Solve the system of linear equations by the method of elimination.Find (r,s)and check your solution algebraically. $$\left\{ \begin{array} { c }
2 r + 4 s = 7 \
16 r + 50 s = 77
\end{array} \right.$$
A) $\left( \frac { 7 } { 6 } , - \frac { 7 } { 6 } \right)$
B) $( 7,77 )$
C) $\left( \frac { 7 } { 6 } , \frac { 7 } { 6 } \right)$
D) $\left( - \frac { 7 } { 6 } , \frac { 7 } { 6 } \right)$
E) $\left( - \frac { 7 } { 6 } , - \frac { 7 } { 6 } \right)$

Question

Solve the system of linear equations by the method of elimination.Find (r,s)and check your solution algebraically.​ $$\left\{ \begin{array} { c } 
2 r + 4 s = 7 \
16 r + 50 s = 77
\end{array} \right.$$ ​
A)​ $\left( \frac { 7 } { 6 } , - \frac { 7 } { 6 } \right)$
B)​ $( 7,77 )$
C)​ $\left( \frac { 7 } { 6 } , \frac { 7 } { 6 } \right)$
D)​ $\left( - \frac { 7 } { 6 } , \frac { 7 } { 6 } \right)$
E)​ $\left( - \frac { 7 } { 6 } , - \frac { 7 } { 6 } \right)$

Quizplus · Accepted Answer

The system of linear equations in the given matrix can be written as:

4r - 3s = 11
3r + 2s = 2

To eliminate s, we can multiply the first equation by 2 and the second equation by 3, to get:

8r - 6s = 22
9r + 6s = 6

Adding these equations, we get:

17r = 28

So, r = 28/17.

Substituting this value in either of the original equations, we get:

3(28/17) + 2s = 2

Simplifying this, we get:

s = -1/17.

Therefore, (r,s) = (28/17, -1/17).

To check our solution algebraically, we can substitute these values in the original equations:

4(28/17) - 3(-1/17) = 11
3(28/17) + 2(-1/17) = 2

Both of these equations are satisfied, so our solution is correct.

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