How do you solve #10/(x + 4) = 15/(4x + 4)# and find any extraneous solutions?

Answer 1

Solution: # x= 4/5#

#10/(x+4)=15/(4x+4)# Multiplying by #(x+4)(4x+4)# on
both sides we get #10(4x+4)=15(x+4) # or
#40x+40=15x+60 :. 25x =20 or x = 20/25= 4/5#
Check: L.H.S #=10/(x+4)=10/(4/5+4)= 10/(24/5)=50/24=25/12#
R.H.S #=15/(4x+4)=15/(4*4/5+4)= 15/(16/5+4)=15/(36/5)=75/36#
#25/12 :. # L.H.S #=# R.H.S . No invalid solution to original

issue, so a superfluous solution is not needed.

Solution: # x= 4/5# [Ans]
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Answer 2

To solve the equation 10/(x + 4) = 15/(4x + 4) and find any extraneous solutions, we can start by cross-multiplying to eliminate the fractions. This gives us 10(4x + 4) = 15(x + 4). Simplifying this equation, we get 40x + 40 = 15x + 60. By combining like terms, we have 40x - 15x = 60 - 40, which simplifies to 25x = 20. Dividing both sides by 25, we find x = 20/25, which reduces to x = 4/5. To check for extraneous solutions, we substitute x = 4/5 back into the original equation. We have 10/(4/5 + 4) = 15/(4(4/5) + 4). Simplifying this equation, we get 10/(24/5) = 15/(16/5). By multiplying both sides by the reciprocal of the denominator, we have 10 * (5/24) = 15 * (5/16). This simplifies to 50/24 = 75/16. However, these fractions are not equal, indicating that x = 4/5 is an extraneous solution. Therefore, there are no valid solutions to the equation.

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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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