How do you solve #(r+4)/3=r/5# and find any extraneous solutions?
To ' eliminate' the fractions on both sides of the equation, multiply by the lowest common multiple (LCM ) of 3 and 5 which is 15
subtract 3r from both sides.
subtract 20 from both sides.
To solve for r, divide both sides by 2
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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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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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