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

Answer 1

#x=1/8(-25+sqrt385)# or #x=1/8(-25-sqrt385)#

To solve the ensuing quadratic equation, we will first write each side as a single fraction, multiply it out, and then solve it.

#5/(x+1)-1/2=(2x)/3+3# #10/(2(x+1))-(x+1)/(2(x+1))=(2x)/3+9/3# #(11-x)/(2x+2)=(2x+9)/3#

Here, we multiply crosswise to obtain:

#3(11-x)=(2x+2)(2x+9)# #33-3x=4x^2+22x+18# #4x^2+25x-15=0#

Now you can either fill in the square or use The Formula; The Formula is what I would suggest.

#x=(-25+-sqrt(25^2-4xx4xx(-15)))/(2xx4)#
#x=(-25+-sqrt(625-240))/8# #x=(-25+-sqrt385)/8#
#:. x=1/8(-25+sqrt385)# or #x=1/8(-25-sqrt385)# #(x~~-0.672, x~~-5.58#)
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Answer 2

To solve the equation ( \frac{5}{x+1} - \frac{1}{2} = \frac{2}{3}x + 3 ), follow these steps:

  1. Multiply every term by the least common denominator, which is (6(x+1)).
  2. Simplify the resulting equation.
  3. Solve for (x).
  4. Check for extraneous solutions.

After solving, you'll find (x = -\frac{5}{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.

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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