91Ó°ÊÓ

Equation - $$\begin{gathered} x+\frac{1}{2}y=\frac{3}{2} \\ \frac{1}{5}x-\frac{1}{5}y=3 \end{gathered}$$

Solve a System of Equations by Elimination and find the value of x and y.

Equation -

$$\begin{gathered} x+\frac{1}{2}y=\frac{3}{2} \\ \frac{1}{5}x-\frac{1}{5}y=3 \end{gathered}$$

Write the equation in the standard form by Clear the fractions by multiplying the First equation by 2 and second equation by 5 -

$$\begin{gathered} (x+\frac{1}{2}y)×2=\frac{3}{2}×2 \\ (\frac{1}{5}x-\frac{1}{5}y)×5=3×5 \\ 2x+y=3 \\ x-y=15 \end{gathered}$$

The equation become -

$$\begin{gathered} 2x+y=3 \\ x-y=15 \end{gathered}$$

We Add the x's and y's and constants -

$\begin{array}{l}2x+y=3\\ \frac{x-y=15}{3x=18}\end{array}$

Solve for x -

$x=6$

Substitute $\mathit{x}\mathbf{=}\mathbf{6}$into the second equation $\frac{\mathbf{1}}{\mathbf{5}}\mathit{x}\mathbf{-}\frac{\mathbf{1}}{\mathbf{5}}\mathit{y}\mathbf{=}\mathbf{3}$, Then Solve for y -

$$\begin{gathered} \frac{1}{5}x-\frac{1}{5}y=3 \\ \frac{1}{5}×6-\frac{1}{5}y=3 \\ \frac{6}{5}-\frac{1}{5}y=3 \\ \frac{1}{5}y=\frac{6}{5}-3 \\ y=-9 \end{gathered}$$

Written as $(x,y)$will be $(6,-9)$

Check that the ordered pair is a solution to both original equations -

Put $(6,-9)$in the equations -

$$\begin{gathered} x+\frac{1}{2}y=\frac{3}{2} \\ 6+\frac{1}{2}×(-9)=\frac{3}{2} \\ 6-\frac{9}{2}=\frac{3}{2} \\ \frac{3}{2}=\frac{3}{2} \\ \frac{1}{5}x-\frac{1}{5}y=3 \\ \frac{1}{5}×6-\frac{1}{5}(-9)=3 \\ \frac{6}{5}+\frac{9}{5}=3 \\ 3=3 \end{gathered}$$

The Solution is - $\mathbf{(}\mathbf{6}\mathbf{,}\mathbf{-}\mathbf{9}\mathbf{)}$