Question 983972
<pre>
Instead of doing your problem for you, I will do one exactly in every detail
like yours and you can use it as a model to do yours by.  I will do this
one instead:
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Rectangle of perimeter 38 cm. if its length decreases 2 cm & its width increases
5 cm it will be square. find area of square.



>>Rectangle of perimeter 38 cm<<
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P = 2L + 2W

So we substitute 38 for P

38 = 2L + 2W   <-- that's one equation
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>>if its length decreases 2 cm<<
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Then its new length will be L-2
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>>& its width increases 5 cm<<  
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Then its new width will be W+5
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>>it will be square<<
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Then its length and width will be equal, so

{{{(matrix(2,1,NEW,LENGTH))}}}{{{""=""}}}{{{(matrix(2,1,NEW,WIDTH))}}}

L-2 = W+5      <-- That's another equation.

So we have this system of two equations and 2 unknowns to solve 
by substitution:

{{{system(38 = 2L + 2W, L-2=W+5)}}}

We can simplify the first equation,

38 = 2L + 2W

by dividing it through by 2, getting,

19 = L + W

We can simplify the second equation,

L - 2 = W + 5

by adding 2 to both sides:

    L = W + 7

Now the system is

{{{system(19 = L + W, L=W+7)}}}

Since the second equation is already solved for L, we
substitute (W + 7) for L in the first equation, and get:

19 = (W + 7) + W

Remove the parentheses:

19 = W + 7 + W

Combine like terms on the right

19 = 2W + 7

Subtract 7 from both sides

12 = 2W

Divide both sides by 2

 6 = W

Substitute 6 for W in

    L = W + 7
    L = 6 + 7
    L = 13

So the original length was 13 cm and the original width 6 cm
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>>find area of square<<
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So the new length will be L-2 or 13-2 = 11 cm
And its new width will be W+5 or 6+5 = 11 cm
So a side of the square will be 11 cm

So its area will be 

A = (side)<sup>2</sup> = 11<sup>2</sup> = 121 square centimeters.

Now do yours which is done exactly the same way, step by step.

Edwin</pre>