Question 164488
Solve each system by substitution:
[r+3s=9
[3r+2s=13


Let [r+3s=9 be Equation A
Let [3r+2s=13 be Equation B

I will solve for r in Equation A.

r + 3s = 9

r = -3s + 9

I will now plug this "value" for r into Equation B and solve for s.


3(-3s + 9) + 2s = 13

-9s + 27 = 13

-9s = -27 + 13

-9s = -14

s = -14/-9

s = 14/9....This is the true value of s.

Now, to find r, replace s with 14/9 in EITHER Equation A or B (your choice which equation) and then simplify.

Can you take it from here?

If not, write back.