therefore, we have a vertical line, x = 1. x = 1.
Find the equation of the line passing through ( −5 , 2 ) ( −5 , 2 ) and ( 2 , 2 ) . ( 2 , 2 ) .
Determining Whether Graphs of Lines are Parallel or Perpendicular
Parallel lines have the same slope and different y- intercepts. Lines that are parallel to each other will never intersect. For example, Figure 4 shows the graphs of various lines with the same slope, m = 2. m = 2.
All of the lines shown in the graph are parallel because they have the same slope and different y- intercepts.
Lines that are perpendicular intersect to form a 90° 90° -angle. The slope of one line is the negative reciprocal of the other. We can show that two lines are perpendicular if the product of the two slopes is −1 : m 1 ⋅ m 2 = −1. −1 : m 1 ⋅ m 2 = −1. For example, Figure 5 shows the graph of two perpendicular lines. One line has a slope of 3; the other line has a slope of − 1 3 . − 1 3 .
Graphing Two Equations, and Determining Whether the Lines are Parallel, Perpendicular, or Neither
Graph the equations of the given lines, and state whether they are parallel, perpendicular, or neither: 3 y = − 4 x + 3 3 y = − 4 x + 3 and 3 x − 4 y = 8. 3 x − 4 y = 8.
The first thing we want to do is rewrite the equations so that both equations are in slope-intercept form.
First equation:
Second equation:
See the graph of both lines in Figure 6
From the graph, we can see that the lines appear perpendicular, but we must compare the slopes.
The slopes are negative reciprocals of each other, confirming that the lines are perpendicular.
Graph the two lines and determine whether they are parallel, perpendicular, or neither: 2 y − x = 10 2 y − x = 10 and 2 y = x + 4. 2 y = x + 4.
Writing the Equations of Lines Parallel or Perpendicular to a Given Line
As we have learned, determining whether two lines are parallel or perpendicular is a matter of finding the slopes. To write the equation of a line parallel or perpendicular to another line, we follow the same principles as we do for finding the equation of any line. After finding the slope, use the point-slope formula to write the equation of the new line.
Given an equation for a line, write the equation of a line parallel or perpendicular to it.
- Find the slope of the given line. The easiest way to do this is to write the equation in slope-intercept form.
- Use the slope and the given point with the point-slope formula.
- Simplify the line to slope-intercept form and compare the equation to the given line.
Writing the Equation of a Line Parallel to a Given Line Passing Through a Given Point
Write the equation of line parallel to a 5 x + 3 y = 1 5 x + 3 y = 1 and passing through the point ( 3 , 5 ) . ( 3 , 5 ) .
First, we will write the equation in slope-intercept form to find the slope.
The slope is m = − 5 3 . m = − 5 3 . The y- intercept is 1 3 , 1 3 , but that really does not enter into our problem, as the only thing we need for two lines to be parallel is the same slope. The one exception is that if the y- intercepts are the same, then the two lines are the same line. The next step is to use this slope and the given point with the point-slope formula.
The equation of the line is y = − 5 3 x + 10. y = − 5 3 x + 10. See Figure 7 .
Find the equation of the line parallel to 5 x = 7 + y 5 x = 7 + y and passing through the point ( −1 , −2 ) . ( −1 , −2 ) .
Finding the Equation of a Line Perpendicular to a Given Line Passing Through a Given Point
Find the equation of the line perpendicular to 5 x − 3 y + 4 = 0 5 x − 3 y + 4 = 0 and passing through the point ( − 4 , 1 ) . ( − 4 , 1 ) .
The first step is to write the equation in slope-intercept form.
We see that the slope is m = 5 3 . m = 5 3 . This means that the slope of the line perpendicular to the given line is the negative reciprocal, or − 3 5 . − 3 5 . Next, we use the point-slope formula with this new slope and the given point.
Access these online resources for additional instruction and practice with linear equations.
- Solving rational equations
- Equation of a line given two points
- Finding the equation of a line perpendicular to another line through a given point
- Finding the equation of a line parallel to another line through a given point
2.2 Section Exercises
What does it mean when we say that two lines are parallel?
What is the relationship between the slopes of perpendicular lines (assuming neither is horizontal nor vertical)?
How do we recognize when an equation, for example y = 4 x + 3 , y = 4 x + 3 , will be a straight line (linear) when graphed?
What does it mean when we say that a linear equation is inconsistent?
When solving the following equation:
2 x − 5 = 4 x + 1 2 x − 5 = 4 x + 1
explain why we must exclude x = 5 x = 5 and x = −1 x = −1 as possible solutions from the solution set.
For the following exercises, solve the equation for x . x .
7 x + 2 = 3 x − 9 7 x + 2 = 3 x − 9
4 x − 3 = 5 4 x − 3 = 5
3 ( x + 2 ) − 12 = 5 ( x + 1 ) 3 ( x + 2 ) − 12 = 5 ( x + 1 )
12 − 5 ( x + 3 ) = 2 x − 5 12 − 5 ( x + 3 ) = 2 x − 5
1 2 − 1 3 x = 4 3 1 2 − 1 3 x = 4 3
x 3 − 3 4 = 2 x + 3 12 x 3 − 3 4 = 2 x + 3 12
2 3 x + 1 2 = 31 6 2 3 x + 1 2 = 31 6
3 ( 2 x − 1 ) + x = 5 x + 3 3 ( 2 x − 1 ) + x = 5 x + 3
2 x 3 − 3 4 = x 6 + 21 4 2 x 3 − 3 4 = x 6 + 21 4
x + 2 4 − x − 1 3 = 2 x + 2 4 − x − 1 3 = 2
For the following exercises, solve each rational equation for x . x . State all x -values that are excluded from the solution set.
3 x − 1 3 = 1 6 3 x − 1 3 = 1 6
2 − 3 x + 4 = x + 2 x + 4 2 − 3 x + 4 = x + 2 x + 4
3 x − 2 = 1 x − 1 + 7 ( x − 1 ) ( x − 2 ) 3 x − 2 = 1 x − 1 + 7 ( x − 1 ) ( x − 2 )
3 x x − 1 + 2 = 3 x − 1 3 x x − 1 + 2 = 3 x − 1
5 x + 1 + 1 x − 3 = − 6 x 2 − 2 x − 3 5 x + 1 + 1 x − 3 = − 6 x 2 − 2 x − 3
1 x = 1 5 + 3 2 x 1 x = 1 5 + 3 2 x
For the following exercises, find the equation of the line using the point-slope formula. Write all the final equations using the slope-intercept form.
( 0 , 3 ) ( 0 , 3 ) with a slope of 2 3 2 3
( 1 , 2 ) ( 1 , 2 ) with a slope of − 4 5 − 4 5
x -intercept is 1, and ( −2 , 6 ) ( −2 , 6 )
y -intercept is 2, and ( 4 , −1 ) ( 4 , −1 )
( −3 , 10 ) ( −3 , 10 ) and ( 5 , −6 ) ( 5 , −6 )
( 1 , 3 ) and ( 5 , 5 ) ( 1 , 3 ) and ( 5 , 5 )
parallel to y = 2 x + 5 y = 2 x + 5 and passes through the point ( 4 , 3 ) ( 4 , 3 )
perpendicular to 3 y = x − 4 3 y = x − 4 and passes through the point ( −2 , 1 ) ( −2 , 1 ) .
For the following exercises, find the equation of the line using the given information.
( − 2 , 0 ) ( − 2 , 0 ) and ( −2 , 5 ) ( −2 , 5 )
( 1 , 7 ) ( 1 , 7 ) and ( 3 , 7 ) ( 3 , 7 )
The slope is undefined and it passes through the point ( 2 , 3 ) . ( 2 , 3 ) .
The slope equals zero and it passes through the point ( 1 , −4 ) . ( 1 , −4 ) .
The slope is 3 4 3 4 and it passes through the point ( 1 , 4 ) ( 1 , 4 ) .
( –1 , 3 ) ( –1 , 3 ) and ( 4 , –5 ) ( 4 , –5 )
For the following exercises, graph the pair of equations on the same axes, and state whether they are parallel, perpendicular, or neither.
y = 2 x + 7 y = − 1 2 x − 4 y = 2 x + 7 y = − 1 2 x − 4
3 x − 2 y = 5 6 y − 9 x = 6 3 x − 2 y = 5 6 y − 9 x = 6
y = 3 x + 1 4 y = 3 x + 2 y = 3 x + 1 4 y = 3 x + 2
x = 4 y = −3 x = 4 y = −3
For the following exercises, find the slope of the line that passes through the given points.
( 5 , 4 ) ( 5 , 4 ) and ( 7 , 9 ) ( 7 , 9 )
( −3 , 2 ) ( −3 , 2 ) and ( 4 , −7 ) ( 4 , −7 )
( −5 , 4 ) ( −5 , 4 ) and ( 2 , 4 ) ( 2 , 4 )
( −1 , −2 ) ( −1 , −2 ) and ( 3 , 4 ) ( 3 , 4 )
( 3 , −2 ) ( 3 , −2 ) and ( 3 , −2 ) ( 3 , −2 )
For the following exercises, find the slope of the lines that pass through each pair of points and determine whether the lines are parallel or perpendicular.
( −1 , 3 ) and ( 5 , 1 ) ( −2 , 3 ) and ( 0 , 9 ) ( −1 , 3 ) and ( 5 , 1 ) ( −2 , 3 ) and ( 0 , 9 )
( 2 , 5 ) and ( 5 , 9 ) ( −1 , −1 ) and ( 2 , 3 ) ( 2 , 5 ) and ( 5 , 9 ) ( −1 , −1 ) and ( 2 , 3 )
For the following exercises, express the equations in slope intercept form (rounding each number to the thousandths place). Enter this into a graphing calculator as Y1, then adjust the ymin and ymax values for your window to include where the y -intercept occurs. State your ymin and ymax values.
0.537 x − 2.19 y = 100 0.537 x − 2.19 y = 100
4,500 x − 200 y = 9,528 4,500 x − 200 y = 9,528
200 − 30 y x = 70 200 − 30 y x = 70
Starting with the point-slope formula y − y 1 = m ( x − x 1 ) , y − y 1 = m ( x − x 1 ) , solve this expression for x x in terms of x 1 , y , y 1 , x 1 , y , y 1 , and m m .
Starting with the standard form of an equation A x + B y = C A x + B y = C solve this expression for y y in terms of A , B , C A , B , C and x x . Then put the expression in slope-intercept form.
Use the above derived formula to put the following standard equation in slope intercept form: 7 x − 5 y = 25. 7 x − 5 y = 25.
Given that the following coordinates are the vertices of a rectangle, prove that this truly is a rectangle by showing the slopes of the sides that meet are perpendicular.
( – 1 , 1 ) , ( 2 , 0 ) , ( 3 , 3 ) ( – 1 , 1 ) , ( 2 , 0 ) , ( 3 , 3 ) and ( 0 , 4 ) ( 0 , 4 )
Find the slopes of the diagonals in the previous exercise. Are they perpendicular?
Real-World Applications
The slope for a wheelchair ramp for a home has to be 1 12 . 1 12 . If the vertical distance from the ground to the door bottom is 2.5 ft, find the distance the ramp has to extend from the home in order to comply with the needed slope.
If the profit equation for a small business selling x x number of item one and y y number of item two is p = 3 x + 4 y , p = 3 x + 4 y , find the y y value when p = $ 453 and x = 75. p = $ 453 and x = 75.
For the following exercises, use this scenario: The cost of renting a car is $45/wk plus $0.25/mi traveled during that week. An equation to represent the cost would be y = 45 + .25 x , y = 45 + .25 x , where x x is the number of miles traveled.
What is your cost if you travel 50 mi?
If your cost were $ 63.75 , $ 63.75 , how many miles were you charged for traveling?
Suppose you have a maximum of $100 to spend for the car rental. What would be the maximum number of miles you could travel?
This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.
Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.
Access for free at https://openstax.org/books/college-algebra-2e/pages/1-introduction-to-prerequisites
- Authors: Jay Abramson
- Publisher/website: OpenStax
- Book title: College Algebra 2e
- Publication date: Dec 21, 2021
- Location: Houston, Texas
- Book URL: https://openstax.org/books/college-algebra-2e/pages/1-introduction-to-prerequisites
- Section URL: https://openstax.org/books/college-algebra-2e/pages/2-2-linear-equations-in-one-variable
© Jun 28, 2024 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.
Gina Wilson All Things Algebra Answer Key | Gina Wilson All things Algebra 2015
In the realm of mathematics education, finding reliable resources that support effective learning can be a challenging task. However, Gina Wilson All Things Algebra is a comprehensive platform that provides educators and students with valuable tools to enhance their mathematical knowledge. One of the key features of this platform is the availability of the answer key, which serves as a vital resource for learners seeking to validate their solutions and progress in their mathematical journey. In this article, we will delve into the benefits of Gina Wilson All Things Algebra and explore how the answer key can be accessed and utilized effectively.
Expansions factorisations
Expansions factorisations printable math worksheet
Print here >
Linear equations
Linear equations printable math worksheet
Logs printable math worksheet
Order of operations
Order of operations printable math worksheet
Quadratic formular
Quadratic formular printable math worksheet
Remainder theorem
Remainder theorem printable math worksheet
Simultaneous equations
Simultaneous equations printable math worksheet
Subject of formula
Subject of formula printable math worksheet
Financial arithmetic
Financial arithmetic printable math worksheet
Converting decimals to fractions
Converting decimals to fractions printable math worksheet
Converting fractions to decimals
Converting fractions to decimals printable math worksheet
Converting fractions to percents
Converting fractions to percents printable math worksheet
Converting percentage to decimals
Converting percentage to decimals printable math worksheet
Decimal addition
Decimal addition printable math worksheet
Decimal division
Decimal division printable math worksheet
Decimals multiplication
Decimals multiplication printable math worksheet
Decimals subtraction
Decimals subtraction printable math worksheet
Pre algebra adition decimals
Pre algebra adition decimals printable math worksheet
Pre algebra adition decimals 3
Pre algebra adition decimals 3 printable math worksheet
Pre algebra adition decimals2
Pre algebra adition decimals2 printable math worksheet
Adding fractions
Adding fractions printable math worksheet
Equivalent fractions
Equivalent fractions printable math worksheet
Fractions addition
Fractions addition printable math worksheet
Fractions multiplication
Fractions multiplication printable math worksheet
Fractions simplification
Fractions simplification printable math worksheet
Fractions subtraction
Fractions subtraction printable math worksheet
Impropper fraction comparisons
Impropper fraction comparisons printable math worksheet
Circumference area
Circumference area printable math worksheet
Complementary supplementary angles
Complementary supplementary angles printable math worksheet
L shapes perimeter area
L shapes perimeter area printable math worksheet
Perimeter area of squares
Perimeter area of squares printable math worksheet
Surface area of complex figures
Surface area of complex figures printable math worksheet
Triangle perimeter area
Triangle perimeter area printable math worksheet
Volume of cylinder
Volume of cylinder printable math worksheet
Linear inequalities
Linear inequalities printable math worksheet
Absolute values
Absolute values printable math worksheet
Add divide multiply intergers
Add divide multiply intergers printable math worksheet
Adding integers
Adding integers printable math worksheet
Comparisons
Comparisons printable math worksheet
Integer equations
Integer equations printable math worksheet
Ordering intergers
Ordering intergers printable math worksheet
Cm mm scale
Cm mm scale printable math worksheet
Metric system converting scales
Metric system converting scales printable math worksheet
Us metric system
Us metric system printable math worksheet
Decimal number patterns
Decimal number patterns printable math worksheet
Mixed decimal number patterns
Mixed decimal number patterns printable math worksheet
Mixed decimal number patterns2
Mixed decimal number patterns2 printable math worksheet
Mixed patterns
Mixed patterns printable math worksheet
Number patterns
Number patterns printable math worksheet
Number patterns higher
Number patterns higher printable math worksheet
Greatest common factor
Greatest common factor printable math worksheet
Least common multiple
Least common multiple printable math worksheet
Number system
Number system printable math worksheet
Percents of numbers
Percents of numbers printable math worksheet
Ratio percent decimals fractions convertions
Ratio percent decimals fractions convertions printable math worksheet
Ratios printable math worksheet
Powers printable math worksheet
Powers exponents
Powers exponents printable math worksheet
Scientific notation 2
Scientific notation 2 printable math worksheet
Scientific notation 3
Scientific notation 3 printable math worksheet
Scientific notation 1
Scientific notation 1 printable math worksheet
Square roots
Square roots printable math worksheet
Number problems
Number problems printable math worksheet
Pre algebra adition
Pre algebra adition printable math worksheet
Pre algebra division decimals
Pre algebra division decimals printable math worksheet
Pre algebra multiplication addition
Pre algebra multiplication addition printable math worksheet
Pre algebra subtraction 1
Pre algebra subtraction 1 printable math worksheet
Probability
Probability printable math worksheet
Sets printable math worksheet
Triangle sides pythagorean theoream 6
Triangle sides pythagorean theoream 6 printable math worksheet
Triangle sides pythagorean theorem 1
Triangle sides pythagorean theorem 1 printable math worksheet
Triangle sides pythagorean theorem 2
Triangle sides pythagorean theorem 2 printable math worksheet
Triangle sides pythagorean theorem 3
Triangle sides pythagorean theorem 3 printable math worksheet
Triangle sides pythagorean theorem 4
Triangle sides pythagorean theorem 4 printable math worksheet
Triangle sides pythagorean theorem 5
Triangle sides pythagorean theorem 5 printable math worksheet
Triangle sides pythagorean theorem 7
Triangle sides pythagorean theorem 7 printable math worksheet
Percents and ratios
Percents and ratios printable math worksheet
Coordinate geometry
Coordinate geometry printable math worksheet
Coordinates 1
Coordinates 1 printable math worksheet
Coordinates 2
Coordinates 2 printable math worksheet
Coordinates 3
Coordinates 3 printable math worksheet
Data on graph
Data on graph printable math worksheet
Graphing linear equations
Graphing linear equations printable math worksheet
Graphs locate in x y
Graphs locate in x y printable math worksheet
Ploting graphs
Ploting graphs printable math worksheet
Table of data 1
Table of data 1 printable math worksheet
Table of data 2
Table of data 2 printable math worksheet
Table of data 3
Table of data 3 printable math worksheet
What is Gina Wilson All Things Algebra?
Gina Wilson All Things Algebra is an educational platform developed by Gina Wilson, an experienced mathematics educator. It offers a wide range of resources, including curriculum materials, lesson plans, activities, and assessments, designed to promote a deeper understanding of algebraic concepts. The platform caters to both teachers and students, providing them with the necessary tools to excel in algebraic reasoning and problem-solving.
Benefits of Using Gina Wilson All Things Algebra
- Comprehensive Content: Gina Wilson All Things Algebra covers a vast array of algebraic topics, ensuring that learners have access to a rich collection of materials that encompass various levels of difficulty.
- Clear Explanations: The resources provided by Gina Wilson are known for their clarity and concise explanations. Students can easily grasp complex concepts and apply them to solve mathematical problems.
- Engaging Activities: The platform incorporates interactive activities that foster student engagement and promote active learning. These activities make the learning process enjoyable and encourage students to develop a deeper interest in algebra.
- Differentiated Instruction: Gina Wilson All Things Algebra offers materials that cater to learners with different abilities. This ensures that each student can progress at their own pace and receive the appropriate level of support.
- Aligned with Standards: The resources provided by Gina Wilson are aligned with common core standards and state-specific curriculum frameworks, making them a reliable choice for educators seeking to meet educational requirements.
How to Access the Answer Key
To access the answer key on Gina Wilson All Things Algebra, users need to have an account on the platform. Once logged in, they can navigate to the desired resource or worksheet and locate the answer key section. The answer key provides step-by-step solutions to the exercises, allowing students to verify their work and gain a better understanding of the mathematical concepts involved.
Exploring the Answer Key Features
The answer key on Gina Wilson All Things Algebra offers various features that enhance the learning experience. Some notable features include:
- Detailed Solutions: The answer key provides comprehensive and detailed solutions to the exercises, enabling students to identify any errors and learn from them.
- Multiple Approaches: In many cases, the answer key offers alternative approaches to solving problems, encouraging students to think critically and explore different problem-solving strategies.
- Common Mistakes: The answer key highlights common mistakes made by students, helping them identify potential pitfalls and misconceptions.
- Additional Notes: Alongside the solutions, the answer key may include additional notes or explanations to clarify key concepts and provide extra guidance.
How to Make the Most of Gina Wilson All Things Algebra
To maximize the benefits of Gina Wilson All Things Algebra, here are some tips to consider:
- Regular Practice: Consistent practice using the resources available on the platform will reinforce mathematical skills and boost confidence.
- Collaborative Learning: Encourage students to work in groups or pairs, discussing and solving problems together. This fosters collaborative learning and the exchange of ideas.
- Utilize Feedback: When using the answer key, pay attention to the feedback provided. Understand the mistakes made and use them as learning opportunities to improve problem-solving skills.
- Seek Clarification: If any concepts or solutions remain unclear, reach out to teachers or fellow students for clarification. Effective communication is key to resolving doubts and gaining a deeper understanding of algebra.
Frequently Asked Questions (FAQs)
- The cost varies depending on the subscription plan chosen. It is best to visit the official website for detailed pricing information.
- Absolutely! The platform caters to both classroom use and self-study, providing learners with the flexibility to learn at their own pace.
- Yes, most resources on the platform have accompanying answer keys to facilitate self-assessment and understanding.
- Yes, the platform is accessible on various devices, including smartphones and tablets, ensuring convenience and flexibility.
- Yes, Gina Wilson All Things Algebra provides technical support to address any issues or concerns users may encounter. Reach out to their support team for prompt assistance.
Gina Wilson All Things Algebra is a valuable resource that empowers both educators and learners in the realm of algebra. The answer key, with its comprehensive solutions and additional features, serves as a powerful tool to validate understanding and promote mathematical growth. By utilizing the platform effectively, students can enhance their problem-solving skills, deepen their conceptual knowledge, and unlock the path to mathematical success.
The Birth of All Things Algebra 2015
All Things Algebra 2015 was born out of Gina Wilson's desire to provide teachers with a comprehensive and easy-to-use curriculum that would help them engage their students and promote deep understanding of mathematical concepts. Recognizing the need for high-quality resources, Gina Wilson set out to create a platform that would serve as a one-stop-shop for educators seeking effective teaching materials.
Key Features of All Things Algebra 2015
1. comprehensive curriculum.
All Things Algebra 2015 offers a comprehensive curriculum that covers a wide range of topics in mathematics. From basic algebra to advanced calculus, Gina Wilson's resources cater to various grade levels and learning objectives. The curriculum is carefully designed to ensure a logical progression of concepts, allowing students to build a solid foundation in mathematics.
2. Engaging Activities and Worksheets
One of the standout features of All Things Algebra 2015 is its collection of engaging activities and worksheets. Gina Wilson understands the importance of hands-on learning and provides educators with a wealth of interactive resources that make math come alive in the classroom. These activities and worksheets not only reinforce concepts but also promote critical thinking and problem-solving skills.
3. Differentiated Instruction
Recognizing that students have different learning styles and abilities, Gina Wilson has integrated differentiated instruction into All Things Algebra 2015. Teachers can easily adapt the resources to meet the diverse needs of their students, ensuring that everyone has the opportunity to succeed. Whether it's through tiered assignments or alternative assessments, Gina Wilson's approach to differentiation empowers teachers to create inclusive learning environments.
4. Online Support and Community
All Things Algebra 2015 goes beyond just providing resources. Gina Wilson has fostered a strong online community where educators can connect, collaborate, and seek support. Through forums, discussion boards, and social media groups, teachers can share ideas, ask questions, and gain valuable insights from their peers. This sense of community enhances the overall teaching experience and encourages professional growth.
5. Continuous Updates and Improvements
To stay at the forefront of mathematics education, Gina Wilson continuously updates and improves All Things Algebra 2015. She actively seeks feedback from teachers and students, incorporating their suggestions into future releases. This commitment to ongoing development ensures that the resources remain relevant, aligned with current standards, and reflect the evolving needs of educators.
Success Stories and Testimonials
All Things Algebra 2015 has garnered praise from educators and students worldwide. Teachers have reported increased student engagement, improved test scores, and a deeper understanding of mathematical concepts. Students have expressed appreciation for the clarity of the resources and the opportunity to learn at their own pace. These success stories and testimonials serve as a testament to the impact of Gina Wilson's work.
In conclusion, Gina Wilson and her creation, All Things Algebra 2015, have revolutionized mathematics education. Through a comprehensive curriculum, engaging activities, differentiated instruction, online support, and continuous updates, Gina Wilson has provided teachers with the tools they need to inspire and empower their students. The impact of All Things Algebra 2015 extends beyond the classroom, shaping the way mathematics is taught and learned.
1. Can All Things Algebra 2015 be used in homeschooling?
Absolutely! All Things Algebra 2015 is a versatile resource that can be used in various educational settings, including homeschooling. Its comprehensive curriculum and engaging activities make it an ideal choice for homeschooling parents.
2. Are the resources in All Things Algebra 2015 aligned with curriculum standards?
Yes, all resources in All Things Algebra 2015 are meticulously aligned with curriculum standards. Gina Wilson ensures that the content remains up-to-date and meets the requirements of various educational frameworks.
3. Is there a free trial available for All Things Algebra 2015?
Unfortunately, there is no free trial available for All Things Algebra 2015. However, you can access a wide range of sample resources on the website to get a sense of the quality and effectiveness of the materials.
4. Can I customize the resources in All Things Algebra 2015 to suit my students' needs?
Yes, you can easily customize the resources in All Things Algebra 2015 to meet the specific needs of your students. The differentiated instruction approach allows for flexibility and adaptation.
5. How often are new resources added to All Things Algebra 2015?
Gina Wilson is dedicated to continuous improvement and regularly adds new resources to All Things Algebra 2015. Updates are released periodically to enhance the curriculum and address emerging educational trends.
We offer PDF printables in the highest quality.
- Preschool/kindergarten
- Grade 1 worksheets.
- Grade 2 - 6 Worksheets
Fun Games for Teaching Maths
- Penalty shooting game
- En Garde Duel Game
- Fling the teacher fun game
- More More Games.
Parents, teachers and educators can now present the knowledge using these vividly presented short videos. Simply let the kids watch and learn.
Quizzes are designed around the topics of addition, subtraction, geometry, shapes, position, fractions, multiplication, division, arithmetic, algebra etc.
Access the materials by looking at topics - Addition, Subtraction, Multiplication, Geometry, Trigonometry, algebra, Decimals, Division and more.
Math Printables by levels
Math practice for kids.
- Math Worksheets
- Math Video Slides
- Math Quizzes
- Math Downloads
PRINTABLE EXERCISES
- Multiplication
- Algebra & More
Interactive Math
- Subtraction Games
- Multiplication Quizzes
- Geometry Exercises
- Video Lessons
- CBSE Class 10
CBSE Class 10 Maths Competency-Based Questions With Answer Key 2024-25:Chapter 3 Pair of Linear Equations in Two Variables Free PDF Download
Cbse class 10 maths chapter 3 practice questions 2025: cbse class 10 students check and download the competency-focused practice questions for maths chapter 3 pair of linear equations in two variables. .
CBSE 2024-25 Competency Based Questions With Answers: The Central Board of Secondary Education (CBSE) is a national level education body. It has released the competency-based practice questions for students of classes 10 and 12. Students can download them from the school section of Jagran Josh. Here we have given the CBSE Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables competency-focused questions in pdf format.
The practice questions will be of great help to the students from the point of examinations. It will help cultivate the much needed skills of analysis, critical thinking and problem solving in students. These competency‐based questions are designed specifically to test conceptual understanding and application of concepts. Students can also use it to understand different kinds of questions and practice specific concepts and competencies.
CBSE Class 10 Maths Chapter 3 Competency Based-Questions
Anjali solved the following equations for the value of x. |
State true or false for the below statement and justify your answer |
A pair of linear equations is shown below. |
The equations of the lines l 1 ,l 2 , and l 3 are given by 5x + 3y = 2p ,35x + 21y = pq , and 100x + 4qy = 240, respectively, where p and q are real numbers. |
Shown below is a parallelogram with ∠ABC = 70°. |
Anisha lives 15 km away from her school. She walks to the bus stop and takes a bus to school everyday. |
CBSE Video Courses for Class 10 Students
Class 10 students can study effectively for the exams with the help of video courses prepared by the subject matter experts. These video courses will explain the concepts in a simple and interactive manner which will help learners to understand clearly.
CBSE Class 10 Video Courses
Also, check
CBSE Class 10 Syllabus 2024-25: Download Latest and Revised FREE PDFs
NCERT Books for Class 10 All Subjects PDF (2024-25)
NCERT Solutions for Class 10 (2024-2025) All Subjects & Chapters: Download in PDF
NCERT Rationalised Syllabus Class 10: List of Chapters & Topics Deleted from Books
Polynomial Equations: Concepts, Definition, Formula and Types
Relations and Functions
Profit And Loss
Get here latest School , CBSE and Govt Jobs notification and articles in English and Hindi for Sarkari Naukari , Sarkari Result and Exam Preparation . Download the Jagran Josh Sarkari Naukri App .
- India Post GDS Merit List 2024
- NDA Question Paper 2024
- RBI Grade B Admit Card 2024
- UP Police Constable Admit Card 2024
- SSC CGL Admit Card 2024
- UP Police Constable Question Paper 2024 PDF
- CDS Question Paper 2024
- UP Police Exam Analysis 2024 Live Updates
- Sri Krishna
- Janmashtami Wishes in Hindi
Latest Education News
Paris Paralympics 2024 India Medals list: भारत ने जीते कितने पदक,पढ़ें सबके नाम
UAE Visa Amnesty Program: दुबई में फंसे भारतीयों के लिए राहत, यूएई वीज़ा माफी योजना का ऐसे उठाये लाभ
India's First Vande Bharat Sleeper Train: पहली वंदे भारत स्लीपर ट्रेन कब और किसी रूट पर दौड़ेगी? जानें
eShram Card: क्या है ई-श्रम कार्ड? लाभ, पात्रता और ऑनलाइन अप्लाई की सभी डिटेल्स यहां देखें, e-shram Card Download का तरीका
वंदे भारत ट्रेन नेटवर्क में आते है उत्तर प्रदेश के कितने शहर? पढ़ें सबके नाम
उत्तर प्रदेश के 8 रेलवे स्टेशनों को मिले नए नाम, यहां देखें नई लिस्ट
Top 10 Weekly Current Affairs in Hindi: 27 अगस्त से 01 सितंबर 2024
NDA Exam Analysis 2024: Check Today's Maths & GAT Paper Review, Difficulty Level, Good Attempts, and Questions Asked
NDA Question Paper 2024: Download NDA 2 Maths and GAT Set Wise Paper PDF Here
NDA 2 Answer Key 2024: Download Unofficial Maths and GAT Solutions
Optical Illusion: What you see first reveals the deepest desires you have in life
Personality Test: Your Eyelash Reveals Your Hidden Personality Traits
Personality Test: Your Toes Reveal Your Hidden Personality Traits
Picture Puzzle IQ Test: Only 1% Highly Observational Can Spot The Mouse In 12 Seconds!
Brain Teaser IQ Test: How Smart Are You? Find The Correct Keyhole In 8 Seconds!
Optical Illusion IQ Test: Test Your Visual Sharpness! Can You Spot the Tiger in 8 Seconds?
Optical Illusion: Are You Among the 2% High IQ Who Can Spot All 4 Faces in 8 Seconds?
Picture Puzzle: Which Button Rings The Bell? Test Your IQ in 8 Seconds!
Brain Teaser IQ Test: Only 1 in 10 Can Solve This Burger Puzzle in 9 Seconds!
Brain Teaser IQ Test: Which Line is Longer? Only 1% Geniuses Answer It Correctly In 6 Seconds!
IMAGES
VIDEO
COMMENTS
Unit 4: Linear Equations Homework 10: Parallel & Perpendicular Lines (Day 2) Write an equation passing through the point and PARALLEL to the given line. + 6 5.1 = +15 6. (-5, -1); 2x-4 5. (-10, 1); 21 + 9 = 15 Directions: Write an equation passing through the point and PERPENDICULAR to the given line. +10 11. (10, 7); 5x-6y= 18
2.1 Use a General Strategy to Solve Linear Equations; 2.2 Use a Problem Solving Strategy; 2.3 Solve a Formula for a Specific Variable; 2.4 Solve Mixture and Uniform Motion Applications; 2.5 Solve Linear Inequalities; 2.6 Solve Compound Inequalities; 2.7 Solve Absolute Value Inequalities
Introduction to Systems of Equations and Inequalities; 7.1 Systems of Linear Equations: Two Variables; 7.2 Systems of Linear Equations: Three Variables; 7.3 Systems of Nonlinear Equations and Inequalities: Two Variables; 7.4 Partial Fractions; 7.5 Matrices and Matrix Operations; 7.6 Solving Systems with Gaussian Elimination; 7.7 Solving Systems with Inverses; 7.8 Solving Systems with Cramer's Rule
2.1 Use a General Strategy to Solve Linear Equations; 2.2 Use a Problem Solving Strategy; 2.3 Solve a Formula for a Specific Variable; 2.4 Solve Mixture and Uniform Motion Applications; 2.5 Solve Linear Inequalities; 2.6 Solve Compound Inequalities; 2.7 Solve Absolute Value Inequalities
Title: Unit 2 - Linear Functions & Systems Author: Amanda Appel Created Date: 8/17/2018 5:17:05 PM
Algebra 2 -25 - Functions, Equations, and Graphs WARM UP Solve each equation for y. 1) 12y=3x 2) −10y=5x 3) 3 4 y=15x y= 1 4 x y=− 1 2 x y=20x KEY CONCEPTS AND VOCABULARY Direct Variation- a linear function defined by an equation of the form y=kx, where k ≠ 0. Constant of Variation - k, where k = y/x GRAPHS OF DIRECT VARIATIONS
Algebra questions and answers; Unit 2: Linear Functions Date: Bell: Homework 3: Writing Linear Equations, Applications, & Linear Regression **This is a 2-page documenti ** Point Slope & Two Points: Write a linear equation in slope-intercept form with the given Information 1. slope = -6; passes through (-4,1) 2. slope = passes through (-5, -6) 3 ...
y = 2. y = −x + 3. y = −2x + 5. This page titled 2.4: Graphing Linear Equations- Answers to the Homework Exercises is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Darlene Diaz ( ASCCC Open Educational Resources Initiative) via source content that was edited to the style and standards of the LibreTexts ...
Homework Answer Keys Final Exam Materials Calculator TIps Answer keys are listed ... Solving Equations Review KEY: File Size: 794 kb: File Type: pdf: Download File. Unit 1 Practice Test KEY: ... SOLVING SYSTEMS OF LINEAR EQUATIONS. Homework 1: File Size: 1055 kb: File Type: pdf: Download File. Homework 2: File Size:
Find step-by-step solutions and answers to Algebra 2: Homework Practice Workbook - 9780078908620, as well as thousands of textbooks so you can move forward with confidence. ... Section 2-4: Writing Linear Equations. Section 2-5: Scatter Plots and Lines of Regressions. Section 2-6: Special Functions. Section 2-7: Parent Functions and ...
2-5 Writing Linear Functions Homework. 2-5 Writing and Interpreting Functions ANSWERS. 2-5 Writing and Interpreting Functions. 2-4 Homework ANSWERS. 2-4 Homework. 2-4 Writing Linear Equations ANSWERS. 2-4 Writing Linear Equations. 2-3 Function Notation Practice 2 ANSWERS. 2-3 Function Notation Practice 2.
Unit 2: Linear Funcöons Homework 5: Solving Systems of Equations ... ** This is a 2-page document! ** Directions: Solve each system of equaäons by elimination. Clearly identify your soluton. ... 8y+26 -3 = 46) 6. 8. 14x+7y=-7 l.ÐI.þ Gina Wilson (All 2015 . Title: Unit 2 - Linear Functions & Systems Author: Amanda Appel Created Date: 9/1 ...
Answer Key to Lesson 2 - Homework 1) (a) −5 (b) 2 (c) − 5 8 (d) For linear functions the average rate of change is constant. Since the average rate of change is different for each of these intervals, this is not a linear function. 2) (a) 2 (b) 6 (c) 10 (d) As we move from left to right, the graph is getting steeper.
Free worksheet(pdf) and answer key on the solving word problems based on linear equations and real world linear models. Scaffolded questions that start relatively easy and end with some real challenges. Plus model problems explained step by step
Find step-by-step solutions and answers to College Algebra - 9780134217451, as well as thousands of textbooks so you can move forward with confidence. ... Section 2.5: Equations of Lines and Linear Models. Section 2.6: Graphs of Basic Functions. Section 2.7: Graphing Techniques. Section 2.8: Function Operations and Composition. Page 296: Review ...
Free printable worksheet (pdf) and Answer Key on slope includes visual aides, model problems, exploratory activities, practice problems, and an online component
the following linear equations to have. Transform the e. a simpler form if necessary.8. ( + )= +If I use the distributive property on the left side, I notice that the coefficients of are. e same, and the constants are the same. Therefore, thi. infinitely many solutions.9. ( + )= −If I use the distributive property on the left side, I notice ...
Lesson 1 - Intro to Systems of Equations. Lesson 2 - Consistent & Inconsistent Solutions. Lesson 3 - Solving by Substitution. Lesson 4 - Solving by Elimination. Lesson 5a - Word Problems. Lesson 5b - Word Problems. Lesson 6 - Arithmetic Sequences. Practice Test. Practice Test Answer Key.
We develop general methods for solving linear equations using properties of equality and inverse operations. Thorough review is given to review of equation solving from Common Core 8th Grade Math. Solutions to equations and inequalities are defined in terms of making statements true. This theme is emphasized throughout the unit.
Free printable worksheets with answer keys on linear equations including finding slope, slope intercept form, equation from 2 points,from 1 point and slope and more. ... as well as challenge questions at the sheets end. Plus each one comes with an answer key. Mixed Problems on Writing Equations of Lines Slope Intercept Form Worksheet ...
Solving Linear Equations in One Variable. A linear equation is an equation of a straight line, written in one variable. The only power of the variable is 1. Linear equations in one variable may take the form a x + b = 0 a x + b = 0 and are solved using basic algebraic operations. We begin by classifying linear equations in one variable as one ...
The answer key on Gina Wilson All Things Algebra offers various features that enhance the learning experience. Some notable features include: Detailed Solutions: The answer key provides comprehensive and detailed solutions to the exercises, enabling students to identify any errors and learn from them. Multiple Approaches: In many cases, the ...
CBSE Class 10 Maths Competency-Based Questions With Answer Key 2024-25:Chapter 3 Pair of Linear Equations in Two Variables Free PDF Download . CBSE Class 10 Maths Chapter 3 Practice Questions 2025 ...