Solve the system of equations by the elimination method.
-$$\begin{array} { l }
\frac { 3 } { 2 } x - \frac { 1 } { 3 } y = - 18 \
\frac { 3 } { 4 } x + \frac { 2 } { 9 } y = - 9
\end{array}$$
A) $( - 12,0 )$
B) $( 0 , - 12 )$
C) $( 0,12 )$
D) $( 12,0 )$

Question

Solve the system of equations by the elimination method.
-$$\begin{array} { l } 
\frac { 3 } { 2 } x - \frac { 1 } { 3 } y = - 18 \
\frac { 3 } { 4 } x + \frac { 2 } { 9 } y = - 9
\end{array}$$
A) $( - 12,0 )$
B) $( 0 , - 12 )$
C) $( 0,12 )$
D) $( 12,0 )$

Quizplus · Accepted Answer

To solve the system of equations by the elimination method, we first make the coefficients of either $x$ or $y$ the same in both equations. Multiplying the first equation by 2 gives us $3x - \frac{2}{3}y = -36$, and the second equation remains unchanged as $\frac{3}{4}x + \frac{2}{9}y = -9$. To eliminate $y$, we can multiply the second equation by 3 to get $\frac{9}{4}x + \frac{2}{3}y = -27$. Adding this to the modified first equation ($3x - \frac{2}{3}y = -36$) eliminates $y$ and gives $3x + \frac{9}{4}x = -36 - 27$. Solving this equation for $x$ yields $x = -12$. Substituting $x = -12$ back into one of the original equations allows us to solve for $y$, which turns out to be $0$. Therefore, the solution to the system of equations is $(-12, 0)$.

Solve the System of Equations by the Elimination Method 32x−13y=−1834x+29y=−9\begin{array} { l } \frac { 3 } { 2 } x - \frac { 1 } { 3 } y = - 18 \\\frac { 3 } { 4 } x + \frac { 2 } { 9 } y = - 9\end{array}23​x−31​y=−1843​x+92​y=−9​

Solve the System of Equations by the Elimination Method $\begin{array} { l } \frac { 3 } { 2 } x - \frac { 1 } { 3 } y = - 18 \\\frac { 3 } { 4 } x + \frac { 2 } { 9 } y = - 9\end{array}$