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How do you solve #3(2x + 1) = 7-2 (x+5)#?

Answer 1

Eli Assouline

#color(blue)(x=-3/4#

#3(2x+1)=7-2(x+5)#

#6x+3=7-2x -10#

#6x+2x=7-10-3#

#8x=-6#

#color(blue)(x=-6/8 = -3/4#

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Answer 2

Julia Adams

First, distribute the terms inside the parentheses: 3(2x + 1) = 7 - 2(x + 5) 6x + 3 = 7 - 2x - 10

Combine like terms: 6x + 3 = -2x - 3

Move the terms involving x to one side and constants to the other side: 6x + 2x = -3 - 3 8x = -6

Divide both sides by 8 to solve for x: x = -6/8

Simplify the fraction: x = -3/4

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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.