Solve the system ½x + ¼y = 4 and −4x − 3y = −2. Give the answer as an ordered pair.

Step-by-step solution

1. 1

   Clear the fractions in the first equation. The denominators are $2$ and $4$, so multiply the whole equation by $4$:

   $4\left(\tfrac12 x\right)+4\left(\tfrac14 y\right)=4(4)\quad\Longrightarrow\quad 2x+y=16$

   Every term, including the right-hand side, must be multiplied — forgetting the $4\times 4$ is the usual slip.

2. 2

   Isolate the variable with coefficient 1. From $2x+y=16$,

   $y=16-2x$

   This makes substitution cheap; elimination would work too, but here $y$ is already alone up to a sign.

3. 3

   Substitute into the second equation.

   $-4x-3(16-2x)=-2$

   $-4x-48+6x=-2$

   Note the $-3$ multiplies both terms of $(16-2x)$, turning $-2x$ into $+6x$.

4. 4

   Solve for x.

   $2x-48=-2\quad\Longrightarrow\quad 2x=46\quad\Longrightarrow\quad x=23$

5. 5

   Back-substitute for y.

   $y=16-2(23)=16-46=-30$

   $\boxed{(23,\,-30)}$

6. 6

   Check in both original equations. First: $\tfrac12(23)+\tfrac14(-30)=11.5-7.5=4$ ✓. Second: $-4(23)-3(-30)=-92+90=-2$ ✓. Both hold, so the pair is correct — checking only one equation would not catch an error made after the substitution.