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 can multiply the first equation by 3 and the second equation by 5 to make the coefficients of $u$ the same:$$3(5u + 6v) = 3(28)$$$$5(3u + 5v) = 5(20)$$This simplifies to:$$15u + 18v = 84$$$$15u + 25v = 100$$Subtracting the first equation from the second gives:$$7v = 16$$Solving for $v$ gives:$$v = \frac{16}{7}$$Substituting $v = \frac{16}{7}$ into one of the original equations to solve for $u$, for example, the first equation:$$5u + 6\left(\frac{16}{7}\right) = 28$$$$5u + \frac{96}{7} = 28$$$$5u = 28 - \frac{96}{7}$$$$5u = \frac{196}{7} - \frac{96}{7}$$$$5u = \frac{100}{7}$$$$u = \frac{20}{7}$$Therefore, the solution is $\left(\frac{20}{7}, \frac{16}{7}\right)$.

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