Chapter 6.3, Problem 28PPS

To determine

To solve the system of equations using elimination method

Answer to Problem 28PPS

$x=12$ and $y=-\frac{39}{8}$

Explanation of Solution

Given:

  $\frac{1}{2}x+\frac{2}{3}y=2\frac{3}{4}$

  $\frac{1}{4}x-\frac{2}{3}y=6\frac{1}{4}$

Calculation:

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

  $\frac{1}{2}x+\frac{2}{3}y=2\frac{3}{4}$

Least Common Multiplier of (2, 3) is 6.

  $\begin{array}{l}\Rightarrow \frac{3}{6}x+\frac{4}{6}y=\frac{11}{4}\\ \Rightarrow \frac{3x}{6}+\frac{4y}{6}=\frac{11}{4}\\ \Rightarrow \frac{3x+4y}{6}=\frac{11}{4}\\ \Rightarrow 4(3x+4y)=6(11)\\ \Rightarrow 12x+16y=66\end{array}$

Let, equation $12x+16y=66\to \text{(1)}$

Now,

  $\frac{1}{4}x-\frac{2}{3}y=6\frac{1}{4}$

Least Common Multiple of (4, 3) is 12.

  $\begin{array}{l}\Rightarrow \frac{3}{12}x-\frac{8}{12}y=\frac{25}{4}\\ \Rightarrow \frac{3x}{12}-\frac{8y}{12}=\frac{25}{4}\\ \Rightarrow \frac{3x-8y}{12}=\frac{25}{4}\\ \Rightarrow 4(3x-8y)=12(25)\\ \Rightarrow 12x-32y=300\end{array}$

Let, equation $12x-32y=300\to \text{(2)}$

Subtract equation (1) and equation (2) so that x is eliminated.

  $\begin{array}{l}\Rightarrow \text{(1}2x-\text{32y)}-\text{(12}x+16y)=300-66\\ \Rightarrow 12x-12x-32y-16y=234\\ \Rightarrow -48y=234\\ \Rightarrow y=-\frac{39}{8}\end{array}$

Substitute the value of y in equation (2),

  $\begin{array}{l}\Rightarrow 12x-32(-\frac{39}{8})=300\\ \Rightarrow 12x+4(39)=300\\ \Rightarrow 12x+156=300\\ \Rightarrow 12x=300-156\\ \Rightarrow 12x=144\\ \Rightarrow x=12\end{array}$

Conclusion:

Therefore, the solutions of the system of equation is $x=12$ and $y=-\frac{39}{8}$ .