How do you solve #(x+2)/3+(x-3)/4=1#?

Answer 1

See the entire solution process below:

By multiplying both sides of the equation by the common denominator of the two fractions, we can first eliminate the fractions.

#3 x 4 = color(red)(12)# is a common denominator.
#color(red)(12)((x + 2)/3 + (x - 3)/4) = color(red)(12) xx 1#
#(color(red)(12) xx (x + 2)/3) + (color(red)(12) xx (x - 3)/4) = 12#
#(cancel(color(red)(12)) 4 xx ((x + 2))/color(red)(cancel(color(black)(3)))) + (cancel(color(red)(12)) 3 xx ((x - 3))/color(red)(cancel(color(black)(4)))) = 12#
#4(x + 2) + 3(x - 3) = 12#

The terms in parenthesis can now be expanded, and on the left side of the equation, we can group and combine like terms:

#(4 xx x) + (4 xx 2) + (3 xx x) - (3 xx 3) = 12#
#4x + 8 + 3x - 9 = 12#
#4x + 3x + 8 - 9 = 12#
#7x - 1 = 12#
Next, we can add #color(red)(1)# to each side of the equation to isolate the #x# term while keeping the equation balanced:
#7x - 1 + color(red)(1) = 12 + color(red)(1)#
#7x - 0 = 13#
#7x = 13#
Now, divide each side of the equation by #color(red)(7)# to solve for #x# while keeping the equation balanced:
#(7x)/color(red)(7) = 13/color(red)(7)#
#(color(red)(cancel(color(black)(7)))x)/cancel(color(red)(7)) = 13/7#
#x = 13/7#
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Answer 2

To solve the equation (x+2)/3 + (x-3)/4 = 1, you can follow these steps:

  1. Multiply every term in the equation by the least common denominator (LCD) of 3 and 4, which is 12. This will eliminate the denominators.

12 * [(x+2)/3] + 12 * [(x-3)/4] = 12 * 1

4(x+2) + 3(x-3) = 12

  1. Distribute the multiplication on both sides of the equation.

4x + 8 + 3x - 9 = 12

  1. Combine like terms on both sides.

7x - 1 = 12

  1. Isolate the variable term by adding 1 to both sides.

7x = 13

  1. Finally, solve for x by dividing both sides by 7.

x = 13/7

Therefore, the solution to the equation (x+2)/3 + (x-3)/4 = 1 is x = 13/7.

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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.

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.

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.

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.

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