Learn on PengiYoshiwara Intermediate AlgebraChapter 1: Linear Models

Lesson 2: Graphs and Equations

In this Grade 7 lesson from Yoshiwara Intermediate Algebra (Chapter 1: Linear Models), students learn to identify solutions of linear equations and inequalities, including the rule for reversing the inequality sign when multiplying or dividing by a negative number. The lesson also introduces equations in two variables, where students practice determining whether ordered pairs satisfy a given equation.

Section 1

πŸ“˜ Graphs and Equations

New Concept

This lesson reveals the link between algebraic equations and their visual graphs. You'll learn that a graph is a picture of all possible solutions, allowing you to solve equations and inequalities like Ax+By=CAx + By = C graphically.

What’s next

Next, you’ll work through interactive examples of solving equations graphically, followed by a sequence of practice cards to master this new skill.

Section 2

Solution of an equation

Property

A solution of an equation is a value of the variable that makes the equation true.

Examples

x=βˆ’2x = -2 is a solution of the equation 4x3+5x2βˆ’8x=44x^3 + 5x^2 - 8x = 4 because substituting βˆ’2-2 for xx results in 4(βˆ’8)+5(4)βˆ’8(βˆ’2)=βˆ’32+20+16=44(-8) + 5(4) - 8(-2) = -32 + 20 + 16 = 4.

To solve 3(2xβˆ’5)βˆ’4x=2xβˆ’(6βˆ’3x)3(2x - 5) - 4x = 2x - (6 - 3x), simplify both sides to 2xβˆ’15=5xβˆ’62x - 15 = 5x - 6. Isolating the variable gives βˆ’3x=9-3x = 9, so the solution is x=βˆ’3x = -3.

Section 3

Solving a linear inequality

Property

Although a linear equation can have at most one solution, a linear inequality can have many solutions.
If we multiply or divide both sides of an inequality by a negative number, we must reverse the direction of the inequality.

Examples

To solve 5βˆ’2x<25 - 2x < 2, first subtract 5 from both sides to get βˆ’2x<βˆ’3-2x < -3. When you divide by βˆ’2-2, you must reverse the inequality to get the solution x>32x > \frac{3}{2}.

To solve the inequality βˆ’2x>βˆ’8-2x > -8, we divide both sides by βˆ’2-2. Because we are dividing by a negative number, we must reverse the inequality symbol, which gives the solution x<4x < 4.

Section 4

Equations in two variables

Property

An equation in two variables, such as y=4xβˆ’2y = 4x - 2, has many solutions. Each solution consists of an ordered pair of values, one for xx and one for yy, that together satisfy the equation (make the equation true.)

Examples

The ordered pair (1,3)(1, 3) is not a solution of y=4xβˆ’2y = 4x - 2 because substituting the values gives 4(1)βˆ’2=24(1) - 2 = 2. Since the y-value is 3, and 3β‰ 23 \ne 2, the pair is not a solution.

Section 5

Equations in two variables

Property

An equation in two variables, such as βˆ’3x+4y=24-3x + 4y = 24, has many solutions. Each solution consists of an ordered pair of values, one for xx and one for yy, that together satisfy the equation (make the equation true.)

Examples

The ordered pair (4,3)(4, 3) is not a solution of βˆ’3x+4y=24-3x + 4y = 24 because substituting the values gives βˆ’3(4)+4(3)=βˆ’12+12=0-3(4) + 4(3) = -12 + 12 = 0, which does not equal 24.

Section 6

Graph of an equation

Property

The graph of an equation in two variables is just a picture of all its solutions. If a point lies on the graph of an equation, it is a solution of the equation.

Examples

If a point (p,q)(p, q) does not lie on the graph of Ax+By=CAx + By = C, it means that the ordered pair (p,q)(p, q) is not a solution, and the statement Ap+Bq=CAp + Bq = C is not true.

Section 7

Graphical solution of equations

Property

We can use graphs to find solutions to equations in one variable.

Examples

To solve the equation 150=285βˆ’15x150 = 285 - 15x using the graph of y=285βˆ’15xy = 285 - 15x, find the point on the graph where the y-coordinate is 150. The x-coordinate of that point is x=9x=9, which is the solution.

To solve the inequality 285βˆ’15xβ‰₯150285 - 15x \ge 150 using the graph of y=285βˆ’15xy = 285 - 15x, find all points where the y-coordinate is 150 or more. The x-coordinates for these points are all values less than or equal to 9, so x≀9x \le 9.

Section 8

Graphical solution of equations

Property

We can use graphs to find solutions to equations and inequalities in one variable.

Examples

Using the graph of y=3xβˆ’2y = 3x - 2:

  • To solve the equation 10=3xβˆ’210 = 3x - 2: Find the point on the line where the y-coordinate is 10. The corresponding x-coordinate, x=4x=4, is the solution.

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 1: Linear Models

  1. Lesson 1

    Lesson 1: Linear Models

  2. Lesson 2Current

    Lesson 2: Graphs and Equations

  3. Lesson 3

    Lesson 3: Intercepts

  4. Lesson 4

    Lesson 4: Slope

  5. Lesson 5

    Lesson 5: Equations of Lines

  6. Lesson 6

    Lesson 6: Chapter Summary and Review

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

πŸ“˜ Graphs and Equations

New Concept

This lesson reveals the link between algebraic equations and their visual graphs. You'll learn that a graph is a picture of all possible solutions, allowing you to solve equations and inequalities like Ax+By=CAx + By = C graphically.

What’s next

Next, you’ll work through interactive examples of solving equations graphically, followed by a sequence of practice cards to master this new skill.

Section 2

Solution of an equation

Property

A solution of an equation is a value of the variable that makes the equation true.

Examples

x=βˆ’2x = -2 is a solution of the equation 4x3+5x2βˆ’8x=44x^3 + 5x^2 - 8x = 4 because substituting βˆ’2-2 for xx results in 4(βˆ’8)+5(4)βˆ’8(βˆ’2)=βˆ’32+20+16=44(-8) + 5(4) - 8(-2) = -32 + 20 + 16 = 4.

To solve 3(2xβˆ’5)βˆ’4x=2xβˆ’(6βˆ’3x)3(2x - 5) - 4x = 2x - (6 - 3x), simplify both sides to 2xβˆ’15=5xβˆ’62x - 15 = 5x - 6. Isolating the variable gives βˆ’3x=9-3x = 9, so the solution is x=βˆ’3x = -3.

Section 3

Solving a linear inequality

Property

Although a linear equation can have at most one solution, a linear inequality can have many solutions.
If we multiply or divide both sides of an inequality by a negative number, we must reverse the direction of the inequality.

Examples

To solve 5βˆ’2x<25 - 2x < 2, first subtract 5 from both sides to get βˆ’2x<βˆ’3-2x < -3. When you divide by βˆ’2-2, you must reverse the inequality to get the solution x>32x > \frac{3}{2}.

To solve the inequality βˆ’2x>βˆ’8-2x > -8, we divide both sides by βˆ’2-2. Because we are dividing by a negative number, we must reverse the inequality symbol, which gives the solution x<4x < 4.

Section 4

Equations in two variables

Property

An equation in two variables, such as y=4xβˆ’2y = 4x - 2, has many solutions. Each solution consists of an ordered pair of values, one for xx and one for yy, that together satisfy the equation (make the equation true.)

Examples

The ordered pair (1,3)(1, 3) is not a solution of y=4xβˆ’2y = 4x - 2 because substituting the values gives 4(1)βˆ’2=24(1) - 2 = 2. Since the y-value is 3, and 3β‰ 23 \ne 2, the pair is not a solution.

Section 5

Equations in two variables

Property

An equation in two variables, such as βˆ’3x+4y=24-3x + 4y = 24, has many solutions. Each solution consists of an ordered pair of values, one for xx and one for yy, that together satisfy the equation (make the equation true.)

Examples

The ordered pair (4,3)(4, 3) is not a solution of βˆ’3x+4y=24-3x + 4y = 24 because substituting the values gives βˆ’3(4)+4(3)=βˆ’12+12=0-3(4) + 4(3) = -12 + 12 = 0, which does not equal 24.

Section 6

Graph of an equation

Property

The graph of an equation in two variables is just a picture of all its solutions. If a point lies on the graph of an equation, it is a solution of the equation.

Examples

If a point (p,q)(p, q) does not lie on the graph of Ax+By=CAx + By = C, it means that the ordered pair (p,q)(p, q) is not a solution, and the statement Ap+Bq=CAp + Bq = C is not true.

Section 7

Graphical solution of equations

Property

We can use graphs to find solutions to equations in one variable.

Examples

To solve the equation 150=285βˆ’15x150 = 285 - 15x using the graph of y=285βˆ’15xy = 285 - 15x, find the point on the graph where the y-coordinate is 150. The x-coordinate of that point is x=9x=9, which is the solution.

To solve the inequality 285βˆ’15xβ‰₯150285 - 15x \ge 150 using the graph of y=285βˆ’15xy = 285 - 15x, find all points where the y-coordinate is 150 or more. The x-coordinates for these points are all values less than or equal to 9, so x≀9x \le 9.

Section 8

Graphical solution of equations

Property

We can use graphs to find solutions to equations and inequalities in one variable.

Examples

Using the graph of y=3xβˆ’2y = 3x - 2:

  • To solve the equation 10=3xβˆ’210 = 3x - 2: Find the point on the line where the y-coordinate is 10. The corresponding x-coordinate, x=4x=4, is the solution.

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 1: Linear Models

  1. Lesson 1

    Lesson 1: Linear Models

  2. Lesson 2Current

    Lesson 2: Graphs and Equations

  3. Lesson 3

    Lesson 3: Intercepts

  4. Lesson 4

    Lesson 4: Slope

  5. Lesson 5

    Lesson 5: Equations of Lines

  6. Lesson 6

    Lesson 6: Chapter Summary and Review