Chapter 6.3, Problem 29PPS

To determine

To solve the system of equations using elimination method

Answer to Problem 29PPS

  $x=\frac{500}{11}$ and $y=-\frac{1250}{33}$

Explanation of Solution

Given:

  $\frac{3}{5}x+\frac{1}{2}y=8\frac{1}{3}$

  $-\frac{3}{5}x+\frac{3}{4}y=8\frac{1}{3}$

Calculation:

To simplify the equations, first find the Least Common Multiplier.

  $\frac{3}{5}x+\frac{1}{2}y=8\frac{1}{3}$

Least Common Multiplier of (5, 2) is 10.

  $\begin{array}{l}\Rightarrow \frac{6}{10}x+\frac{5}{10}y=\frac{25}{3}\\ \Rightarrow \frac{6x}{10}+\frac{5y}{10}=\frac{25}{3}\\ \Rightarrow \frac{6x+5y}{10}=\frac{25}{3}\\ \Rightarrow 3(6x+5y)=10(25)\\ \Rightarrow 18x+15y=250\end{array}$

Let, equation $18x+15y=250\to \text{(1)}$

Now,

  $-\frac{3}{5}x+\frac{3}{4}y=8\frac{1}{3}$

Least Common Multiple of (5, 4) is 20.

  $\begin{array}{l}\Rightarrow -\frac{3}{20}x-\frac{8}{20}y=\frac{25}{3}\\ \Rightarrow -\frac{3x}{20}-\frac{8y}{20}=\frac{25}{3}\\ \Rightarrow \frac{-3x-8y}{20}=\frac{25}{3}\\ \Rightarrow 3(-3x-8y)=20(25)\\ \Rightarrow -9x-24y=500\end{array}$

Let, equation $-9x-24y=500\to \text{(2)}$

Multiply equation (2) with 2,

  $-18x-48y=1000\to \text{(3)}$

Add equation (1) and equation (3), so that x is eliminated.

  $\begin{array}{l}\Rightarrow \text{(18}x+15\text{y)+(}-18x-48y)=250+1000\\ \Rightarrow 18x-18x+15y-48y=1250\\ \Rightarrow -33y=1250\\ \Rightarrow y=-\frac{1250}{33}\end{array}$

Substitute the value of y in equation (1),

  $\begin{array}{l}\Rightarrow 18x+15(\frac{-1250}{33})=250\to \text{(1)}\\ \Rightarrow \frac{594x-18750}{33}=250\\ \Rightarrow 594x-18750=8250\\ \Rightarrow 594x=8250+18750\\ \Rightarrow 594x=27000\\ \Rightarrow x=\frac{500}{11}\end{array}$

Conclusion:

Therefore, the solutions of the system of equation is $x=\frac{500}{11}$ and $y=-\frac{1250}{33}$ .