Introductory Algebra for College Students (7th Edition)

The solution is $(-5, -2)$.
The first thing we want to do is to rewrite the equations so that all the variables are on one side and the constants are on the other. For the first equation, we subtract $3y$ from both sides of the equation. For the second equation, we leave it alone because the variables are already on one side of the equation. $2x - 3y = -4$ $-6x + 12y = 6$ We first need to modify the two equations so that the coefficients of one of the variables differs only in sign. With this modification, we will be able to cancel out one of the variables and solve for the other one. We see that if we multiply the first equation by $3$ and leave the second equation unchanged, the coefficients of the $x$ term will be $6$ and $-6$. The coefficients will now only differ in sign, so we can cancel them out when we add the two equations. Let us do the multiplication: $3(2x - 3y) = 3(-4)$ $-6x + 12y = 6$ Use distributive property: $3(2x) - 3(3y) = 3(-4)$ $-6x + 12y = 6$ Let's multiply out the terms: $6x - 9y = -12$ $-6x + 12y = 6$ We can now cancel out the $x$ terms to get: $-9y = -12$ $12y = 6$ Now we add both sides of the two equations to get: $3y = -6$ Divide both sides of the equation by $3$ to solve for $y$: $y = -2$ Now that we have the value for $y$, we can plug this value into one of the equations to solve for $x$. Let's use the first equation: $2x = 3(-2) - 4$ Multiply first: $2x = -6 - 4$ Do the subtraction on the right-hand side of the equation: $2x = -10$ Divide both sides by $2$ to solve for $x$: $x = -5$ The solution is $(-5, -2)$.