The answer is
q=35 and r=28
I'm not totally sure cause I'm only in 5th grade but...
12q 12,q 6,2,q 3,2,2,q
12q - 4q = 8
Get the 'q' together on one side and everything else on the other side. Add 12q to both sides --> 12q - 12q + 4 = 12q + 8q -6 --> 4 = 20q -6 Add 6 to both sides --> 6 + 4 = 6 + 20q - 6 --> 10 = 20q Then divide both sides by 20 --> 10/20 = 20q / 20 ---> 1/2 = q Substitute q = 1/2 into the original equation to make sure you did it right.
It's 12q -5
-12q+4=q-6 4-6=q+12q -2=13q q=-2/13
6p + 12q + 18r = 6 (p + 2q + 3r)
12q 2,6q 2,2,3q 2,2,3,q
steps to solve 4(3q-q): =(4)(3q+-q) =(4)(3q)+(4)(-q) =12q-4 answer: 8q
12q 12,q 6,2,q 3,2,2,q
composite
4(7p + 3q - 2)
2p+4q=16 (now divide the equation by two) p+2q=8 (now subtract 2q) p=8-2q 7p+12q=52 (substitue the answer you got for p in the previous equation) 7(8-2q)+12q=52 (multiply the first equation by 7) 56-14q+12q=52 (subtract 14q from 12q) 56-2q=52 (subtract the 56 from 52) -2q=-4 (multiply by -1) 2q=4 (divide by 2) q=2 p=8-2q (substitute the value of q) p=8-2(2) (multiply) p=8-4 (subtract) p=4 2p+4q=16 (check your answers with the new values of p and q) 2(4)+4(2)=16 8+8=16 true 7p+12q=52 7(4)+12(2)=52 28+24=52 true