Question

How do you solve the system of equations $10x+8y=2$ and $-2x-4y=6$?

Answer 1

Solve the system of equations $\mathbf{10}\mathit{x}\mathbf{+}\mathbf{8}\mathit{y}\mathbf{=}\mathbf{2}$ and $\mathbf{-}\mathbf{2}\mathit{x}\mathbf{-}\mathbf{4}\mathit{y}\mathbf{=}\mathbf{6}$.

To solve using the substitution method:

$10x+8y=2...\left(\mathrm{i}\right)$

$-2x-4y=6...\left(\mathrm{ii}\right)$

Step 1: Solve the equation for $\mathit{y}$.

From the equation $\left(\mathrm{i}\right)$:

$\begin{array}{rcl}10x+8y& =& 2\\ & \Rightarrow & y=\frac{2-10x}{8}\\ & \Rightarrow & y=\frac{1-5x}{4}\left(\mathrm{iii}\right)\end{array}$

Step 2: Find the solution to the equations.

Substitute the value $y=\frac{1-5x}{4}$ in the equation $\left(\mathrm{ii}\right)$:

$\begin{array}{rcl}-2x-4y& =& 6\\ & \Rightarrow & -2x-4\left(\frac{1-5x}{4}\right)=6\\ & \Rightarrow & -2x-1+5x=6\\ & \Rightarrow & 3x=7\\ & \Rightarrow & x=\frac{7}{3}\end{array}$

Substitute $x=\frac{7}{3}$in (iii) equation:

$\begin{array}{rcl}y& =& \frac{1-5x}{4}\\ & \Rightarrow & y=\frac{1-5\left({\displaystyle \frac{7}{3}}\right)}{4}\\ & \Rightarrow & y=\frac{3-35}{12}\\ & \Rightarrow & y=-\frac{8}{3}\end{array}$

Hence, the solution for the system of equations $\mathbf{10}\mathit{x}\mathbf{+}\mathbf{8}\mathit{y}\mathbf{=}\mathbf{2}$ and $\mathbf{-}\mathbf{2}\mathit{x}\mathbf{-}\mathbf{4}\mathit{y}\mathbf{=}\mathbf{6}$ is $\mathit{x}\mathbf{=}\frac{\mathbf{7}}{\mathbf{3}}$ and $\mathit{y}\mathbf{=}\mathbf{-}\frac{\mathbf{8}}{\mathbf{3}}$.