# Two-Step Equations and Properties of Equality

## Maintain balance of an equation while solving using addition, subtraction, multiplication, or division.

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Solving Two-Step Equations in Algebra

The Cell-U-Lar cell phone company has a data plan where 2.5 GB are included for $79.00 per month. For every 10 MB (there are 1024 MB in 1 GB) over that amount, the company charges an additional$3.00. Last month, Taylor's cell phone bill was 115. How many megabytes of data was he over the limit? ### Two-Step Equations Now, we will compound steps to solve slightly more complicated equations. We will continue to undo the operations that are in an equation. Remember that order matters. To undo the operations, work in the opposite order of the Order of Operations. Let's solve the following equations. 1. Solve \begin{align*}4x-9=11\end{align*}. Notice that we need to undo two operations. Therefore, we need to get everything on the other side of the equals sign to solve for \begin{align*}x\end{align*}. First, add 9 to both sides. \begin{align*}& 4x - \bcancel{9} = 11\\ & \underline{\quad \ + \bcancel{9} \ \ +9}\\ & \quad \ \ 4x=20\end{align*} Now, undo the multiplication. Divide both sides by 4 to solve for \begin{align*}x\end{align*}. \begin{align*}\frac{\cancel{4}x}{\cancel{4}} &= \frac{20}{4}\\ x &= 5 \end{align*} Check your answer: \begin{align*}4(5) - 9 = 20 - 9 = 11\end{align*} 1. Solve \begin{align*}-\frac{1}{2}g+5=14\end{align*}. First, subtract 5 from both sides. \begin{align*}& -\frac{1}{2}g+\bcancel{5}=14\\ & \underline{\qquad \ \ -\bcancel{5} \ \ -5 \; \;}\\ & \quad \ -\frac{1}{2}g=9\end{align*} Multiplying by one-half is the same as dividing by 2. To undo this fraction, multiply both sides by –2. \begin{align*} -\frac{1}{2}g &= 9\\ -\bcancel{2} \cdot -\frac{1}{\bcancel{2}}g &= 9 \cdot -2\\ g &= -18\end{align*} Check your answer: \begin{align*}-\frac{1}{2}(-18)+5=9+5=14\end{align*} 1. Solve \begin{align*}\frac{3}{4}x-\frac{2}{3}=\frac{7}{3}\end{align*}. Even though this problem involves fractions, it is still solved the same way. First, add \begin{align*}\frac{2}{3}\end{align*} to both sides and then multiply both sides by the reciprocal of \begin{align*}\frac{3}{4}\end{align*}. \begin{align*}& \frac{3}{4}x-\bcancel{\frac{2}{3}}=\frac{7}{3}\\ & \underline{\quad \ + \bcancel{\frac{2}{3}} \ +\frac{2}{3} \; \; \; \; \; \; \;}\\ & \quad \ \ \frac{3}{4}x=\frac{9}{3}=3\\ & \ \ \xcancel{\frac{4}{3} \cdot \frac{3}{4}} x=\bcancel{3} \cdot \frac{4}{\bcancel{3}}\\ & \qquad \ \ x=4\end{align*} Check your answer: \begin{align*}\frac{3}{4}(4)-\frac{2}{3}=3-\frac{2}{3}=\frac{9}{3}-\frac{2}{3}=\frac{7}{3}\end{align*}. To summarize, for two-step equations you must do the following: Step 1: Undo any addition or subtraction. Step 2: Undo any multiplication or division. ### Examples #### Example 1 Earlier, you were asked to find how many megabytes of data Taylor is over the limit. Writing an equation for the phone bill, we have: \begin{align*}C=79+3m\end{align*} where C is the cost, and m is the number of 10 megabytes over the 2.5 GB. So, when we get our final answer, we need to multiply it by 10 to get the total number of megabytes over the included 2.5 GB amount. Set C =115 and solve for m.

\begin{align*} \bcancel{79} + 3m &= 115\\ \bcancel{-79} + 3m &= -79 \\ 3m &=36\\ \frac{\cancel{3}m}{\cancel{3}} &= \frac{36}{3}\\ m &= 12\\ 12 \cdot 10 = 120 \end{align*}

Taylor is 120 MB over his data limit for the month.

#### Example 2

Solve \begin{align*}\frac{x}{-6}-7=-11\end{align*}.

Follow Steps 1 and 2 from above. First, add 7 to both sides and then multiply both sides by –6.

\begin{align*}& \ \frac{x}{-6} -\bcancel{7} = -11\\ & \underline{\qquad +\bcancel{7} \quad \ +7 \; \; \; \; \; \; \; \;}\\ & \qquad \frac{x}{-6}=-4\\ & -\bcancel{6} \cdot \frac{x}{-\bcancel{6}}=-4 \cdot -6\\ & \qquad \quad x=24\end{align*}

Check your answer: \begin{align*}\frac{24}{-6}-7=-4-7=-11 \end{align*}

#### Example 3

Solve \begin{align*}18=6-\frac{2}{5}y\end{align*}

Again, follow Steps 1 and 2. Here, we will subtract 6 and then multiply by \begin{align*}-\frac{5}{2}\end{align*}.

\begin{align*}& \qquad \quad 18=\bcancel{6}-\frac{2}{5}y\\ & \qquad \ \ \underline{-6 \ -\bcancel{6} \; \; \; \; \; \; \; \; \; \; \;}\\ & \qquad \ \ 12=-\frac{2}{5}y\\ & -\frac{5}{2} \cdot 12=-\frac{\bcancel{2}}{\cancel{5}}y \cdot - \frac{\cancel{5}}{\bcancel{2}}\\ & \quad \ -30=y\end{align*}

Check your answer: \begin{align*}6-\frac{2}{5}(-30)=6+12=18 \end{align*}

### Review

1. \begin{align*}2x-5=-17\end{align*}
2. \begin{align*}-4x+3=-5\end{align*}
3. \begin{align*}-1=\frac{x}{2}-6\end{align*}
4. \begin{align*}\frac{1}{3}x+11=-2\end{align*}
5. \begin{align*}\frac{3}{4}=\frac{1}{2}-\frac{1}{8}x\end{align*}
6. \begin{align*}-18=\frac{x}{-5}-3\end{align*}
7. \begin{align*}\frac{5}{6}x+4=29\end{align*}
8. \begin{align*}-11x+4=125\end{align*}
9. \begin{align*}6-x=-22\end{align*}
10. \begin{align*}\frac{2}{7}x+8=20\end{align*}
11. \begin{align*}\frac{4}{5}=-\frac{2}{5}+\frac{3}{2}x\end{align*}
12. \begin{align*}15-\frac{x}{-9}=21\end{align*}
13. \begin{align*}1.4x-5.6=2.8\end{align*}
14. \begin{align*}14.4=-2.7x - 1.8\end{align*}
15. \begin{align*}-\frac{5}{4}=\frac{1}{6}x+\frac{3}{12}\end{align*}

When dealing with fractions, one way to “get rid” of them is to multiply everything in the equation by the Least Common Denominator (LCD) for the entire equation. Try this technique with the following equations. We will get you started in #16.

1. \begin{align*}\frac{3}{8}x-\frac{2}{5}=\frac{7}{4}\end{align*} The LCD of 8, 5, and 4 is 40. \begin{align*}40 \left(\frac{3}{8}x-\frac{2}{5}=\frac{7}{4}\right)\end{align*}. Distribute the 40 and solve.
2. \begin{align*}\frac{10}{3}x+\frac{3}{4}=\frac{5}{2}\end{align*}
3. \begin{align*}\frac{9}{10}x-\frac{1}{2}=\frac{2}{3}\end{align*}
4. Solve #15 using the LCD method.
5. Which method is easier for you to solve equations with fractions?

To see the Review answers, open this PDF file and look for section 1.8.

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