Chapter 9.5, Problem 31WE

To determine

To find: The value of $a\text{ and }b$ by using multiplication with the addition-or-subtraction method.

Answer to Problem 31WE

The value of $a$ is $4$ , and the value of $b$ is $3$ .

Explanation of Solution

Given information:

The given equations are,

  $\begin{array}{c}\frac{a}{4}+\frac{b}{3}=2\\ \frac{a}{2}-b=-1\end{array}$

Calculation:

Calculate the equations.

  $\begin{array}{c}\frac{a}{4}+\frac{b}{3}=2\\ 12\left(\frac{a}{4}+\frac{b}{3}\right)=12\left(2\right)\text{{LCD is 12}\\ 3a+4b=24\end{array}$ … (1)

  $\begin{array}{c}\frac{a}{2}-b=-1\\ 2\left(\frac{a}{2}-b\right)=2\left(-1\right)\text{{LCD is 2}\\ a-2b=-2\end{array}$ … (2)

Multiply by $2$ in equation $\left(2\right)$ .

  $\begin{array}{c}2\left(a-2b\right)=2\left(-2\right)\\ 2a-4b=-4\end{array}$ … (3)

Add equation $\left(3\right)$ and equation $\left(1\right)$ .

  $\begin{array}{c}3a+4b=24\\ 2a-4b=-4\end{array}$

  $\text{5a =20}$

Solve the result,

  $a=4$

Substitute the value of $a$ in $\frac{a}{2}-b=-1$ .

  $\begin{array}{c}\frac{4}{2}-b=-1\\ -b=-1-2\\ b=3\end{array}$

Therefore,

  $\begin{array}{l}a=4\\ b=3\end{array}$

Therefore, the value of $a$ is $4$ , and the value of $b$ is $3$ .