Solve the system of linear equations by the method of elimination.Find (x,y)and check your solution algebraically. $$\left\{ \begin{array} { l }
2 x + 7 y = \frac { 112 } { 9 } \\
7 x + 8 y = \frac { 62 } { 9 }
\end{array} \right.$$
A) $\left( \frac { - 14 } { 9 } , \frac { 20 } { 9 } \right)$
B) $\left( \frac { 20 } { 9 } , \frac { - 14 } { 9 } \right)$
C) $\left( \frac { 20 } { 9 } , \frac { 14 } { 9 } \right)$
D) $\left( \frac { 14 } { 9 } , \frac { 20 } { 9 } \right)$
E) $\left( \frac { 14 } { 9 } , \frac { - 20 } { 9 } \right)$

Question

Solve the system of linear equations by the method of elimination.Find (x,y)and check your solution algebraically.​ $$\left\{ \begin{array} { l } 
2 x + 7 y = \frac { 112 } { 9 } \\
7 x + 8 y = \frac { 62 } { 9 }
\end{array} \right.$$ ​
A) $\left( \frac { - 14 } { 9 } , \frac { 20 } { 9 } \right)$
B)​ $\left( \frac { 20 } { 9 } , \frac { - 14 } { 9 } \right)$
C)​ $\left( \frac { 20 } { 9 } , \frac { 14 } { 9 } \right)$
D)​ $\left( \frac { 14 } { 9 } , \frac { 20 } { 9 } \right)$
E)​ $\left( \frac { 14 } { 9 } , \frac { - 20 } { 9 } \right)$

Quizplus · Accepted Answer

To solve the system of equations by elimination, we first multiply each equation by a number that makes the coefficients of either $x$ or $y$ the same in both equations. Multiplying the first equation by 7 and the second by 2 gives:$$14x + 49y = \frac{784}{9}$$$$14x + 16y = \frac{124}{9}$$Subtracting the second equation from the first to eliminate $x$ gives:$$33y = \frac{660}{9}$$Solving for $y$ gives:$$y = \frac{20}{9}$$Substituting $y = \frac{20}{9}$ into one of the original equations to solve for $x$, using the first equation:$$2x + 7\left(\frac{20}{9}\right) = \frac{112}{9}$$$$2x + \frac{140}{9} = \frac{112}{9}$$$$2x = \frac{112}{9} - \frac{140}{9} = -\frac{28}{9}$$$$x = -\frac{14}{9}$$Therefore, the solution is $\left( \frac{-14}{9}, \frac{20}{9} \right)$.

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