Hi, I need help.

If 4r+7s=23 and r−2s=17 then 3r+3s=

(A) 8 (B) 24 (C) 32 (D) 40 (E) 16

Explanation:

Add the left hand sides of the two equations to get (4r+7s)+(r-2s) = 5r+5s

Do the same for the right hand sides of the two equations 23+17 = 40

After adding the equations, we end up with 5r+5s = 40 which is the same as 5(r+s) = 40

Next, divide both sides by 5 to lead to r+s = 8

Finally, multiply both sides by 3 to end up with 3(r+s) = 24 which is the same as 3r+3s = 24

An alternative is to solve r-2s = 17 for r to get r = 2s+17. Then plug this into 4r+7s = 23 getting 4(2s+17)+7s = 23. This solves to...

4(2s+17)+7s = 23

8s+68+7s = 23

15s+68 = 23

15s = 23-68

15s = -45

s = -45/15

s = -3

Which means,

r = 2s+17

r = 2(-3)+17

r = 11

and therefore,

3r+3s = 3(-3)+3(11) = -9+33 = 24

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3 is greater then both of those numbers