How do you solve #3/(x+6)=5/(x-2)# and find any extraneous solutions?

Answer 1

#x=-18#

#"multiply both sides by "(x+6)(x-2)#
#cancel((x+6))(x-2)xx3/cancel((x+6))=cancel((x-2))(x+6)xx5/cancel((x-2))#
#rArr3(x-2)=5(x+6)larrcolor(blue)"no fractions"#
#"distribute brackets on both sides"#
#3x-6=5x+30#
#"subtract 5x from both sides"#
#3x-5x-6=cancel(5x)cancel(-5x)+30#
#rArr-2x-6=30#
#"add 6 to both sides"#
#-2xcancel(-6)cancel(+6)=30+6#
#rArr-2x=36#
#"divide both sides by "-2#
#(cancel(-2) x)/cancel(-2)=36/(-2)#
#rArrx=-18#
#color(blue)"As a check"#

This value is the solution if you substitute it into the equation and both sides come out equal.

#"left "=3/(-12)=-1/4#
#"right "=5/(-20)=-1/4#
#rArrx=-18" is the only solution"#
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Answer 2

To solve the equation 3/(x+6)=5/(x-2) and find any extraneous solutions, we can start by cross-multiplying to eliminate the fractions. This gives us 3(x-2) = 5(x+6). Expanding and simplifying, we get 3x - 6 = 5x + 30. Rearranging the equation, we have 2x = -36. Dividing both sides by 2, we find x = -18.

To check for extraneous solutions, we substitute x = -18 back into the original equation. We have 3/(-18+6) = 5/(-18-2), which simplifies to 3/(-12) = 5/(-20). Evaluating further, we get -1/4 = -1/4. Since both sides are equal, there are no extraneous solutions in this case. Therefore, the solution to the equation is x = -18.

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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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