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How do you solve for x: #5(x-3)+2(3-x)=12#?

Answer 1

Eli Assouline

You first work away the parentheses

#->(5*x-5*3)+(2*3-2*x)=12# (these brackets only used for clarification)

#->5x-15+6-2x=12->3x-9=12#

Add #9# to both sides: #->3x-cancel9+cancel9=12+9# #->3x=21->x=7#

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Answer 2

Nathan Axel

Just a different approach

#x=7#

Given:#" "5(x-3)+2(3-x)=12#

If we can change #2(3-x)" such that "(3-x)" became "(x-3)# then we have a common factor.

Note that #-2(x-3) = -2x+6" "->" "+2(3-x)=-2x+6#

Write the given equation as:

#5(x-3)-2(x-3)=12#

Factor out #(x-3)#

#(x-3)(5-2)=12#

#3(x-3)=12#

Divide both sides by 3

#x-3=4#

Add 3 to both sides

#x=7#

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Answer 3

Eva Abramson

To solve for x:

5(x - 3) + 2(3 - x) = 12 5x - 15 + 6 - 2x = 12 5x - 2x - 15 + 6 = 12 3x - 9 = 12 3x = 12 + 9 3x = 21 x = 21/3 x = 7

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.