Solve by elimination.
-$x+\frac{1}{4} y=-\frac{23}{30}$

$\frac{1}{4} x-y=\frac{371}{60}$

A) $\left(\frac{11}{15},-6\right)$

B) $\left(6, \frac{11}{15}\right)$

C) $\left(-6,-\frac{11}{15}\right)$

D) $\left(\frac{11}{15}, 6\right)$

Question

Solve by elimination.
-$x+\frac{1}{4} y=-\frac{23}{30}$

$\frac{1}{4} x-y=\frac{371}{60}$

A) $\left(\frac{11}{15},-6\right)$

B) $\left(6, \frac{11}{15}\right)$

C) $\left(-6,-\frac{11}{15}\right)$

D) $\left(\frac{11}{15}, 6\right)$

Quizplus · Accepted Answer

To solve by elimination, we first make the coefficients of $y$ in both equations the same. Multiplying the first equation by 4 gives us $4x + y = -\frac{92}{30}$ or simplified to $4x + y = -\frac{46}{15}$. The second equation is already in a form where the coefficient of $y$ is -1, so we can add the two equations directly:$$4x + y + \frac{1}{4}x - y = -\frac{46}{15} + \frac{371}{60}$$Combining like terms and simplifying:$$\frac{17}{4}x = \frac{-46*4 + 371}{60}$$$$\frac{17}{4}x = \frac{-184 + 371}{60}$$$$\frac{17}{4}x = \frac{187}{60}$$$$x = \frac{187}{60} * \frac{4}{17}$$$$x = \frac{11}{15}$$Substitute $x = \frac{11}{15}$ back into one of the original equations to find $y$. Using the first equation:$$\frac{11}{15} + \frac{1}{4}y = -\frac{23}{30}$$Multiplying everything by 60 (the least common multiple of 15, 4, and 30) to clear the denominators:$$44 + 15y = -46$$$$15y = -46 - 44$$$$15y = -90$$$$y = -6$$Therefore, the solution is $x = \frac{11}{15}, y = -6$, which corresponds to choice A.