## College Algebra 7th Edition

(a) The two equations are linear. (b) The solution is $(-2, 3)$
(a) This system of equations is linear because all of the terms are, at most, first-degree terms. (b) We want to solve this equation by the elimination method. We want to get both equations to the point that one of the terms is the same so we can eliminate that one term in both equations. Then, we can solve the two systems for one variable only. We keep one equation as-is and just change the other one. We multiply the first equation by 5 to make the $x$ terms in both equations the same. The first equation is: $x + 3y = 7$ We multiply each term by 5 to get an equation that is equivalent to the original equation: $5x + 3(5)(y) = 7(5)$ $5x + 15y = 35$ We now place the first equation that has been modified together with the original second equation: $5x +15y = 35$ $5x +2y = -4$ We can eliminate the $5x$ term by subtracting the second equation from the first equation. We now have: $13y = 39$ We now have only one variable in the equation and can solve for that variable: $y = 3$ Now that we have the $y$ term, we can substitute this value into one of the equations to get a value for $x$. We will use the first equation because it is easier to work with: $x + 3(3) = 7$ $x + 9 = 7$ We subtract 9 from both sides to isolate $x$: $x = -2$ These two equations intersect at point $(-2, 3)$.