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Ivan
4 years ago
5

Solve each system of equations.

Mathematics
2 answers:
sweet-ann [11.9K]4 years ago
6 0
1. y=-3
2. y=33
y=-1
( sorry if number 2 is wrong)
Leto [7]4 years ago
5 0
1. 2(4) - 4y = 20

8 -4y = 20
8-8 -4y = 20 - 8
-4/-4y = 12/-4
y = -3
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Which expression is equivalent?
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Answer:

Third choice from the top is the one you want

Step-by-step explanation:

This whole concept relies on the fact that if the index of a radical exactly matches the power under the radical, both the radical and the power cancel each other out.  For example:

\sqrt[6]{x^6} =x and another example:

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Let's take this step by step.  First we will rewrite both the numerator and the denominator in rational exponential equivalencies:

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In order to do anything with this, we need to make the index (ie. the denominators of each of those rational exponents) the same number.  The LCM of 3 and 4 is 12.  So we rewrite as

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Now we will put it back into radical form so we can rationalize the denominator:

\frac{\sqrt[12]{6^3} }{\sqrt[12]{2^4} }

In order to rationalize the denominator, we need the power on the 2 to be a 12.  Right now it's a 4, so we are "missing" 8.  The rule for multiplying like bases is that you add the exponents.  Therefore,

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We will rationalize by multiplying in a unit multiplier equal to 1 in the form of

\frac{\sqrt[12]{2^8} }{\sqrt[12]{2^8} }

That looks like this:

\frac{\sqrt[12]{6^3} }{\sqrt[12]{2^4} }*\frac{\sqrt[12]{2^8} }{\sqrt[12]{2^8} }

This simplifies down to

\frac{\sqrt[12]{216*256} }{\sqrt[12]{2^{12}} }

Since the index and the power on the 2 are both 12, they cancel each other out leaving us with just a 2!  Doing the multiplication of those 2 numbers in the numerator gives us, as a final answer:

\frac{\sqrt[12]{55296} }{2}

Phew!!!

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