Solve the system.
-$ \left\{\begin{array}{l}3 x-5 y=-4 \ 5 x+3 y=-18 \ \end{array}\right. $

A) $ (3,1) $
B) $ (-3,1) $
C) $ (-3,-1) $
D) $ (3,-1) $

Question

Solve the system.
-$ \left\{\begin{array}{l}3 x-5 y=-4 \ 5 x+3 y=-18 \ \end{array}\right. $

A) $ (3,1) $
B) $ (-3,1) $
C) $ (-3,-1) $
D) $ (3,-1) $

Quizplus · Accepted Answer

To solve the system of equations, we can use either substitution or elimination method. Here, using the elimination method seems efficient:1. Multiply the first equation by 5 and the second equation by 3 to make the coefficients of $y$ opposites:$$ \begin{align*}(3x - 5y) \cdot 5 &= -4 \cdot 5 \(5x + 3y) \cdot 3 &= -18 \cdot 3\end{align*} $$This gives:$$ \begin{align*}15x - 25y &= -20 \15x + 9y &= -54\end{align*} $$2. Add the two equations to eliminate $y$:$$ \begin{align*}15x - 25y + 15x + 9y &= -20 - 54 \30x - 16y &= -74\end{align*} $$However, I made a mistake in the calculation process. Let's correct that and directly solve the original system correctly:Original system:$$ \begin{align*}3x - 5y &= -4 \5x + 3y &= -18\end{align*} $$To eliminate one of the variables, we can multiply the first equation by 3 and the second by 5:$$ \begin{align*}9x - 15y &= -12 \25x + 15y &= -90\end{align*} $$Adding these equations to eliminate $y$:$$ \begin{align*}34x &= -102 \x &= -3\end{align*} $$Substitute $x = -3$ into one of the original equations to find $y$:$$ \begin{align*}3(-3) - 5y &= -4 \-9 - 5y &= -4 \-5y &= 5 \y &= -1\end{align*} $$Therefore, the solution is $x = -3, y = -1$, which corresponds to choice C $(-3, -1)$.

Solve the System {3x−5y=−45x+3y=−18 \left\{\begin{array}{l}3 x-5 y=-4 \\ 5 x+3 y=-18 \\ \end{array}\right. {3x−5y=−45x+3y=−18​

Solve the System $\left\{\begin{array}{l}3 x-5 y=-4 \\ 5 x+3 y=-18 \\ \end{array}\right.$