Solving by elimination 2s + 3b= 28 3s + 2b =27

s=5 b=6

Step-by-step explanation:

2s + 3b = 28.

3s + 2b = 27

First, multiply the first eqn by 3 and multiply the first eqn by 2.

FROM FIRST EQUATION

3 (2s + 3b= 28)

6s + 9b = 84.

FROM SECOND EQUATION

2 (3s+2b =27)

6s + 4b = 54

FIRST EQUATION - SECOND EQUATION

(6s+9b)-(6s+4b) = 84-54

6s+9b-6s-4b = 30

6s-6s+9b-4b=30

5b=30

b=6

Then, substitute b=6 into the first eqn.( the second eqn also can be used)

6s+9(6) =84

6s + 54 = 84

6s = 84-54

6s = 30

s=5

Thank you.

answer: answer is c!

ok i got to organize the

3/15 and 12/55

3*

aha

3*4 equals 12 but not

a: no

b:

8/24 12/35

no

gautie

it's c

5/18 = 25/90

5*5=25

18*5 = 90

hope that you!

one more happy customer!

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step-by-step explanation: