# Difference between revisions of "1950 AHSME Problems/Problem 14"

## Problem

For the simultaneous equations $$2x-3y=8$$ $$6y-4x=9$$

$\textbf{(A)}\ x=4,y=0\qquad\textbf{(B)}\ x=0,y=\frac{3}{2}\qquad\textbf{(C)}\ x=0,y=0\qquad\\ \textbf{(D)}\ \text{There is no solution}\qquad\textbf{(E)}\ \text{There are an infinite number of solutions}$

## Solution

Try to solve this system of equations using the elimination method.

\begin{align*} -2(2x-3y)&=-2(8)\\ 6y-4x&=-16\\ 6y-4x&=9\\ 0&=-7 \end{align*}

Something is clearly contradictory so $\boxed{\mathrm{(D)}\text{ There is no solution}.}$

Alternatively, note that the second equation is a multiple of the first except that the constants don't match up. So, there is no solution.

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