SOLUTION: 15-a/3=-9

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Question 216540: 15-a/3=-9
Found 3 solutions by ichudov, vleith, drj:
Answer by ichudov(507) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: EXPLAIN simplification of an expression
Your Result:


YOUR ANSWER


  • This is an equation! Solutions: a=72.
  • Graphical form: Equation 15-a%2F3=-9 was fully solved.
  • Text form: 15-a/3=-9 simplifies to 0=0
  • Cartoon (animation) form: simplify_cartoon%28+15-a%2F3=-9+%29
    For tutors: simplify_cartoon( 15-a/3=-9 )
  • If you have a website, here's a link to this solution.

DETAILED EXPLANATION

Look at highlight_red%28+15+%29-a%2F3=-9.
Moved 15 to the right of expression
It becomes -a%2F3%2Bhighlight_green%28+15+%29=-9.
Look at -a%2F3%2B15=highlight_red%28+-9+%29.
Moved these terms to the left highlight_green%28+9+%29
It becomes -a%2F3%2B15%2Bhighlight_green%28+9+%29=0.
Look at -a%2F3%2Bhighlight_red%28+15+%29%2Bhighlight_red%28+9+%29=0.
Added fractions or integers together
It becomes -a%2F3%2Bhighlight_green%28+24+%29=0.
Look at highlight_red%28+-a%2F3%2B24+%29=0.
Solved linear equation highlight_red%28+-a%2F3%2B24=0+%29 equivalent to -0.333333333333333*a+24 =0
It becomes highlight_green%28+0+%29=0.
Result: 0=0
This is an equation! Solutions: a=72.

Universal Simplifier and Solver


Done!

Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
is that %2815-a%29%2F3+=+9 or 15+-+%28a%2F3%29+=+9??
The way you stated it, it is 15+-+%28a%2F3%29+=+9 so we'll solve that one
15+-+%28a%2F3%29+=+9
%2845+-+a%29%2F3+=+9
45+-+a+=+27
45+-+27+=+a
18+=+a
If the problem was %2815-a%29%2F3+=+9
Then
%2815-a%29%2F3+=+9
%2815-a%29+=+27
15-27+=+a
-12+=+a

Answer by drj(1380) About Me  (Show Source):
You can put this solution on YOUR website!
15-a%2F3=-9

Step 1. Multiply three to both sides of the equation

3%2A15-3%2Aa%2F3=-9%2A3

45-a=-27

Step 2. Add a+27 to both sides

45-a%2Ba%2B27=-27%2Ba%2B27

a=72

Check...15-72%2F3=-9...which is a true statement.

Step 3. ANSWER is a=72

I hope the above steps were helpful.

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Good luck in your studies!

Respectfully,
Dr J