the tables represent two linear functions in a system.\n| x | y |\n| -6 | -22 |\n| -3 | -10 |\n| 0 | 2 |\n|…

Explanation:

Step1: Find the equation of the first line

The slope - intercept form of a line is $y = mx + b$, where $m$ is the slope and $b$ is the y - intercept. For the first table, when $x = 0$, $y=2$, so $b = 2$. The slope $m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{-10-(-22)}{-3 - (-6)}=\frac{-10 + 22}{-3+6}=\frac{12}{3}=4$. So the equation of the first line is $y = 4x+2$.

Step2: Find the equation of the second line

For the second table, when $x = 0$, $y=-12$, so $b=-12$. The slope $m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{-21-(-30)}{-3 - (-6)}=\frac{-21 + 30}{-3 + 6}=\frac{9}{3}=3$. So the equation of the second line is $y=3x - 12$.

Step3: Solve the system of equations

Set the two equations equal to each other: $4x+2=3x - 12$. Subtract $3x$ from both sides: $4x-3x+2=3x-3x - 12$, which gives $x+2=-12$. Then subtract 2 from both sides: $x=-12 - 2=-14$.

Step4: Find the y - value

Substitute $x = - 14$ into the first equation $y = 4x+2$. So $y=4\times(-14)+2=-56 + 2=-54$.

Answer:

$(-14,-54)$