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Lesson 15Solving Equations with Rational Numbers

Let’s solve equations that include negative values.

Learning Targets:

  • I can solve equations that include rational numbers and have rational solutions.

15.1 Number Talk: Opposites and Reciprocals

The variables a through h all represent different numbers. Mentally find numbers that make each equation true.

\frac35 \boldcdot \frac53 = a

7 \boldcdot b = 1

c \boldcdot d = 1

\text-6 + 6 = e

11 + f = 0

g + h = 0

15.2 Match Solutions

Match each equation to a value that makes it true by dragging the answer to the corresponding equation. Be prepared to explain your reasoning.

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15.3 Trip to the Mountains

The Hiking Club is on a trip to hike up a mountain.

  1. The members increased their elevation 290 feet during their hike this morning. Now they are at an elevation of 450 feet.

    1. Explain how to find their elevation before the hike.
    2. Han says the equation e + 290 = 450 describes the situation. What does the variable e represent?
    3. Han says that he can rewrite his equation as e=450 + (\text-290) to solve for e . Compare Han's strategy to your strategy for finding the beginning elevation.

  2. The temperature fell 4 degrees in the last hour. Now it is 21 degrees. Write and solve an equation to find the temperature it was 1 hour ago.

  3. There are 3 times as many students participating in the hiking trip this year than last year. There are 42 students on the trip this year.

    1. Explain how to find the number of students that came on the hiking trip last year.
    2. Mai says the equation 3s=42 describes the situation. What does the variable s represent?
    3. Mai says that she can rewrite her equation as s=\frac13 \boldcdot 42 to solve for s . Compare Mai's strategy to your strategy for finding the number of students on last year’s trip.

  4. The cost of the hiking trip this year is \frac23 of the cost of last year's trip. This year's trip cost $32. Write and solve an equation to find the cost of last year's trip.

Are you ready for more?

A number line is shown below. The numbers 0 and 1 are marked on the line, as are two other rational numbers a and b .

A number line with 4 tick marks. The number 0 is indicated. Rational number a is to the right of 0 and 1 is slightly more than halfway between 0 and a. Rational number b is to the far left of 0, where b is a greater distance from 0 than a is from 0.

Decide which of the following numbers are positive and which are negative.

a-1

a-2

\text-b

a+b

a-b

ab+1

15.4 Card Sort: Matching Inverses

Your teacher will give you a set of cards with numbers on them.

  1. Match numbers with their additive inverses.
  2. Next, match numbers with their multiplicative inverses.
  3. What do you notice about the numbers and their inverses?

Lesson 15 Summary

To solve the equation x + 8 = \text-5 , we can add the opposite of 8, or -8, to each side:

\begin{align} x + 8 &= \text-5\\ (x+ 8) + \text-8&=(\text-5)+ \text-8\\ x&=\text-13 \end{align}

Because adding the opposite of a number is the same as subtracting that number, we can also think of it as subtracting 8 from each side.

We can use the same approach for this equation:

\begin{align} \text-12 & = t +\text- \frac29\\ (\text-12)+ \frac29&=\left( t+\text-\frac29\right) + \frac29\\\text-11\frac79& = t\end{align}

To solve the equation 8x = \text-5 , we can multiply each side by the reciprocal of 8, or \frac18 :

\begin{align} 8x & = \text-5\\ \frac18 ( 8x )&= \frac18 (\text-5)\\ x&=\text-\frac58 \end{align}

Because multiplying by the reciprocal of a number is the same as dividing by that number, we can also think of it as dividing by 8. We can use the same approach for this equation:

\begin{align} \text-12& =\text-\frac29 t\\ \text-\frac92\left( \text-12\right)&= \text-\frac92 \left(\text-\frac29t\right) \\ 54& = t\end{align}

Lesson 15 Practice Problems

  1. Solve.

    1. \frac25t=6

    2. \text-4.5 = a-8

    3. \frac12+p= \text-3

    4. 12=x \boldcdot 3

    5. \text-12 = \text-3y

  2. Evaluate each expression if x is \frac{2}{5} , y is \text-4 , and z is -0.2.

    1. x+y

    2. 2x-z

    3. x+y+z

    4. y \boldcdot x

  3. Match each equation to a step that will help solve the equation.

    1. 5x=0.4
    2. \frac{x}{5}=8
    3. 3=\frac {\text{-}x}{5}
    4. 7=\text-5x
    1. Multiply each side by 5.
    2. Multiply each side by -5.
    3. Multiply each side by \frac15 .
    4. Multiply each side by \frac {\text{-}1}{5} .
    1. Write an equation where a number is added to a variable, and a solution is -8.
    2. Write an equation where a number is multiplied by a variable, and a solution is \frac {\text{-}4}{5} .

  5. The markings on the number line are evenly spaced. Label the other markings on the number line.

    A blank number line with 9 evenly spaced tick marks. Starting on the left, the first tick mark is labeled negative 2 point 5, the fourth tick mark is labeled negative 1, and the sixth tick mark is labeled 0.

  6. In 2012, James Cameron descended to the bottom of Challenger Deep in the Marianas Trench; the deepest point in the ocean. The vessel he rode in was called DeepSea Challenger.

    Challenger Deep is 35,814 feet deep at its lowest point

    1. DeepSea Challenger’s descent was a change in depth of (\text-4) feet per second. We can use the equation y=\text-4x to model this relationship, where y is the depth and x is the time in seconds that have passed. How many seconds does this model suggest it would take for DeepSea Challenger to reach the bottom?
    2. To end the mission DeepSea Challenger made a one-hour ascent to the surface. How many seconds is this?
    3. The ascent can be modeled by a different proportional relationship y=kx . What is the value of k in this case?