- anonymous

Medal for this please
Part A: Solve A = 9 over 2 (x + 23) for x.
Part B: Determine the value of x when A = 108.
Part C: Solve -np - 90 > 30 for n. Show your work

- katieb

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- anonymous

- anonymous

any idea?

- anonymous

Part A: \[A = \frac{ 9 }{ 2(x+23) }\]

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## More answers

- anonymous

Okay so first lets distribute the 2 to x and 23. What would you get?

- anonymous

2x and 46

- anonymous

is this for part B?

- anonymous

No, this is for Part A. We have to solve for x. We haven't done that yet.

- anonymous

Anyways so yes you get 2x + 46

- anonymous

So now it looks like:
\[A = \frac{ 9 }{ 2x + 46 }\]

- anonymous

so do we add 46 and 9?

- anonymous

No don't add.

- anonymous

multiply

- anonymous

so 414?

- anonymous

then divide it by 2

- anonymous

Wait hold up. Don't multiply either. I'll tell you.

- anonymous

okay sorry haha

- anonymous

Okay so lets multiply 2x + 46 on both sides.
\[A(2x+46) = \frac{ 9 }{ 2x+46 }(2x+46)\]

- anonymous

This cancels out 2x + 46 on the right side. And we have to then distribute the A to 2x and 46.
\[2xA + 46A = 9\]

- anonymous

So far so good?

- anonymous

We have to find a common factor between 2xA and 46A. The common factor would be 2A since both coefficients are divisible by 2 and A. It would look like this now:
\[2A(x+23) = 9\]

- anonymous

We have to divide 2A on both sides.
\[\frac{ 2A(x+23) }{ 2A } = \frac{ 9 }{ 2A }\]

- anonymous

okay got it so far

- anonymous

This cancels out 2A on the left side. Now the equation looks like this:
\[x + 23 = \frac{ 9 }{ 2A }\]
Now we can isolate x by subtracting 23 from both sides.
\[x + 23 -23 = \frac{ 9 }{ 2A }-23\]
Now, x solved is
\[x = \frac{ 9 }{ 2A }-23\], which is the answer to PART A.

- anonymous

For Part B, just substitute 108 for A in
\[x = \frac{ 9 }{ 2A }-23\]
What would you get for the value of x?

- anonymous

its like doing part A but plugging in 108 instead of (x + 23)?

- anonymous

Yes basically

- anonymous

So what is the value of x for Part B?

- anonymous

okay i can do it from here thank you :)

- anonymous

Alright then. No problem:)

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