Solve the system by the addition method. $$\begin{array} { l }
\frac { 1 } { 6 } x - \frac { 1 } { 14 } y = 6 \
- \frac { 1 } { 14 } x + \frac { 1 } { 2 } y = 4
\end{array}$$
A) $( 42,14 )$
B) $( - 56,0 )$
C) $( - 42,28 )$
D) $( 0 , - 84 )$

Question

Solve the system by the addition method. $$\begin{array} { l } 
\frac { 1 } { 6 } x - \frac { 1 } { 14 } y = 6 \
- \frac { 1 } { 14 } x + \frac { 1 } { 2 } y = 4
\end{array}$$
A) $( 42,14 )$
B) $( - 56,0 )$
C) $( - 42,28 )$
D) $( 0 , - 84 )$

Quizplus · Accepted Answer

To solve the system by the addition method, first make the coefficients of one of the variables the same in both equations. Multiply the first equation by 14 and the second equation by 6 to eliminate $y$:$$14(\frac{1}{6}x - \frac{1}{14}y) = 14(6)$$$$6(-\frac{1}{14}x + \frac{1}{2}y) = 6(4)$$This simplifies to:$$14/6x - y = 84$$$$-3/7x + 3y = 24$$Multiplying the first equation by 6 and the second by 14 to get common coefficients for $y$, we get:$$14x - 6y = 504$$$$-6x + 42y = 504$$Adding these equations to eliminate $x$, we get:$$36y = 504$$Solving for $y$, we find $y = 14$.Substitute $y = 14$ into one of the original equations to solve for $x$:$$\frac{1}{6}x - \frac{1}{14}(14) = 6$$Simplifying:$$\frac{1}{6}x - 1 = 6$$$$\frac{1}{6}x = 7$$$$x = 42$$Therefore, the solution to the system is $(42, 14)$.

Solve the System by the Addition Method 16x−114y=6−114x+12y=4\begin{array} { l } \frac { 1 } { 6 } x - \frac { 1 } { 14 } y = 6 \\- \frac { 1 } { 14 } x + \frac { 1 } { 2 } y = 4\end{array}61​x−141​y=6−141​x+21​y=4​

Solve the System by the Addition Method $\begin{array} { l } \frac { 1 } { 6 } x - \frac { 1 } { 14 } y = 6 \\- \frac { 1 } { 14 } x + \frac { 1 } { 2 } y = 4\end{array}$