## Algebra 1: Common Core (15th Edition)

$x = 6$ $y = 0$
To solve a system of equations using substitution, we plug in an expression in place of a variable. In this system of equations, we already have one variable expressed in terms of the other variable, so let's use the second equation to substitute for $y$ in the first equation: $4x + 9(-\frac{1}{3}x + 2) = 24$ Use the distributive property on the right side of the equation: $4x - \frac{9}{3}x + 18 = 24$ Simplify the fraction by dividing both the numerator and denominator by their greatest common factor, $3$: $4x - 3x + 18 = 24$ Combine like terms on the left side of the equation: $x + 18 = 24$ Subtract $18$ from each side of the equation to move constants to the right side of the equation: $x = 6$ Now that we have the value for $x$, we can substitute it into one of the equations to solve for $y$: $y = (-\frac{1}{3})(6) + 2$ Multiply first: $y = -\frac{6}{3} + 2$ Simplify the fraction by dividing the numerator and denominator by their greatest common factor, $3$: $y = -2 + 2$ Add to solve for $y$: $y = 0$