## anonymous one year ago 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

• This Question is Open
1. anonymous

@triciaal

2. anonymous

any idea?

3. anonymous

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

4. anonymous

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

5. anonymous

2x and 46

6. anonymous

is this for part B?

7. anonymous

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

8. anonymous

Anyways so yes you get 2x + 46

9. anonymous

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

10. anonymous

so do we add 46 and 9?

11. anonymous

12. anonymous

multiply

13. anonymous

so 414?

14. anonymous

then divide it by 2

15. anonymous

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

16. anonymous

okay sorry haha

17. anonymous

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

18. anonymous

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

19. anonymous

So far so good?

20. 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: $2A(x+23) = 9$

21. anonymous

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

22. anonymous

okay got it so far

23. 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.

24. anonymous

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

25. anonymous

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

26. anonymous

Yes basically

27. anonymous

So what is the value of x for Part B?

28. anonymous

okay i can do it from here thank you :)

29. anonymous

Alright then. No problem:)