Topics
2. Expressions, Equations, & Inequalities
topic badge
United States of AmericaVirginia
Grade 8

2.04 Solve multistep equations

Lesson
Practice

Solve equations using visual models

An equation shows that two expressions are equal in value. Because of this equivalence we can use a balance scale model to visualize how to solve an equation.

Consider the balance scale model for the equation 2(2x-1)=2(x+5) where a blue bag represents x, a sack of balloons tied to the scale represents -x and takes away the same amount of weight that a blue bag adds. A blue block represents +1 and a single balloon tied to the scale represents -1 and takes away the same amount of weight that a blue block adds.

A scale balance. On the left, 2 groups of 2 bags of x and a ballon of -1. On the right, 2 groups of a bag of x and 5 blocks of +1.

Since each side of the balance scale is made up of two equal groups of something, we can eliminate one group from each side. Remember, the scale is balanced so the weight on each side was equal and now we have removed half of the weight from each side so the scale will stay balanced.

This gives us an equation of 2x-1=x+5 which is equivalent to the original equation.

A scale balance. On the left, 2 bags of x and a balloon of -1. On the right, a bag of x and five blocks of +1.

On the left side of the balanced equation, we have a balloon (negative). We can create a zero pair by adding a blue (positive) block. Adding a positive block to the left side will increase the weight of that side and throw off the balance. So we need to also add a blue (positive) block to the right side to keep the scale balanced.

This creates an equation of 2x-1+1=x+5+1 which is equivalent to both of the previous equations.

A scale balance. On the left, 2 bags of x, a balloon of -1, and a block of +1. On the right, a bag of x and six blocks of +1.

We can remove the zero pair (one blue block and one balloon) from the left side of the scale without changing the weight and get an equation of 2x=x+6.

A scale balance. On the left, 2 bags of x. On the right, a bag of x and six blocks of +1.

On the right side of the balanced equation, we have a blue (positive) bag. We can create a zero pair by adding a bag of balloons to the right side of the scale and to keep the scale balanced will add a bag of balloons to the left side as well.

This gives us an equation of 2x-x=x-x+6. This new equation is equivalent to all previous equations.

A scale balance. On the left, 2 bags of x and ballons with cover of a bag of -x. On the right, a bag of x, ballons with cover of a bag of -x, and six blocks of +1.

Removing the zero pairs on the left and right sides of the scale, we get a new equation x=6. This means that one blue bag is equal to six blue (positive) blocks.

A scale balance. On the left, a bag of x. On the right, six blocks of +1.

We have now solved the equation 2(2x-1)=2(x+5) and got x=6 using a balance scale.

Examples

Example 1

Consider the equation 4(x−3)+8=12

a

Represent the equation using algebra tiles.

Worked Solution
Create a strategy

To represent this equation with algebra tiles, represent each part of the equation as a different type of tile. You will need variable titles and positive and negative unit tiles.

Apply the idea

Let's start with the 4(x-3) part of the expression on the left hand side of the equation. The parentheses tell us that we have 4 groups of x-3. To create x-3 we can use one x tile and three -1 tiles. Then we will create 4 groups of this to represent 4(x-3).

The additional 8 unit tiles on the left side represent the +8.

For the right side, use 12 unit tiles to represent the number 12.

4 tiles of +x, 12 tiles of -1, 8 tiles of +1 equals 12 tiles of +1.
b

Use the algebra tiles to solve the equation.

Worked Solution
Create a strategy

Remove any zero pairs, then add tiles as needed to create and remove additional zero pairs until only one x tile remains on the left side of the equation.

Apply the idea

Identify the zero pairs on the left side of the equation. One +1 tile and one -1 tile create a zero pair. There are 8 zero pairs on the left side of the equation.

8 tiles of -1 and 8 tiles of +1.

Removing the 8 zero pairs, we have:

4 tiles of +x, 4 tiles of -1 equals 12 tiles of +1.

To be able to remove the -1 tiles from the left side, we need to add four +1 tiles. To keep the balance we will do the same on the right side of the equation.

4 tiles of +x, 4 tiles of -1, 4 tiles of +1 equals 16 tiles of +1.

Removing the zero pairs we created on the left, we are left with four x tiles on the left and 16 unit tiles on the right.

4 tiles of +x equals 16 tiles of +1.

Let's see if we can divide the 16 unit tiles on the right into 4 equal sized groups so that one group represents one x tile.

4 tiles of +x and 16 tiles of +1.

Removing 3 groups from each side will keep the equation balanced and leave us with one x tile on the left and 4 unit tiles on the right. So x=4.

A tile of +x equals 4 tiles of +1.
Reflect and check

After finding the solution, substitute it into the original equation to verify it.

\displaystyle 4(x−3)+8\displaystyle =\displaystyle 12Write the equation
\displaystyle 4(4-3)+8\displaystyle =\displaystyle 12Substitute x=4
\displaystyle 4(1)+8\displaystyle =\displaystyle 12Evaluate the subtraction
\displaystyle 4+8\displaystyle =\displaystyle 12Evaluate the multiplication
\displaystyle 12\displaystyle =\displaystyle 12Evaluate the addition

This confirms that x=4 is the solution to the equation.

Example 2

A local community library is organizing a book drive. Jordan collects 5 boxes of books and 2 individual books. A local bookstore donates 3 boxes of books and 12 individual books to the drive.

Each box contains an equal number of books, and Jordan and the bookstore both donated the same total number of books.

a

Write an equation to represent this situation.

Worked Solution
Create a strategy

Represent the unknown value as a variable. In this case the unknown is the number of books in each box so we can use the variable b.

The equation should represent how the total number of books collected by Jordan and the bookstore are equal.

Apply the idea

Jordan donates 5 boxes of books. We can represent this algebraically as 5b (or 5 times the number of books in a box).

Jordan also donates 2 more individual books which we can show as 2. So the total number of books Jordan donates is 5b+2.

The bookstore donates 3 boxes of books. We can represent this algebraically as 3b (or 3 times the number of books in a box).

The bookstore also donates 12 more individual books which we can show as 12, for a total of 3b+12.

Since Jordan and the bookstore each donate the same number of books, we can set each donation amount equal to each other.

This gives us the equation 5b+2=3b+12.

b

Represent the equation using a pictorial model.

Worked Solution
Create a strategy

Use an image of a box to represent the number of boxes and an image of a book to represent each book.

Apply the idea

On the left, draw 5 boxes to represent the boxes of books and a separate pile of books to represent the 2 individual books donated by Jordan. On the right, draw 3 boxes to represent the boxes of books and a separate pile of books to represent the 12 individual books donated by the bookstore.

5 boxes and 2 books equals 3 boxes and 12 books.
c

Use the pictorial model to solve the equation.

Worked Solution
Create a strategy

We need to remove the 2 books from the left side of the equation and the 3 boxes on the right side of the equation while still keeping it balanced. Then we need to isolate a single box on the left side of the equation.

Apply the idea

Pictorial model of the equation from part (b).

5 boxes and 2 books equals 3 boxes and 12 books.

We first want to subtract 2 books from each side of the equation. Removing a book is represented by a grayed out book.

5 boxes, and 2 grayed books equals 3 boxes, 10 books, and 2 grayed books.

Now we can see that 5 boxes of books is equal to 3 boxes and 10 books. We can represent this as the equation 5b=3b+10.

5 boxes equals 3 boxes, and 10 books.

We want to subtract 3 boxes from each side of the equation. Removing a box is represented by a grayed out box.

3 grayed boxes, and 2 boxes equals 3 grayed boxes, and 10 books.

Now we see that 2 boxes of books contain a total of 10 books. We can represent this as the equation 2b=10.

2 boxes equals 10 books.

If we separate the books into two groups of the same size, each containing 5 books, we can find how many books will go in one box. Assuming the same number of books is in each box, one box contains 5 books. This relationship can be expressed with the equation b=5.

A box equals 5 books.
Idea summary

We can use visual models to represent and solve equations pictorally. We can use zero pairs and applying the same changes to both sides to maintain the balance or equality of the equation.

Solve equations algebraically

While visual models help us to better understand equations and their solutions, it's not always easy or efficient to solve equations using visual models.

Fortunately, we can use inverse operations, the properties of real numbers, and the properties of equality to solve equations algebraically.

Recall the properties of equality, inverse, and identity:

Symmetric property of equality\text{If } a=b, \text{then } b=a
Transitive property of equality\text{If } a=b \text{ and } b=c, \text{then } a=c
Addition property of equality\text{If } a=b, \text{then } a+c=b+c
Subtraction property of equality\text{If } a=b, \text{then } a-c=b-c
Multiplication property of equality\text{If } a=b, \text{then } ac=bc
Division property of equality\text{If } a=b \text{ and } c \neq0, \text{then } \dfrac{a}{c}=\dfrac{b}{c}
Substitution property of equality\text{If } {a=b}, \text{ then } b \text{ may be substituted for } a \text{ in any expression}
Additive identity\text{If } {a=b}, \text{ then } a+0=b \text{ and } a=b+0
Multiplicative identity\text{If } {a=b}, \text{ then } a\cdot 1=b \text{ and } a=b \cdot 1
Additive inversea+(-a)=0
Multiplicative inversea \cdot \dfrac{1}{a}=1

When we solve equations with variables on both sides, we want to rearrange the equation so that all of the variables are on the same side of the equation and all of the constants are on the other side of the equation. This process is similar to combining like terms, we just do it across the equal sign.

The side you choose to move variables to does not make a difference to the final solution; however, there is usually a way that will make it easier to solve.

For example, if we were solving the equation:

\displaystyle 10x - 4\displaystyle =\displaystyle 6xGiven equation
\displaystyle 10x-6x-4\displaystyle =\displaystyle 6x-6xSubtraction property of equality
\displaystyle 4x-4\displaystyle =\displaystyle 6x-6xCombine like terms
\displaystyle 4x-4\displaystyle =\displaystyle 0Additive inverse
\displaystyle 4x-4+4\displaystyle =\displaystyle 0+4Addition property of equality
\displaystyle 4x\displaystyle =\displaystyle 0+4Additive inverse
\displaystyle 4x\displaystyle =\displaystyle 4Additive identity
\displaystyle \dfrac{4x}{4}\displaystyle =\displaystyle \dfrac{4}{4}Division property of equality
\displaystyle 1 \cdot x\displaystyle =\displaystyle 1Multiplicative inverse
\displaystyle x\displaystyle =\displaystyle 1Multiplicative identity

Examples

Example 3

Solve the following equations:

a

5 \left( 2 y + 2\right) + 3 \left( 4 y - 5\right) +5y= 45

Worked Solution
Create a strategy

Use the distributive property and properties of equality to remove the parentheses and isolate the variable.

Apply the idea
\displaystyle 5 \left( 2 y + 2\right) + 3 \left( 4 y - 5\right) +5y\displaystyle =\displaystyle 45Given equation
\displaystyle 5 \cdot 2 y + 5 \cdot 2 + 3 \cdot 4 y - 3 \cdot 5 +5y\displaystyle =\displaystyle 45Distributive property
\displaystyle 10y+10+12y-15+5y\displaystyle =\displaystyle 45Evaluate the multiplication
\displaystyle 10y+12y+5y+10-15\displaystyle =\displaystyle 45Associative property of addition
\displaystyle 27y-5\displaystyle =\displaystyle 45Combine like terms
\displaystyle 27y-5+5\displaystyle =\displaystyle 45+5Addition property of equality
\displaystyle 27y\displaystyle =\displaystyle 50Combine like terms
\displaystyle \frac{27y}{27}\displaystyle =\displaystyle \dfrac{50}{27}Division property of equality
\displaystyle y\displaystyle =\displaystyle \dfrac{50}{27}Evaluate the division
Reflect and check

We don't always show the steps for all of the properties we are using, notice the use of the additive inverse, multiplicative inverse, and multiplicative identity in between steps. For instance, when we have:

\displaystyle 5 \left( 2 y + 2\right) + 3 \left( 4 y - 5\right) +5y\displaystyle =\displaystyle 45Given equation
\displaystyle 27y-5+5\displaystyle =\displaystyle 45+5Addition property of equality
\displaystyle 27y+0\displaystyle =\displaystyle 45+5Additive inverse
\displaystyle 27y\displaystyle =\displaystyle 45+5Additive identity
\displaystyle 27y\displaystyle =\displaystyle 50Evaluate the addition
\displaystyle \frac{27y}{27}\displaystyle =\displaystyle \dfrac{50}{27}Division property of equality
\displaystyle 1 \cdot y\displaystyle =\displaystyle \dfrac{50}{27}Multiplicative inverse
\displaystyle y\displaystyle =\displaystyle \dfrac{50}{27}Multiplicative identity
b

- \dfrac{3}{2} \left(x + \dfrac{5}{4}\right) + \dfrac{11}{3} = - \dfrac{125}{6}

Worked Solution
Create a strategy

Use the distributive property and properties of equality to remove the parentheses and isolate the variable.

Apply the idea
\displaystyle - \dfrac{3}{2} \left(x + \dfrac{5}{4}\right) + \dfrac{11}{3}\displaystyle =\displaystyle - \dfrac{125}{6}Given equation
\displaystyle - \dfrac{3}{2} \cdot x + - \dfrac{3}{2} \cdot \dfrac{5}{4} + \dfrac{11}{3}\displaystyle =\displaystyle - \dfrac{125}{6}Distributive property
\displaystyle - \dfrac{3}{2}x-\dfrac{15}{8} + \dfrac{11}{3}\displaystyle =\displaystyle - \dfrac{125}{6}Evaluate the multiplication
\displaystyle - \dfrac{3}{2}x+\dfrac{43}{24}\displaystyle =\displaystyle - \dfrac{125}{6}Combine like terms
\displaystyle - \dfrac{3}{2}x+\dfrac{43}{24}-\dfrac{43}{24}\displaystyle =\displaystyle - \dfrac{125}{6}-\dfrac{43}{24}Subtraction property of equality
\displaystyle - \dfrac{3}{2}x\displaystyle =\displaystyle -\dfrac{181}{8}Combine like terms
\displaystyle - \dfrac{3}{2}x \cdot - \dfrac{2}{3}\displaystyle =\displaystyle -\dfrac{181}{8} \cdot - \dfrac{2}{3}Multiplication property of equality
\displaystyle x\displaystyle =\displaystyle \dfrac{181}{12}Evaluate the multiplication
Reflect and check

We can check our work by substituting the solution, x=\dfrac{181}{12}, into the original equation and simplifying.

\displaystyle - \dfrac{3}{2} \left(\dfrac{181}{12}\right)-\dfrac{15}{8} + \dfrac{11}{3}\displaystyle =\displaystyle - \dfrac{125}{6}Substitute x=\dfrac{181}{12}
\displaystyle \dfrac{181}{8}-\dfrac{15}{8} + \dfrac{11}{3}\displaystyle =\displaystyle - \dfrac{125}{6}Evaluate the multiplication
\displaystyle - \dfrac{125}{6}\displaystyle =\displaystyle - \dfrac{125}{6}Evaluate the subtraction and addition

Since the left and right side of the equation is equal, we can confirm our solution of x=\dfrac{181}{12}

c

9x+40-7x=4x

Worked Solution
Create a strategy

Rearrange the equation so that all of the x terms are on the same side.

Apply the idea
\displaystyle 9x+40-7x\displaystyle =\displaystyle 4xGiven equation
\displaystyle 9x+40-7x-4x\displaystyle =\displaystyle 4x-4xSubtraction property of equality
\displaystyle 9x+40-7x-4x\displaystyle =\displaystyle 0Additive inverse
\displaystyle -2x+40\displaystyle =\displaystyle 0Combine like terms
\displaystyle -2x+40-40\displaystyle =\displaystyle 0-40Subtraction property of equality
\displaystyle -2x\displaystyle =\displaystyle -40Combine like terms
\displaystyle \dfrac{-2x}{-2}\displaystyle =\displaystyle \dfrac{-40}{-2}Division property of equality
\displaystyle x\displaystyle =\displaystyle 20Evaluate the division
Reflect and check

Another way we could have solved is by moving the variable terms to the right side of the equation:

\displaystyle 9x+40-7x\displaystyle =\displaystyle 4xGiven equation
\displaystyle 9x+40-7x+7x\displaystyle =\displaystyle 4x+7xAddition property of equality
\displaystyle 9x+40\displaystyle =\displaystyle 11xCombine like terms
\displaystyle 9x-9x+40\displaystyle =\displaystyle 11x-9xSubtraction property of equality
\displaystyle 40\displaystyle =\displaystyle 2xCombine like terms
\displaystyle \dfrac{40}{2}\displaystyle =\displaystyle \dfrac{2x}{2}Division property of equality
\displaystyle 20\displaystyle =\displaystyle xEvaluate the division
\displaystyle x\displaystyle =\displaystyle 20Symmetric property of equality

Even though we used different steps, we still relied on the properties of equality and real numbers and ended up with x=20.

d

12.6x+28.25=1.1x+85.75

Worked Solution
Create a strategy

The coefficent for x will be positive if we rearrange the equation by moving the x values to the left side since 12.6 \gt 1.1.

Apply the idea
\displaystyle 12.6x+28.25\displaystyle =\displaystyle 1.1x+85.75Given equation
\displaystyle 12.6x-1.1x+28.25\displaystyle =\displaystyle 1.1x-1.1x+85.75Subtraction property of equality
\displaystyle 11.5x+28.25\displaystyle =\displaystyle 85.75Combine like terms
\displaystyle 11.5x+28.25-28.25\displaystyle =\displaystyle 85.75-28.25Subtraction property of equality
\displaystyle 11.5x\displaystyle =\displaystyle 57.5Combine like terms
\displaystyle \dfrac{11.5x}{11.5}\displaystyle =\displaystyle \dfrac{57.5}{11.5}Division property of equality
\displaystyle x\displaystyle =\displaystyle 5Evaluate the division
e

2\left(x-3\right)=x-3.

Worked Solution
Create a strategy

Use the distributive property first and then modify the equation so that all of the x terms are on the same side.

Apply the idea
\displaystyle 2\left(x-3\right)\displaystyle =\displaystyle x-3Given equation
\displaystyle 2 \cdot x - 2 \cdot 3\displaystyle =\displaystyle x-3Distributive property
\displaystyle 2x - 6\displaystyle =\displaystyle x-3Evaluate the multiplication
\displaystyle 2x - 6 + 6\displaystyle =\displaystyle x-3 + 6Addition property of equality
\displaystyle 2x\displaystyle =\displaystyle x-3 + 6Additive inverse
\displaystyle 2x\displaystyle =\displaystyle x +3Evaluate the subtraction
\displaystyle 2x-x\displaystyle =\displaystyle x - x +3Subtraction property of equality
\displaystyle 2x-x\displaystyle =\displaystyle 3Additive inverse
\displaystyle x\displaystyle =\displaystyle 3Combine like terms
Reflect and check

Substitute the solution, x=3, back into the original equation to verify its correctness.

\displaystyle 2(3-3)\displaystyle =\displaystyle 3-3Substitute x=3
\displaystyle 6-6\displaystyle =\displaystyle 3-3Evaluate the multipication
\displaystyle 0\displaystyle =\displaystyle 0Simplify

Since the left and right side of the equation is equal, we can confirm our solution of x=3

Idea summary

When solving multi-step equations involving parentheses:

  • Use the distributive property to clear parentheses.
  • Combine like terms to simplify each side.
  • Add or subtract to get the variable term on one side of the equals sign.
  • Multiply or divide to isolate the variable and solve the equation.

To solve an equation with variables on both sides, we can rearrange the equation so that all of the variables are on the same side of the equation and all of the constants are on the other side of the equation.

Outcomes

8.PFA.4

The student will write and solve multistep linear equations in one variable, including problems in context that require the solution of a multistep linear equation in one variable.

8.PFA.4a

Represent and solve multistep linear equations in one variable with the variable on one or both sides of the equation (up to four steps) using a variety of concrete materials and pictorial representations.

8.PFA.4b

Apply properties of real numbers and properties of equality to solve multistep linear equations in one variable (up to four steps). Coefficients and numeric terms will be rational. Equations may contain expressions that need to be expanded (using the distributive property) or require combining like terms to solve.

8.PFA.4c

Write a multistep linear equation in one variable to represent a verbal situation, including those in context.

8.PFA.4e

Solve problems in context that require the solution of a multistep linear equation.

8.PFA.4f

Interpret algebraic solutions in context to linear equations in one variable.

8.PFA.4g

Confirm algebraic solutions to linear equations in one variable.

What is Mathspace

About Mathspace