Question

kenneth and gary both wrote a linear equation to make up a system of equations. kenneth graphed his equation and gary made a table of values.

x y
-4 -1
-2 0
0 1
2 2

(0,-2)
(-1,-1.5)
(-1,-2.5)
(-2,0)

Explanation:

Step1: Find the equation from Gary's table

Use the slope - intercept form $y = mx + b$. First, find the slope $m=\frac{y_2 - y_1}{x_2 - x_1}$. Let $(x_1,y_1)=(-4,-1)$ and $(x_2,y_2)=(-2,0)$. Then $m=\frac{0 - (-1)}{-2-(-4)}=\frac{1}{2}$. Using the point - slope form with the point $(0,1)$ (where $b = 1$), the equation is $y=\frac{1}{2}x + 1$.

Step2: Find the equation from Kenneth's graph

Find two points on the line, say $(0,-2)$ and $(2,-4)$. The slope $m=\frac{-4-(-2)}{2 - 0}=\frac{-2}{2}=-1$. Using the slope - intercept form $y=mx + b$ with the point $(0,-2)$ (so $b=-2$), the equation is $y=-x - 2$.

Step3: Solve the system of equations

Set $\frac{1}{2}x + 1=-x - 2$. Add $x$ to both sides: $\frac{1}{2}x+x+1=-2$. Combine like terms: $\frac{3}{2}x+1=-2$. Subtract 1 from both sides: $\frac{3}{2}x=-3$. Multiply both sides by $\frac{2}{3}$: $x=-2$. Substitute $x = - 2$ into $y=-x - 2$, we get $y=-(-2)-2=0$.

Answer:

(-2,0)