Solve the system of linear equations by the method of elimination.Find (x,y) and check your solution algebraically. $\left\{ \begin{array} { r }
x + 5 y = 11 \
3 x - 10 y = 18
\end{array} \right.$
A) $\left( - 8 , \frac { 3 } { 5 } \right)$
B) $\left( - 8 , - \frac { 3 } { 5 } \right)$
C) $\left( 8 , \frac { 3 } { 5 } \right)$
D) $\left( 8 , - \frac { 3 } { 5 } \right)$
E) $\left( 8 , \frac { 5 } { 3 } \right)$

Question

Solve the system of linear equations by the method of elimination.Find (x,y) and check your solution algebraically.  $\left\{ \begin{array} { r } 
x + 5 y = 11 \
3 x - 10 y = 18
\end{array} \right.$  
A) $\left( - 8 , \frac { 3 } { 5 } \right)$ 
B) $\left( - 8 , - \frac { 3 } { 5 } \right)$  
C) $\left( 8 , \frac { 3 } { 5 } \right)$  
D) $\left( 8 , - \frac { 3 } { 5 } \right)$  
E) $\left( 8 , \frac { 5 } { 3 } \right)$

Quizplus · Accepted Answer

To solve the system of equations by elimination, we first multiply the first equation by 3 (to match the coefficient of $x$ in the second equation) and the second equation by 1 (to keep it unchanged), resulting in:$$3(x + 5y) = 3(11)$$$$3x - 10y = 18$$This simplifies to:$$3x + 15y = 33$$$$3x - 10y = 18$$Subtracting the second equation from the first gives:$$25y = 15$$$$y = \frac{15}{25} = \frac{3}{5}$$Substituting $y = \frac{3}{5}$ into the first original equation:$$x + 5(\frac{3}{5}) = 11$$$$x + 3 = 11$$$$x = 8$$Thus, the solution is $(x, y) = (8, \frac{3}{5})$.

Solve the System of Linear Equations by the Method of Elimination.Find