Chapter 7.2, Problem 18E

To determine

To Find: To find the solutions of the given equations.

Answer to Problem 18E

The solution is $r=\frac{5}{6}\text{and}s=\frac{5}{6}$

Explanation of Solution

Given information:Given equations are $8r+16s=20\text{and}16r+50s=55$

Formula used: Elimination method

Calculation:

Since the given equations are

  $\begin{array}{l}8r+16s=20\mathrm{...}(1)\\ 16r+50s=55\mathrm{...}(2)\end{array}$

Multiplying (1) by 2 and (2) by 1, then

  $\begin{array}{l}16r+32s=40\\ 16r+50s=55\end{array}$

Subtracting these equations

  $\begin{array}{l}16r+32s-(16r+50s)=40-55\\ 16r+32s-16r-50s=-15\\ -18s=-15\\ s=\frac{-15}{-18}\\ s=\frac{5}{6}\end{array}$

Also, multiplying (1) by 25 and (2) by 8

  $\begin{array}{l}200r+400s=500\\ 128r+400s=440\end{array}$

Subtracting these equations

  $\begin{array}{l}200r+400s-(128r+400s)=500-440\\ 200r+400s-128r-400s=60\\ 72r=60\\ r=\frac{60}{72}\\ r=\frac{5}{6}\end{array}$

Thus, the required solution is $r=\frac{5}{6}\text{and}s=\frac{5}{6}$

Verification

  $\begin{array}{l}8r+16s=20\\ \text{Consider}\text{left}\text{hand}\text{side}\\ 8r+16s\\ =8\times \frac{5}{6}+16\times \frac{5}{6}\\ =\frac{20}{3}+\frac{40}{3}\\ =\frac{60}{3}\\ =20\end{array}$

Which is equal to right hand side.

Hence verified.