Lesson 15: Solving Equations with Rational Numbers

Solving Equations with Rational Numbers

Student Summary

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

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.

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

We can use the same approach for this equation:

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

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

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.

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

We can use the same approach for this equation:

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

Visual / Anchor Chart

Standards

Building On

7.NS.A

6.EE.5

6.EE.B.5

Addressing

7.NS.3

7.NS.A.3

7.EE.4

7.EE.B.4

7.NS.3

7.NS.A.3

7.EE.3

7.EE.B.3

7.NS.A

Building Toward

7.EE.B

7.EE.4.a

7.EE.B.4.a