Solve the system of linear equations by the method of elimination.Find (u,v) and check your solution algebraically. $\left\{ \begin{array} { l }
5 u + 6 v = 28 \
3 u + 5 v = 20
\end{array} \right.$
A) $\left( \frac { 7 } { 20 } , \frac { 7 } { 16 } \right)$
B) $\left( - \frac { 20 } { 7 } , \frac { 16 } { 7 } \right)$
C) $\left( \frac { 20 } { 7 } , - \frac { 16 } { 7 } \right)$
D) $\left( - \frac { 20 } { 7 } , - \frac { 16 } { 7 } \right)$
E) $\left( \frac { 20 } { 7 } , \frac { 16 } { 7 } \right)$

Question

Solve the system of linear equations by the method of elimination.Find (u,v) and check your solution algebraically.  $\left\{ \begin{array} { l } 
5 u + 6 v = 28 \
3 u + 5 v = 20
\end{array} \right.$  
A) $\left( \frac { 7 } { 20 } , \frac { 7 } { 16 } \right)$
B) $\left( - \frac { 20 } { 7 } , \frac { 16 } { 7 } \right)$  
C) $\left( \frac { 20 } { 7 } , - \frac { 16 } { 7 } \right)$  
D) $\left( - \frac { 20 } { 7 } , - \frac { 16 } { 7 } \right)$  
E) $\left( \frac { 20 } { 7 } , \frac { 16 } { 7 } \right)$

Quizplus · Accepted Answer

To solve the system of equations by elimination, we first multiply the equations by suitable numbers to make the coefficients of either $u$ or $v$ the same. Multiplying the first equation by 3 and the second by -5 gives:$$15u + 18v = 84$$$$-15u - 25v = -100$$Adding these equations eliminates $u$, resulting in:$$-7v = -16$$Solving for $v$ gives $v = \frac{16}{7}$.Substituting $v = \frac{16}{7}$ into one of the original equations (for simplicity, use the second equation):$$3u + 5\left(\frac{16}{7}\right) = 20$$$$3u + \frac{80}{7} = 20$$$$3u = 20 - \frac{80}{7}$$$$3u = \frac{140 - 80}{7}$$$$3u = \frac{60}{7}$$$$u = \frac{20}{7}$$Thus, the solution is $u = \frac{20}{7}, v = \frac{16}{7}$, which corresponds to choice E.

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