Solve the system of equations by elimination, if a solution exists.
-$$\left\{ \begin{array} { l }
8 x + 8 y = 64 \
3 x - 5 y = - 40
\end{array} \right.$$
A) $x = 0 , y = 9$
B) $x = - 1 , y = 9$
C) No solution
D) $x = 0 , y = 8$

Question

Solve the system of equations by elimination, if a solution exists.
-$$\left\{ \begin{array} { l } 
8 x + 8 y = 64 \
3 x - 5 y = - 40
\end{array} \right.$$
A) $x = 0 , y = 9$
B) $x = - 1 , y = 9$
C) No solution
D) $x = 0 , y = 8$

Quizplus · Accepted Answer

To solve the system of equations by elimination, we first aim to eliminate one of the variables by making the coefficients in front of that variable the same in both equations. We can start by simplifying the first equation:$$8x + 8y = 64$$Divide every term by 8:$$x + y = 8$$  (Equation 1)The second equation remains:$$3x - 5y = -40$$  (Equation 2)To eliminate one of the variables, we can multiply Equation 1 by 5 to match the coefficient of $y$ in Equation 2:$$5(x + y) = 5(8)$$$$5x + 5y = 40$$  (Equation 3)Now, we add Equation 3 to Equation 2 to eliminate $y$:$$3x - 5y + 5x + 5y = -40 + 40$$$$8x = 0$$$$x = 0$$Substitute $x = 0$ back into Equation 1 to find $y$:$$0 + y = 8$$$$y = 8$$Therefore, the solution is $x = 0, y = 8$, which corresponds to choice D.

Solve the System of Equations by Elimination, If a Solution {8x+8y=643x−5y=−40\left\{ \begin{array} { l } 8 x + 8 y = 64 \\3 x - 5 y = - 40\end{array} \right.{8x+8y=643x−5y=−40​

Solve the System of Equations by Elimination, If a Solution $\left\{ \begin{array} { l } 8 x + 8 y = 64 \\3 x - 5 y = - 40\end{array} \right.$