$$\begin{array}{r}3 x+4 y=-23 \\2 y-x=-19\end{array}$$What is the solution $(x, y)$ to the system of equations above?
SAT · Math · previous-year question
- A.$(-5,-2)$
- B.$(3,-8)$correct
- C.$(4,-6)$
- D.$(9,-6)$
Answer
B. $(3,-8)$
Explanation
Choice $\mathbf{B}$ is correct. Adding $x$ and 19 to both sides of $2 y-x=-19$ gives $x=2 y+19$. Then, substituting $2 y+19$ for $x$ in $3 x+4 y=-23$ gives $3(2 y+19)+4 y=-23$. This last equation is equivalent to $10 y+57=-23$. Solving $10 y+57=-23$ gives $y=-8$. Finally, substituting -8 for $y$ in $2 y-x=-19$ gives $2(-8)-x=-19$, or $x=3$. Therefore, the solution $(x, y)$ to the given system of equations is $(3,-8)$.Choices $A, C$, and $D$ are incorrect because when the given values of $x$ and $y$ are substituted in $2 y-x=-19$, the value of the left side of the equation does not equal -19 .
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