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# Solving Real-World Problems with Two-Step Equations

## Equations that require multiple steps solve real world problems. This involves translating words into algebraic equations and solving them.

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Practice Solving Real-World Problems with Two-Step Equations
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Solving Real-World Problems with Two-Step Equations

### Solving Real-World Problems with Two-Step Equations

The hardest part of solving word problems is translating from words to an equation. First, you need to identify the unknown quantity in the word problem. What is the value for which you have to solve? That will be your variable. What operations are described in the problem, and how do they affect the unknown value? Your equation should use mathematics to describe two equal quantities, with the unknown value (in the form of a variable letter) on at least one side of the equation. Once you have restated the information in the word problem as a mathematical statement, the actual computation is often the easy part.

#### Real-World Application: Repair Job

An emergency plumber charges $65 as a call-out fee plus an additional$75 per hour. He arrives at a house at 9:30 and works to repair a water tank. If the total repair bill is 196.25, at what time was the repair completed? In order to solve this problem, we collect the information from the text and convert it to an equation. Unknown: time taken in hours – this will be x\begin{align*}x\end{align*} The bill is made up of two parts: a call out fee and a per-hour fee. The call out is a flat fee, and independent of x\begin{align*}x\end{align*}, it’s the same no matter how many hours the plumber works. The per-hour part depends on the number of hours, x.\begin{align*}x.\end{align*} So the total fee is65 (regardless of the number of hours) plus 75x\begin{align*}\75x\end{align*} (where x\begin{align*}x\end{align*} is the number of hours), or 65+75x\begin{align*}65 + 75x\end{align*}. Looking at the problem again, we also can see that the total bill is196.25. So our statement of equal quantities (our equation) is: 196.25=65+75x\begin{align*}196.25 = 65 + 75x\end{align*}.

Solving for x\begin{align*}x\end{align*}:

196.25131.251.75=65+75xSubtract 65 from both sides.=75xDivide both sides by 75.=xThe job took 1.75 hours.\begin{align*}196.25 &= 65 + 75x \qquad \text{Subtract 65 from both sides.}\\ 131.25 &= 75x \qquad \qquad \text{Divide both sides by 75.}\\ 1.75 &= x \qquad \qquad \quad \! \text{The job took 1.75 hours.}\end{align*}

The repair job was completed 1.75 hours after 9:30, so it was completed at 11:15AM.

#### Real-World Application: Predicting Height

When Asia was young, her Daddy marked her height on the door frame every month. Asia’s Daddy noticed that between the ages of one and three, he could predict her height (in inches) by taking her age in months, adding 75 inches, and multiplying the result by one-third. Use this information to answer the following:

a) Write an equation linking her predicted height, h\begin{align*}h\end{align*}, with her age in months, m\begin{align*}m\end{align*}

To convert the text to an equation, first determine the type of equation we have. We are going to have an equation that links two variables. Our unknown will change, depending on the information we are given. For example, we could solve for height given age, or solve for age given height. However, the text gives us a way to determine height, so our equation will start with “h=\begin{align*}h=\end{align*}”.

The text tells us that we can predict her height by taking her age in months, adding 75, and multiplying by 13\begin{align*}\frac{1}{3} \end{align*}. So our equation is h=(m+75)13\begin{align*}h = (m + 75) \cdot \frac{1}{3} \end{align*}, or h=13(m+75)\begin{align*}h = \frac{1}{3}(m + 75)\end{align*}.

b) Determine her predicted height on her second birthday

To predict Asia’s height on her second birthday, we substitute m=24\begin{align*}m=24\end{align*} into our equation (because 2 years is 24 months) and solve for h\begin{align*}h\end{align*}.

hhh=13(24+75)=13(99)=33\begin{align*} h &= \frac{1}{3}(24 + 75)\\ h &= \frac{1}{3}(99)\\ h &= 33\end{align*}

Asia’s height on her second birthday was predicted to be 33 inches.

c) Determine at what age she is predicted to reach three feet tall.

To determine the predicted age when she reached three feet, substitute h=36\begin{align*}h = 36\end{align*} into the equation and solve for m\begin{align*}m\end{align*}.

3610833=13(m+75)=m+75=m\begin{align*}36 &= \frac{1}{3}(m + 75)\\ 108 &= m + 75\\ 33 &= m\end{align*}

Asia was predicted to be 33 months old when her height was three feet.

#### Real-World Application: Converting Temperatures

To convert temperatures in Fahrenheit to temperatures in Celsius, take the temperature in degrees Fahrenheit, subtract 32, then divide the result by 1.8. This gives the temperature in degrees Celsius.

a) Write an equation that shows the conversion process.

The text gives the process to convert Fahrenheit to Celsius. We can write an equation using two variables. We will use f\begin{align*}f\end{align*} for temperature in Fahrenheit, and c\begin{align*}c\end{align*} for temperature in Celsius.

First we take the temperature in Fahrenheit and subtract 32.Then divide by 1.8.This equals the temperature in Celsius.f32f321.8c=f321.8\begin{align*}&\text{First we take the temperature in Fahrenheit and subtract 32.} && f - 32\\ &\text{Then divide by 1.8.} && \frac{f - 32}{1.8}\\ &\text{This equals the temperature in Celsius.} && c = \frac{f - 32}{1.8}\end{align*}

In order to convert from one temperature scale to another, simply substitute in for whichever temperature you know, and solve for the one you don’t know.

b) Convert 50 degrees Fahrenheit to degrees Celsius.

To convert 50 degrees Fahrenheit to degrees Celsius, substitute f=50\begin{align*}f = 50\end{align*} into the equation.

ccc=50321.8=181.8=10\begin{align*}c &= \frac{50 - 32}{1.8}\\ c &= \frac{18}{1.8}\\ c &= 10\end{align*}

50 degrees Fahrenheit is equal to 10 degrees Celsius.

### Examples

To convert temperatures in Fahrenheit to temperatures in Celsius, take the temperature in degrees Fahrenheitsubtract 32, then divide the result by 1.8.

#### Example 1

Convert 25 degrees Celsius to degrees Fahrenheit.

To convert 25 degrees Celsius to degrees Fahrenheit, substitute c=25\begin{align*}c = 25\end{align*} into the equation:

254577=f321.8=f32=f\begin{align*} 25 &= \frac{f - 32}{1.8}\\ 45 &= f - 32\\ 77 &= f \end{align*}

25 degrees Celsius is equal to 77 degrees Fahrenheit.

#### Example 2

Convert -40 degrees Celsius to degrees Fahrenheit.

To convert -40 degrees Celsius to degrees Fahrenheit, substitute c=40\begin{align*}c = -40\end{align*} into the equation.

407240=f321.8=f32=f\begin{align*}-40 &= \frac{f - 32}{1.8}\\ -72 &= f - 32\\ -40 &= f\end{align*}

-40 degrees Celsius is equal to -40 degrees Fahrenheit. (No, that’s not a mistake! This is the one temperature where they are equal.)

### Review

1. Jade is stranded downtown with only $10 to get home. Taxis cost$0.75 per mile, but there is an additional $2.35 hire charge. Write a formula and use it to calculate how many miles she can travel with her money. 2. Jasmin’s dad, Andrew, is planning a surprise birthday party for her. He will hire a bouncy castle, and will provide party food for all the guests. The bouncy castle costs$150 for the afternoon, and the food will cost $3 per person. Andrew has a budget of$300. Write an equation and use it to determine the maximum number of guests he can invite.

The local amusement park sells summer memberships for $50 each. Normal admission to the park costs$25; admission for members costs $15. 1. If Darren wants to spend no more than$100 on trips to the amusement park this summer, how many visits can he make if he buys a membership with part of that money?
2. How many visits can he make if he does not?
3. If he increases his budget to \$160, how many visits can he make as a member?
4. And how many as a non-member?

For an upcoming school field trip, there must be one adult supervisor for every five children.

1. Write an expression for the number of children and adults who will go on the field trip.
2. If the bus seats 40 people, how many children can go on the trip?
3. How many children can go if a second 40-person bus is added?
4. Four of the adult chaperones decide to arrive separately by car. Now how many children can go in the two buses?

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