How do you solve #(r+4)/3=r/5# and find any extraneous solutions?

Answer 1

#r=-10#

To ' eliminate' the fractions on both sides of the equation, multiply by the lowest common multiple (LCM ) of 3 and 5 which is 15

#cancel(15)^5xx(r+4)/cancel(3)^1=cancel(15)^3xxr/cancel(5)^1#
#rArr5(r+4)=3rlarr" no fractions"#
#rArr5r+20=3r#

subtract 3r from both sides.

#5r-3r+20=cancel(3r)cancel(-3r)#
#rArr2r+20=0#

subtract 20 from both sides.

#2rcancel(+20)cancel(-20)=0-20#
#rArr2r=-20#

To solve for r, divide both sides by 2

#(cancel(2) r)/cancel(2)=(-20)/2#
#rArrr=-10" is the only solution"#
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Answer 2

To solve the equation (r+4)/3=r/5 and find any extraneous solutions, first cross multiply to eliminate the fractions, then solve for r.

(r + 4) * 5 = r * 3 5r + 20 = 3r

Move all terms involving r to one side of the equation:

5r - 3r = -20 2r = -20

Divide both sides by 2:

r = -10

Now, check for extraneous solutions by substituting the found value back into the original equation:

(((-10) + 4) / 3) = (-10) / 5

(-6 / 3) = -2

-2 ≠ -2

Since the equation does not hold true for r = -10, it is an extraneous solution. Therefore, there are no real solutions for 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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