Question

Solve by the elimination method
3r-2s=19
2r+3s=30

Answer 1

29

Answer 2

hello diana

Answer 3

-2(3r-2s=19) --> -6r+4s=-38
3(2r+3s=30) --> 6r+9s=90
Can you solve the rest?

Answer 4

no

Answer 5

i always have trouble with these

Answer 6

5s=128

Answer 7

divide 128divided by 5s

Answer 8

25.6

Answer 9

wait where did you get 5s?
its 13s=52
then you divide by 13 and get s=4

Answer 10

2s+3s

Answer 11

oh. okay.
remember how i multiplied one equation by -2 and the other by 3, well that gives you -6r+4s=-38 and 6r+9s=90. these are your new equations and it allows you to cancel out r and solve for s.

Answer 12

So the solution is what

Answer 13

s=4 and r=9

Answer 14

do you want me to show you the steps?

Answer 15

yes

Answer 16

s=4 and r=9

Answer 17

sorry hold on

Answer 18

step one:
look for a number that can you can multiply each equation to make one of the coefficients in front of the variable cancel out...
-2(3r-2s=19) --> -6r+4s=-38
3(2r+3s=30) --> 6r+9s=90
step 2:
when adding the two new equations -6r and 6r cancels out while 4s+9s=13s and -38+90=52. this gives you 13s=52
step 3:
divide by 13 to both sides and s=4
step 4:
plug in s into one of the original equations and you solve for r soo..
3r-2(4)=19
3r-8=19
step 5:
add 8 to both sides
3r=27
step 6:
divide by 3 on both sides
r=9

there you go. hope that helps.

Answer 19

ty