Solve the system of linear equations by the method of elimination.Find (x,y) and check your solution algebraically. $\left\{ \begin{array} { c }
5 x + 3 y = 6 \
3 x - y = 5
\end{array} \right.$
A) $\left( \frac { 3 } { 2 } , \frac { 1 } { 2 } \right)$
B) $\left( - \frac { 3 } { 2 } , \frac { 1 } { 2 } \right)$
C) $\left( \frac { 3 } { 2 } , - \frac { 9 } { 2 } \right)$
D) $\left( - \frac { 3 } { 2 } , - \frac { 1 } { 2 } \right)$
E) $\left( \frac { 3 } { 2 } , - \frac { 1 } { 2 } \right)$

Question

Solve the system of linear equations by the method of elimination.Find (x,y) and check your solution algebraically.  $\left\{ \begin{array} { c } 
5 x + 3 y = 6 \
3 x - y = 5
\end{array} \right.$  
A)  $\left( \frac { 3 } { 2 } , \frac { 1 } { 2 } \right)$
B)  $\left( - \frac { 3 } { 2 } , \frac { 1 } { 2 } \right)$
C)  $\left( \frac { 3 } { 2 } , - \frac { 9 } { 2 } \right)$
D)  $\left( - \frac { 3 } { 2 } , - \frac { 1 } { 2 } \right)$
E) $\left( \frac { 3 } { 2 } , - \frac { 1 } { 2 } \right)$

Quizplus · Accepted Answer

To solve the system of equations by elimination, we first make the coefficients of $y$ in both equations the same. We can multiply the second equation by 3 to achieve this:Original equations:1) $5x + 3y = 6$2) $3x - y = 5$After multiplying the second equation by 3:1) $5x + 3y = 6$2) $9x - 3y = 15$Adding the two equations together to eliminate $y$:$5x + 9x + 3y - 3y = 6 + 15$$14x = 21$$x = \frac{21}{14} = \frac{3}{2}$Substitute $x = \frac{3}{2}$ into one of the original equations to find $y$. Using the second equation:$3(\frac{3}{2}) - y = 5$$\frac{9}{2} - y = 5$$-y = 5 - \frac{9}{2}$$-y = \frac{10}{2} - \frac{9}{2}$$-y = \frac{1}{2}$$y = -\frac{1}{2}$Thus, the solution is $x = \frac{3}{2}, y = -\frac{1}{2}$, which corresponds to choice E.

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