Solve the system by elimination if possible. $$\left\{ \begin{array} { c }
2 x + 3 y = 29 \
3 x - 2 y = - 15
\end{array} \right.$$
A)(1, 9)
B)(1, $- 9$ )
C)(9, 1)
D)infinitely many solutions, dependent equations
E)no solution, inconsistent system

Question

Solve the system by elimination if possible. $$\left\{ \begin{array} { c } 
2 x + 3 y = 29 \
3 x - 2 y = - 15
\end{array} \right.$$
A)(1, 9)
B)(1, $- 9$ )
C)(9, 1)
D)infinitely many solutions, dependent equations
E)no solution, inconsistent system

Quizplus · Accepted Answer

To solve the system by elimination, we can multiply the equations by appropriate values to make the coefficients of either $x$ or $y$ cancel out when we add or subtract the equations. Multiplying the first equation by 2 and the second equation by 3 gives us:$$4x + 6y = 58$$$$9x - 6y = -45$$Adding these equations together to eliminate $y$, we get:$$13x = 13$$Solving for $x$, we find $x = 1$. Substituting $x = 1$ into the first original equation:$$2(1) + 3y = 29$$$$2 + 3y = 29$$$$3y = 27$$$$y = 9$$Therefore, the solution to the system is $x = 1$, $y = 9$, which corresponds to choice A.

Solve the System by Elimination If Possible {2x+3y=293x−2y=−15\left\{ \begin{array} { c } 2 x + 3 y = 29 \\3 x - 2 y = - 15\end{array} \right.{2x+3y=293x−2y=−15​

Solve the System by Elimination If Possible $\left\{ \begin{array} { c } 2 x + 3 y = 29 \\3 x - 2 y = - 15\end{array} \right.$