Solve the system by elimination.
-$$\begin{array} { l }
6 x + 4 y = 30 \
4 x + 2 y = 20
\end{array}$$
A) $\{ ( 5,0 ) \}$

B) $\varnothing$

C) $\{ ( 4,1 ) \}$

D) $\{ ( 5,1 ) \}$

Question

Solve the system by elimination.
-$$\begin{array} { l } 
6 x + 4 y = 30 \
4 x + 2 y = 20
\end{array}$$
A) $\{ ( 5,0 ) \}$

B) $\varnothing$

C) $\{ ( 4,1 ) \}$

D) $\{ ( 5,1 ) \}$

Quizplus · Accepted Answer

To solve the system by elimination, we can multiply the second equation by 2 to make the coefficients of $y$ in both equations the same. This gives us:$$6x + 4y = 30$$$$8x + 4y = 40$$Subtracting the first equation from the second gives:$$2x = 10$$Solving for $x$ gives $x = 5$. Substituting $x = 5$ into the first equation:$$6(5) + 4y = 30$$$$30 + 4y = 30$$$$4y = 0$$Solving for $y$ gives $y = 0$. Therefore, the solution to the system is $(5, 0)$.

Solve the System by Elimination 6x+4y=304x+2y=20\begin{array} { l } 6 x + 4 y = 30 \\4 x + 2 y = 20\end{array}6x+4y=304x+2y=20​

Solve the System by Elimination $\begin{array} { l } 6 x + 4 y = 30 \\4 x + 2 y = 20\end{array}$