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How do you solve #2(3-x)=16(x+1)#?

Answer 1

Eva Abramson

#x=-5/9#

Let's distribute the #2# and the #16# to their respective terms. We get

#6-2x=16x+16#

Let's subtract #6# from both sides to get

#-2x=16x+10#

Next, we can subtract #16x# from both sides to get

#-18x=10#

Lastly, we can divide both sides by #-18# to get

#x=-10/18#, which can be simplified to

#x=-5/9#

Hope this helps!

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Answer 2

Julia Adams

To solve the equation 2(3-x) = 16(x+1), first distribute the numbers outside the parentheses:

2 * 3 - 2 * x = 16 * x + 16 * 1

This simplifies to:

6 - 2x = 16x + 16

Next, combine like terms by moving all terms involving x to one side of the equation and constants to the other side:

6 - 16 = 16x + 2x + 16

-10 = 18x + 16

Subtract 16 from both sides:

-10 - 16 = 18x + 16 - 16

-26 = 18x

Divide both sides by 18 to solve for x:

-26 / 18 = 18x / 18

x ≈ -1.44

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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.