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1. Equations & Inequalities
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Algebra 1

1.04 Multi-step equations

Lesson
Practice

Multi-step equations

A solution to an equation is any value that can replace the variable and make a true statement. To determine whether or not a value satisfies an equation, we need to check whether the left-hand side of the equation is equal to the right-hand side of the equation. However, an equation may not have just one answer.

Exploration

Complete the table of values by evaluating each side of equation 1:

3x -5 = 3x-5

x3x-53x-5
-1
0
1
2
3

Now, complete the table for equation 2:

9+x=x+5

x9+xx+5
-1
0
1
2
3

Finally, complete the table for equation 3:

35+5x=2x+35

x35+5x2x+35
-1
0
1
2
3
  1. What are the differences between equations 1, 2, and 3?

  2. How many solutions do you think there will be to each equation? Why?

  3. How does the structure of an equation relate to the number of solutions it has?

  4. How many solutions do you think the equation 2\left(x-8\right)=3x+4 will have? Why?

We can determine the solutions to an equation by looking at its structure. To illustrate the difference, we will show how equations look different depending on having one solution, no solution, or infinitely many solutions using the starting equation

3(x+2)=⬚

If the equation simplifies to variables with the same coefficients and the same constants on each side of the equality, it will have infinitely many solutions.

An algebra balance scale with algebra tiles. On left side, there are 3 rows of +x tiles and 3 rows of two +1 tiles. On right side, there are 3 rows of +x tiles and 2 rows of three +1 tiles.

The equation represented is 3(x+2)=x+2(x+3), which has infinite solutions.

When each side is simplified, the equation becomes 3x+6=3x+6, shown on the balance by three +x tiles and six +1 tiles.

Regardless of what value of x is chosen for the +x tile, there are equal numbers of +x tiles on each side that will be added to the constants.

The number of +1 tiles are also equal on each side. This creates a balanced scale regardless of the value chosen for x.

If the equation simplifies to variables with the same coefficients but different constants, it will have no solution.

An algebra balance scale with algebra tiles. On left side, there are 3 rows of +x tiles and 3 rows of two +1 tiles. On right side, there are 3 rows of +x tiles and 2 rows of one -1 tiles.

The equation represented is 3(x+2)=3x-2, which has no solution.

Similar to the previous example, regardless of the value of x chosen, the value represented by the +x tiles on each side of the equality will always be the same.

However, the constants on either side of the equality are not the same. While the value of the +x tiles will always match, the constants keep the two sides from balancing.

If the equation simplifies to variables with different coefficients, there will be a unique solution regardless of constant values.

An algebra balance scale with algebra tiles. On left side, there are 3 rows of +x tiles and 3 rows of two +1 tiles. On right side, there are 1 row of three -1 tiles and 2 rows of +x tiles and 2 rows of two +x tiles.

The equation represented is 3(x+2)=4x-3, which has a unique solution of x=9.

When simplified, the equation is 3x+6=4x-3.

In our other examples, attempting to add or remove tiles to isolate a variable would eliminate the variable entirely.

Each side of this equation has a different number of +x tiles, so we can add and remove tiles equally to work towards isolating the variable to find the solution.

While there is more than one way to algebraically solve an equation, it is important to keep both sides of an equation equivalent by remembering that what you do to one side, you must do to the other.

Examples

Example 1

Verify that the m=-4 is a solution to the equation 14-7m= -3(-2+3m).

Worked Solution
Create a strategy

Both sides of an equation should be equivalent after substituting a solution for the variable and evaluating.

Apply the idea

Substitute m=-4 and use the order of operations to determine if both sides are equivalent.

\displaystyle 14 - 7 \cdot (-4)\displaystyle =\displaystyle -3\left(-2 + 3 \cdot (-4)\right)Substitute m=-4
\displaystyle 14+28\displaystyle =\displaystyle -3(-2-12)Evaluate the multiplication
\displaystyle 14+28\displaystyle =\displaystyle -3 \cdot (-14)Evaluate the subtraction
\displaystyle 42\displaystyle =\displaystyle 42Evaluate the multiplication

Since both sides are equivalent, m=-4 is a solution to 14-7m=3(-2+3m).

Example 2

Use a balance scale and algebra tiles to solve 3(x-4)=2(-2x+1).

Worked Solution
Create a strategy

A balance scale represents the algebraic expressions on each side of an equation using +x,\,-x,\,+1, and -1 tiles. We work towards isolating a varible by adding and removing tiles on each side equally.

Apply the idea

The original equation 3\left(x-4\right)=2\left(-2x+1\right) is represented by the balance scale:

An algebra balance scale with algebra tiles. On left side, there are 3 rows of one +x tiles and 3 rows of four -1 tiles. On right side, there are 2 rows of two -x tiles and 2 rows of one +1 tiles.

Adding +4x to each side creates zero pairs for the -4x on the right side.

A balance scale with algebra tiles. On left side, there are seven row of one +x tiles and three rows of four -1 tiles. On right side, there are 2 rows of two +x tiles, 2 rows of two -x tiles and 2 rows of one +1 tiles.

Next we remove the zero pairs on the right side and the scale stays balanced.

A balance scale with algebra tiles. On left side, there are seven row of one +x tiles and three rows of four -1 tiles. On right side, there are 2 rows of one +1 tiles.

Adding +12 to each side creates zero pairs for the -12 on the left side.

A balance scale with algebra tiles. On left side, there are seven rows of one +x tiles, three rows of four +1 tiles and three rows of four -1 tiles. On right side, there are three rows of four +1 tiles and two rows of one +1 tiles.

Remove the zero pairs from the left and the scale stays balanced. Since there are seven +x tiles on the left side, splitting the +1 tiles on the right into seven equal groups will tell us what each +x tile represents.

A balance scale with algebra tiles. On left side, there are seven +x tiles. On right side, there are fourteen +1 tiles.

Remove +x tiles from the left and two +1 tiles from the right at the same time to keep the scale balanced. Continue doing this until only one +x tile remains.

A balance scale with algebra tiles. On left side, there is one +x tiles. On right side, there are two +1 tiles.

We can see that x=2.

Reflect and check

Balance scales can also be used to verify solutions to equations. For this example, to verify that x=2 is the solution to 3 \left(x - 4 \right) = 2 \left(-2x + 1 \right), each +x will be replaced with two +1 tiles and each -x will be replaced with two -1 tiles.

A balance scale with algebra tiles. On left side, there are three rows of two +x tiles and 2 groups of three rows of two -1 tiles. On right side, there are two rows of three -1 tiles, a group of two rows of one -1 tiles and two rows of one +1 tiles.

After making zero pairs, a solution of an equation should show the same number of tiles left on each side of the balance.

A balance scale with algebra tiles. On left side, there are two rows of three -x tiles. On right side, there are two rows of three -1 tiles.

Since there are six -1 tiles on each side of the balance, x=2 is a solution to 3\left(x-4\right)=2\left(-2x+1\right).

Example 3

Determine how many solutions the following equations have without solving.

a

4\left(x-9\right)=x+6

Worked Solution
Create a strategy

Start by comparing both sides of the equation. We can see both sides are different and that there is an x on both sides. The coefficient on the left side is 4, and the coefficient on the right side is 1.

Apply the idea

The equation will have one unique solution.

Reflect and check

We can verify our answer by solving the equation.

\displaystyle 4\left(x-9\right)\displaystyle =\displaystyle x+6Original equation
\displaystyle 4x-36\displaystyle =\displaystyle x+6Distributive property
\displaystyle 3x-36\displaystyle =\displaystyle 6Subtraction property of equality
\displaystyle 3x\displaystyle =\displaystyle 42Addition property of equality
\displaystyle x\displaystyle =\displaystyle 14Division property of equality

There is only one solution to this equation.

b

2x-5=0.5\left(4x+10\right)

Worked Solution
Create a strategy

The first thing we notice is that both sides seem to have different values. But if we distribute the coefficient on the right side, the first term would be 2x, and the second term would be 5. The equation now has the same variables with the same coefficients on both sides, but the constant values are different.

Apply the idea

No solution

Reflect and check

We can verify our answer by solving the equation.

\displaystyle 2x-5\displaystyle =\displaystyle 0.5\left(4x+10\right)Original equation
\displaystyle 2x-5\displaystyle =\displaystyle 2x+5Distributive property
\displaystyle -5\displaystyle =\displaystyle 5Subtraction property of equality

The resulting statement is not true. This is how we know there is no solution to this equation.

Example 4

Solve the following equations:

a

4\left(5x+1\right)=-3\left(5x-5\right)

Worked Solution
Create a strategy

Looking at the equation, we see that the variables have different coefficients on both sides of the equation. The coefficient on the left side is 4\cdot 5=20, and the coefficient on the right side is -3\cdot 5=-15. This means it will have a unique solution.

Apply the idea
\displaystyle 4\left(5x+1\right)\displaystyle =\displaystyle -3\left(5x-5\right)Given equation
\displaystyle 20x+4\displaystyle =\displaystyle -15x+15Distributive property
\displaystyle 35x+4\displaystyle =\displaystyle 15Addition property of equality
\displaystyle 35x\displaystyle =\displaystyle 11Subtraction property of equality
\displaystyle x\displaystyle =\displaystyle \dfrac{11}{35}Division property of equality
Reflect and check

We can verify the solution by substituting it back into the equation to see if it makes the equation true.

\displaystyle 4\left(5x+1\right)\displaystyle =\displaystyle -3\left(5x-5\right)Original equation
\displaystyle 4\left[5\left(\dfrac{11}{35}\right)+1\right]\displaystyle =\displaystyle -3\left[5\left(\dfrac{11}{35}\right)-5\right]Substitute x=\dfrac{11}{35}
\displaystyle 4\left(\dfrac{11}{7}+1\right)\displaystyle =\displaystyle -3\left(\dfrac{11}{7}-5\right)Multiply 5\cdot\dfrac{11}{35}
\displaystyle 4\left(\dfrac{11}{7}+\dfrac{7}{7}\right)\displaystyle =\displaystyle -3\left(\dfrac{11}{7}-\dfrac{35}{7}\right)Create common denominators
\displaystyle 4\left(\dfrac{18}{7}\right)\displaystyle =\displaystyle -3\left(-\dfrac{24}{7}\right)Evaluate the addition
\displaystyle \dfrac{72}{7}\displaystyle =\displaystyle \dfrac{72}{7}Evaluate the multiplication

What results is a true equation, so we know x=\dfrac{11}{35} is the correct solution.

b

\dfrac{\left(x+3\right)}{2}+1.3x=\dfrac{0.8x}{4}

Worked Solution
Create a strategy

This equation has a mix of rational numbers, and the variables are in multiple terms. If we multiply both sides of the equation by the lowest common denominator, that will remove fractions and only leave us with decimals. Then, we can work on collecting all the variables.

Apply the idea

The denominators are 2 and 4, so the lowest common denominator is 4.

\displaystyle \dfrac{\left(x+3\right)}{2}+1.3x\displaystyle =\displaystyle \dfrac{0.8x}{4}Original equation
\displaystyle 4\left[\dfrac{\left(x+3\right)}{2}+1.3x\right]\displaystyle =\displaystyle 4\left(\dfrac{0.8x}{4}\right)Multiplication property of equality
\displaystyle 2\left(x+3\right)+5.2x\displaystyle =\displaystyle 0.8xDistributive property
\displaystyle 2x+6+5.2x\displaystyle =\displaystyle 0.8xDistributive property
\displaystyle 7.2x+6\displaystyle =\displaystyle 0.8xCombine like terms
\displaystyle 6\displaystyle =\displaystyle -6.4xSubtraction property of equality
\displaystyle -0.9375\displaystyle =\displaystyle xDivision property of equality
Reflect and check

There are many other ways we could have started solving this equation, but eliminating the fractions first made it much easier to solve.

Example 5

Right now, Bianca's father is 48 years older than Bianca.

Two years ago, her father was 5 times older than her.

Solve for y, Bianca's current age.

Worked Solution
Create a strategy

We want to write expressions representing Bianca's age and her father's age. Then we relate them with an equation and solve for y.

Bianca's father is currently y+48 years old.

Two years ago, Bianca was y-2 years old. Her father was y+48-2=y+46 years old.

Her father's age was five times her age at this time, which produces our equation:

y+46=5\left(y-2\right)

Apply the idea
\displaystyle y+46\displaystyle =\displaystyle 5\left(y-2\right)Original equation
\displaystyle y+46\displaystyle =\displaystyle 5y-10Distributive property
\displaystyle y+56\displaystyle =\displaystyle 5yAddition property of equality
\displaystyle 56\displaystyle =\displaystyle 4ySubtraction property of equality
\displaystyle 14\displaystyle =\displaystyle yDivision property of equality

Bianca is currently 14 years old.

Reflect and check

Let's check by referring back to the information given in the problem. Bianca's father is 48 years older, so he is 14+48=62 years old. Two years ago, he was 5 times older. Two years ago, Bianca would have been 12, and her father would have been 60. Is this 5 times Bianca's age? Yes, this is correct.

Idea summary

A solution to an equation is any value that can replace the variable and make a true statement.

A fully simplified equation in one variable will take one of the following three forms, corresponding to how many solutions the equation has:

  • x=a, where a is a number (a unique solution)

  • a=a, where a is a number (infinitely many solutions)

  • a=b, where a and b are different numbers (no solutions)

We can simplify an equation using the distributive property or the multiplication property of equality to eliminate fractions or decimals. After simplifying, we can continue using properties of equality to solve for the unknown.

Outcomes

A.EI.1

The student will represent, solve, explain, and interpret the solution to multistep linear equations and inequalities in one variable and literal equations for a specified variable.

A.EI.1a

Write a linear equation or inequality in one variable to represent a contextual situation.

A.EI.1b

Solve multistep linear equations in one variable including those in contextual situations, by applying the properties of real numbers and/or properties of equality.

A.EI.1e

Determine if a linear equation in one variable has one solution, no solution, or an infinite number of solutions.

A.EI.1f

Verify possible solution(s) to multistep linear equations and inequalities in one variable algebraically, graphically, and with technology to justify the reasonableness of the answer(s). Explain the solution method and interpret solutions for problems given in context.

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