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# Multi-Step Equations

## Maintain balance of an equation throughout all steps needed to solve.

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Practice Multi-Step Equations
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Solving Multi-Step Equations

Let's look at cell phone companies again. Cell-U-Lar includes 500 texts in their $79/month fee and then charges$0.10 for each additional text. We-B-Mobile also includes 500 texts in their $59/month fee and then charges$0.25 for each additional text. At how many texts, over 500, will the two bills be the same?

### Guidance

The types of equations in this concept involve at least three steps. Keep in mind that the last two steps when solving a linear equation will always be the same: add or subtract the number that is on the same side of the equals sign as the variable, then multiply or divide by the number with the variable.

#### Example A

Solve $3(x-5)+4=10$ .

Solution: When solving more complicated equations, start with one side and simplify as much as you can. The left side of this equation looks more complicated, so let’s simplify it by using the Distributive Property and combining like terms.

$3(x-5)+4 &= 10\\3x-15+4 &= 10\\3x-11 &= 10$

Now, this looks like an equation from the previous concept. Continue to solve.

$& \ 3x-\bcancel{11}=10\\& \underline{\quad \ \ + \bcancel{11} \ + 11 \; \;}\\& \qquad \frac{\bcancel{3}x}{\bcancel{3}}=\frac{21}{3}\\& \qquad \ \ x=7$

Check your answer: $3(7-5)+4=3 \cdot 2+4=6+4=10$

#### Example B

Solve $8x-17=4x+23$ .

Solution: This equation has $x$ on both sides of the equals sign. Therefore, we need to move one of the $x$ terms to the other side of the equation. It does not matter which $x$ term you move. We will move the $4x$ to the other side so that, when combined, the $x$ term is positive.

$& 8x-17=\bcancel{4x}+23\\& \underline{-4x \quad \ \ -\bcancel{4x} \; \; \; \; \; \; \; \;}\\& 4x-\bcancel{17}=23\\& \underline{\quad + \bcancel{17} \ +17 \; \; \; \; \; \; \; \; \;}\\& \qquad \frac{\bcancel{4}x}{\bcancel{4}}=\frac{40}{4}\\& \qquad \ \ x=10$

$8(10)-17 &= 4(10)+23\\80-17 &= 40+23 \\63 &= 63$

#### Example C

Solve $2(3x-1)+2x=5-(2x-3)$ .

Solution: This equation combines the two previous examples. First, use the Distributive Property.

$2(3x-1)+2x &= 5-(2x-3)\\6x-2+2x &= 5-2x+3$

Don’t forget to distribute the negative sign in front of the second set of parenthesis. Treat it like distributing a –1. Now, combine like terms and solve the equation.

$& \ \ \ 8x-\bcancel{2}=8-\bcancel{2x}\\& \underline{+2x+\bcancel{2} +2+\bcancel{2x}}\\& \quad \ \frac{\bcancel{10}x}{\bcancel{10}}=\frac{10}{10}\\& \qquad \ x=1$

$2(3(1)-1)+2(1) &= 5-(2(1)-3)\\2 \cdot 2+2 &= 5-(-1) \\4+2 &= 6$

Intro Problem Revisit Let's write an expression for the total cost associated with each cell phone company.

Cell-U-Lar: $79+0.1t$

We-B-Mobile: $59+0.25t$

Where t is the number of texts over 500. Set the two expressions equal to each other to see when the plans are equal.

$79+0.1t&=59+0.25t\\20&=0.15t\\133. \bar 3 &=t$

At 133 additional texts, the bills would be just about the same, but don't forget that each company includes 500 texts at no additional cost. So, you would have to go just over 633 texts to make the two plans equal. If you text less than that, the plan from We-B-Mobile is a better deal. More than 633, then Cell-U-Lar makes more sense.

### Guided Practice

1. $\frac{3}{4}+\frac{2}{3}x=2x+\frac{5}{6}$

2. $0.6(2x-7)=5x-5.1$

1. Use the LCD method introduced in the problem set from the previous concept. Multiply every term by the LCD of 4, 3, and 6.

$& 12 \left(\frac{3}{4}+\frac{2}{3}x = 2x+\frac{5}{6}\right)\\& \qquad 9+8x = 24x+10$

Now, combine like terms. Follow the steps from Example B.

$& \quad 9+\bcancel{8x}=24x+\bcancel{10}\\& \underline{-10-\bcancel{8x} \ \ -8x-\bcancel{10} \; \;}\\& \qquad \frac{-1}{16}=\frac{\bcancel{16}x}{\bcancel{16}}\\& \quad -\frac{1}{16}=x$

$\frac{3}{4}+\frac{2}{3} \left(-\frac{1}{16}\right) &= 2\left(-\frac{1}{16}\right)+\frac{5}{6}\\\frac{3}{4}-\frac{1}{24} &= -\frac{1}{8}+\frac{5}{6}\\\frac{18}{24}-\frac{1}{24} &= -\frac{3}{24}+\frac{20}{24} \\\frac{17}{24}&=\frac{17}{24}$

2. Even though there are decimals in this problem, we can approach it like any other problem. Use the Distributive Property and combine like terms.

$& \ 0.6(2x+7)=4.3x-5.1\\& \ \ 1.2x+\cancel{4.2}=\cancel{4.3x}-5.1\\& \underline{-4.3x-\cancel{4.2} \ -\cancel{4.3x}-4.2 \; \;}\\& \quad \quad \ \frac{-\cancel{3.1}x}{-\cancel{3.1}}=\frac{-9.3}{-3.1}\\& \qquad \qquad \ \ x=3$

$0.6(2(3)+7) &= 4.3(3)-5.1\\0.6 \cdot 13 &= 12.9-5.1 \\7.8 &= 7.8$

### Practice

Solve each equation and check your solution.

1. $-6(2x-5)=18$
2. $2-(3x+7)=-x+19$
3. $3(x-4)=5(x+6)$
4. $x-\frac{4}{5}=\frac{2}{3}x+\frac{8}{15}$
5. $8-9x-5=x+23$
6. $x-12+3x=-2(x+18)$
7. $\frac{5}{2}x+\frac{1}{4}=\frac{3}{4}x+2$
8. $5 \left(\frac{x}{3}+2\right)=\frac{32}{3}$
9. $7(x-20)=x+4$
10. $\frac{2}{7}\left(x+\frac{2}{3}\right)=\frac{1}{5}\left(2x-\frac{2}{3}\right)$
11. $\frac{1}{6}(x+2)=2\left(\frac{3x}{2}-\frac{5}{4}\right)$
12. $-3(-2x+7)-5x=8(x-3)+17$

Challenge Solve the equations below. Check your solution.

1. $\frac{3x}{16}+\frac{x}{8}=\frac{3x+1}{4}+\frac{3}{2}$
2. $\frac{x-1}{11}=\frac{2}{5} \cdot \frac{x+1}{3}$
3. $\frac{3}{x}=\frac{2}{x+1}$