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ss7ja [257]
3 years ago
11

Solve xy -5= k for x

Mathematics
2 answers:
Virty [35]3 years ago
5 0

Hey there!

To solve, let's first add 5 to both sides.

xy=k+5

Now we divide both sides by y.

x=\frac{k+5}{y}

I hope this helps!

ad-work [718]3 years ago
5 0

Hey there! :)

xy - 5 = k

We need to isolate x, so we can start out by adding 5 to both sides.

xy - 5 + 5 = k + 5

Simplify.

xy = k + 5

Now, we can simply divide both sides by y.

xy ÷ y = k + 5 ÷ y

Simplify.

x = \frac{k + 5}{y}

~hope I helped~

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  First "order of operations" mistake: step 2

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Step-by-step explanation:

As we understand Rena's work, she wants to simplify ...

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for x = -1 and y = 2.

Her work seems to be ...

<u>Step 1</u>

  \text{Substitute $x=-1$ and $y=2$ into the expression}\\\\\left(\dfrac{(-1)^{-3}2^{-2}}{2(-1)^42^{-4}}\right)^{-3}\qquad\text{no error}

<u>Step 2</u>

  \text{Simplify the parentheses}\\\\\left(\dfrac{2^4}{2(-1)^4(-1)^32^2}\right)^{-3}=\left(\dfrac{2^2}{2(-1)^7}\right)^{-3}\qquad\text{order of operations error}

<u>Step 3</u>

  \text{Evaluate the power to a power}\\\\\dfrac{2^{-6}}{2^{-3}(-1)^{21}}\qquad\text{no error}

<u>Step 4</u>

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_____

So, the first arithmetic error is in Step 4. However, the order of operations requires exponents be evaluated first. Doing that makes step 2 look like ...

  \left(\dfrac{-\dfrac{1}{4}}{2(1)\dfrac{1}{16}}\right)^{-3}=(-2)^{-3}\qquad\text{proper Step 2}

__

We expect your answer is supposed to be Step 4.

7 0
3 years ago
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