what is the solution (q, r) to this system of linear equations?\n12q + 3r = 15\n-4q - 4r = -44\n(-18…

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 $-12q-12r=-132$: $$(12q + 3r)+(-12q-12r)=15+( - 132)$$ $$12q + 3r-12q-12r=15 - 132$$ $$-9r=-117$$

Step3: Solve for $r$

Divide both sides of $-9r=-117$ by $-9$: $r=\frac{-117}{-9}=13$

Step4: Substitute $r$ into the first equation

Substitute $r = 13$ into the first equation $12q+3r = 15$: $12q+3\times13 = 15$ $12q+39 = 15$ $12q=15 - 39$ $12q=-24$

Step5: Solve for $q$

Divide both sides of $12q=-24$ by 12: $q=\frac{-24}{12}=-2$

Answer:

$(-2,13)$