Answer
The solution to this system of equations is $(-3, 0)$.
Work Step by Step
We can solve for $y$ in terms of $x$ in the second equation so that we can use this value for $y$ to substitute into the first equation. Solve for $y$ in the second equation by adding $3x$ to both sides of the equation to get:
$y = 3x + 9$
Substitute this expression for $y$ in the first equation:
$4y-6=2x\\
4(3x + 9) - 6 = 2x$
Distribute and multiply to simplify:
$(12x + 36) - 6 = 2x$
$12x + 30 = 2x$
Subtract $30$ from both sides of the equation to move constants to the right of the equation:
$12x = 2x - 30$
Subtract $2x$ from both sides of the equation to move variables to the left side of the equation:
$10x = -30$
Divide both sides by $10$ to solve for $x$:
$x = -3$
Now, we use this value for $x$ to substitute into the first equation:
$4y-6=2x\\
4y - 6 = 2(-3)$
$4y - 6 = -6$
Add $6$ to both sides of the equation to move constants to the right side of the equation:
$4y = 0$
Divide each side by $4$ to solve for $y$:
$y = 0$
The solution to this system of equations is $(-3, 0)$.
