SOLUTION: The perimeter of a rectangle is 64 meters. the sum of the length and width is 32 meters. Find the length and width of the rectangle. Tried 2x+2y= 64 x+y=32...now what? Thanks so

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Take note that the same 'x's and 'y's are being used here. So we can use this fact to eliminate one variable so we can solve for the other. There are many ways to do this, but I'll use substitution.

x%2By=32

Start with the second equation

y=32-x

Subtract 'x' from both sides.

2x%2B2y=64

Move onto the first equation.

2x%2B2highlight%28%2832-x%29%29=64

Plug in

. In other words, replace each 'y' with 32-x

(this is possible since the two are equal).

So we now have the equation 2x%2B2%2832-x%29=64

which now consists of one variable. Simply solve this equation to find 'x'. I'll let you do that (let me know if you still need help). Once you find 'x', plug that value into y=32-x

to find 'y'.

You can put this solution on YOUR website!
The perimeter of a rectangle is 64 meters. The sum of the length and width is 32 meters. Find the length and width of the rectangle.

Step 1. Perimeter P means adding up all 4 sides of a rectangle.

Step 2. Let w be the width.

Step 3. Let 32-w be the length.

Step 4. Then, P=w+w+w-32+w-32=4w-64=64

Solved by pluggable solver: EXPLAIN simplification of an expression

Your Result:
Simplify 4w-64=64
Your Answer:

Solutions: w=32.
Graphical form: Equation was fully solved.
Text form: 4w-64=64 simplifies to 0=0
Cartoon (animation) form:
simplify_cartoon( 4w-64=64 )

Detailed explanation:

Look at

.
Moved these terms to the left highlight_green%28+-64+%29

It becomes 4%2Aw-64-highlight_green%28+64+%29=0

Look at

4%2Aw-highlight_red%28+64+%29-highlight_red%28+64+%29=0

.
Added fractions or integers together
It becomes 4%2Aw%2Bhighlight_green%28+-128+%29=0

Look at

.
Removed extra sign in front of

It becomes 4%2Aw-highlight_green%28+128+%29=0

Look at

.
Solved linear equation highlight_red%28+4%2Aw-128=0+%29

equivalent to 4*w-128 =0
It becomes highlight_green%28+0+%29=0

.
Result: 0=0

Solutions: w=32.

Done!

With

and

. And

which is a true statement. But the answer is not realistic.

Step 5. ANSWER: The width is 32 meters and the length is 0 meters to satisfy the problem statement. So you were on the right track but note your equations are also identical if you multiply one of them by 2.

I hope the above steps were helpful.

For FREE Step-By-Step videos in Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.

Good luck in your studies!

Respectfully,
Dr J

You can put this solution on YOUR website!
The perimeter of a rectangle is 64 meters. the sum of the length and width is 32 meters. Find the length and width of the rectangle.
Tried 2x+2y= 64
x+y=32...now what? Thanks so much!

Let width of rectangle be W

Let length of rectangle be L

Since its perimeter = 64, then we can say that 2W + 2L = 64

Since it's also given that the sum of its width and length is 32, then we can say that W + L = 32

We now have 2 equations, as follows:
W + L = 32 -------- (i)
2W + 2L = 64 ------ (ii)
2W + 2L = 64 -------- (iii) ------ Multiplying eq (i) by 2

We can clearly see that the 2 equations are identical, which means that there are an infinite number of REAL numbers that can represent the length or the width of this rectangle. However, whatever length and width are given to the rectangle, they both have to be GREATER than 0, and their sum has to add to 32. In other words, if width = 10, then length = 22, or vice-versa. Likewise, if width were to be 1, then length = 31, and so on.

Therefore, we can say that the width of this rectangle = highlight_green%280+%3C+W+%3C+32%29

, and that:

the length of this rectangle = highlight_green%28L+=+32+-+W%29

, with W, or width being greater than 0.