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

Answer 1

Given:

#a/(a-5)+2=(3a)/(a+5)#
Multiply through by #(a^2-25) = (a-5)(a+5)# to get:
#a(a+5)+2(a^2-25) = 3a(a-5) = 3a^2-15a#
Subtract #3a^2-15a# from both sides to get:
#0 = a(a+5)+2(a^2-25) - 3a^2+15a#
#=a^2+5a+2a^2-50-3a^2+15a#
#=(a^2+2a^2-3a^2)+(5a+15a)-50#
#=20a-50#
Add #50# to both sides to get
#20a=50#
Divide both sides by #50# to get:
#a=5/2#

Check by substituting back in the original equation:

LHS #= a/(a-5)+2 = (5/2)/(5/2-5)+2#
#=5/(5-10)+2 = 5/-5+2 = -1 + 2 = 1#
RHS #= (3a)/(a+5) = (3(5/2))/(5/2+5) = (3(5/2))/(3(5/2)) = 1#
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Answer 2

To solve the equation (a/(a-5)) + 2 = (3a)/(a+5), you can follow these steps:

  1. Start by multiplying both sides of the equation by (a-5)(a+5) to eliminate the denominators.
  2. Simplify the equation by distributing and combining like terms.
  3. Rearrange the equation to isolate the variable, a.
  4. Solve for a by continuing to simplify and solve the resulting equation.
  5. Check your solution by substituting the value of a back into the original equation to ensure it satisfies the equation.

The detailed steps are as follows:

  1. Multiply both sides of the equation by (a-5)(a+5): (a/(a-5)) * (a-5)(a+5) + 2(a-5)(a+5) = (3a/(a+5)) * (a-5)(a+5)

  2. Simplify the equation by distributing and combining like terms: a(a+5) + 2(a-5)(a+5) = 3a(a-5)

  3. Rearrange the equation to isolate the variable, a: a^2 + 5a + 2(a^2 - 25) = 3a^2 - 15a

  4. Solve for a by continuing to simplify and solve the resulting equation: a^2 + 5a + 2a^2 - 50 = 3a^2 - 15a 3a^2 - a^2 - 5a + 15a - 2a^2 + 50 = 0 0a^2 + 10a + 50 = 0 10a + 50 = 0 10a = -50 a = -5

  5. Check your solution by substituting the value of a back into the original equation: (-5/(-5-5)) + 2 = (3(-5)/(-5+5)) (-5/-10) + 2 = (-15/0)

    Simplifying further, we get: 1/2 + 2 = undefined

    Since the right side is undefined, the equation has no solution.

Therefore, the equation (a/(a-5)) + 2 = (3a)/(a+5) has no solution.

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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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