SOLUTION: If 4A = I and 8A = U, show algebraically the number of I's needed to equal one U.

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Question 1044855: If 4A = I and 8A = U, show algebraically the number of I's
needed to equal one U.
Found 3 solutions by josgarithmetic, Edwin McCravy, MathTherapy:
Answer by josgarithmetic(39620) About Me  (Show Source):
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
If 4A = I and 8A = U, show algebraically the number of I's
needed to equal one U.
Let n = the number of I's needed to equal one U.

Then

nI = U

Substitute 4A for I and substitute 8A for U

n(4A) = 8A

Solve for n:

Simplifying:

4nA = 8A

Divide both sides by 4A

4nA%2F%284A%29%22%22=%22%228A%2F%284A%29

n%22%22=%22%222

So we need 2 I's to equal one U.

Edwin

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

If 4A = I and 8A = U, show algebraically the number of I's
needed to equal one U.
8A =  U

2(4A) = U

2I = U ----- Substituting I for 4A

highlight_green%28matrix%281%2C5%2C+2%2C+Is%2C+for%2C+one%2C+U%29%29