How do you solve #4( 2s - 3) = 3( s + 1) }#?

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Answer 1 · 2017-09-25T03:15:49

#s= 3#

In order to solve for s, I would do it step-by-step.

#4(2s-3) = 3(s+1)#

Use the distributive property to get rid of the parentheses
#8s-12= 3s+3#

Isolate the ones with the variable in the left and the nonvariables to the right. You could do it the other way because it does not matter at all.

#8s-12+12= 3s+3+12#

#8s= 3s+15#

#8s-3s= 3s-3s+15#

#5s= 15#

Divide 5 on both sides
#5s/5= 15/5#

Final
#s= 3#