# Chapter 7 - Section 7.1 - Systems of Linear Equations in Two Variables - Exercise Set - Page 819: 53

Desired linear system is $y=x-4,y=-\frac{1}{3}x+4$.

#### Work Step by Step

We know that if the equation satisfies the value of $x$ and $y$ then the set is the solution of the equation. Therefore, put the value of $x=6$ and $y=2$ in the equation \begin{align} & x-y=4 \\ & 6-2=4 \\ & 4=4 \end{align} So, $\left( 6,2 \right)$ is the solution set of the equation $x-y=4$. Then, for the slope–intercept form write the equation in the form of $y=mx+c$: And add –x both sides in the equation $x-y=4$: \begin{align} & x-y-x=4-x \\ & -y=4-x \end{align} And multiply -1 both sides in the equation $x-y=4$: \begin{align} & \left( -1\cdot -y \right)=-1\cdot 4-\left( -1\cdot x \right) \\ & y=x-4 \end{align} And put the value of $x=6$ and $y=2$ in the equation: \begin{align} & x+3y=12 \\ & 6+3\cdot \left( 2 \right)=12 \\ & 12=12 \end{align} So, $\left( 6,2 \right)$ is the solution set of the equation $x+3y=12$. And for the slope–intercept form write the equation in the form of $y=mx+c$: And multiply $\frac{1}{3}$ both sides in the equation $x+3y=12$ \begin{align} & \frac{1}{3}\cdot x+\frac{1}{3}\cdot 3y=\frac{1}{3}\cdot 12 \\ & \frac{x}{3}+y=4 \end{align} And add $-\frac{x}{3}$ both sides in the equation $x+3y=12$: \begin{align} & \frac{x}{3}+y-\frac{x}{3}=4-\frac{x}{3} \\ & y=-\frac{x}{3}+4 \end{align} But $x-3y=-6$ and $x-3y=6$ does not satisfy the solution set $\left( 6,2 \right)$. Hence, the equation in slope–intercept form that satisfies the solution set $\left( 6,2 \right)$ $y=x-4,\ y=-\frac{1}{3}x+4$.

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