what is the solution (q, r) to this system of linear equations? 12q + 3r = 15 -4q - 4r = -44 (-18, 29) (-2…

Explanation:

Step1: Multiply second - equation

Multiply the second equation $-4q - 4r=-44$ by 3 to make the coefficient of $q$ opposite of that in the first equation. We get $-12q-12r = - 132$.

Step2: Add the two equations

Add the first equation $12q + 3r=15$ and $-12q-12r=-132$: [ \begin{align*} (12q + 3r)+(-12q-12r)&=15+( - 132)\ 12q+3r - 12q-12r&=15 - 132\ -9r&=-117 \end{align*} ] Solve for $r$: $r=\frac{-117}{-9}=13$.

Step3: Substitute $r$ into the first equation

Substitute $r = 13$ into the first equation $12q+3r = 15$: [ \begin{align*} 12q+3\times13&=15\ 12q+39&=15\ 12q&=15 - 39\ 12q&=-24\ q&=-2 \end{align*} ]

Answer:

B. $(-2,13)$