Solve each system by using either the substitution or the elimination-by-addition method, whichever seems more appropriate. $$\left( \begin{array} { c }
\frac { 3 } { 4 } x + \frac { 1 } { 9 } y = 0 \
\frac { 4 } { 3 } x - \frac { 4 } { 27 } y = - 28
\end{array} \right)$$
A) ( $-$ 12, 81)
B) ( 81, $-$ 12)
C) ( $-$ 12, $-$ 81)
D) no solutions
E) infinitely many solutions

Question

Solve each system by using either the substitution or the elimination-by-addition method, whichever seems more appropriate. $$\left( \begin{array} { c } 
\frac { 3 } { 4 } x + \frac { 1 } { 9 } y = 0 \
\frac { 4 } { 3 } x - \frac { 4 } { 27 } y = - 28
\end{array} \right)$$
A) ( $-$ 12, 81)
B) ( 81, $-$ 12)
C) ( $-$ 12, $-$ 81)
D) no solutions
E) infinitely many solutions

Quizplus · Accepted Answer

To solve the system, the elimination method seems more appropriate due to the coefficients of $x$ and $y$ in both equations. Multiplying the first equation by $4/3$ and the second by $3/4$ makes the coefficients of $x$ in both equations equal, allowing for easy elimination of $x$. After simplification and elimination, solving for $y$ and then substituting back to find $x$ yields the solution $(-12, 81)$.

Solve Each System by Using Either the Substitution or the Elimination-By-Addition