SOLUTION: The perimeter of a rectangle is 900 yards. What are the dimensions of the rectangle if the length is 30 yards more than the width?

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Question 223646: The perimeter of a rectangle is 900 yards. What are the dimensions of the rectangle if the length is 30 yards more than the width?
Answer by drj(1380) About Me  (Show Source):
You can put this solution on YOUR website!
The perimeter of a rectangle is 900 yards. What are the dimensions of the rectangle if the length is 30 yards more than the width?

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

Step 2. Let w be the width.

Step 3. Let w+30 be the length.

Step 4. Then, P=w+w+w+30+w+30=4w+60=900

Solved by pluggable solver: EXPLAIN simplification of an expression
Your Result:


YOUR ANSWER


  • This is an equation! Solutions: w=210.
  • Graphical form: Equation 4w%2B60=900 was fully solved.
  • Text form: 4w+60=900 simplifies to 0=0
  • Cartoon (animation) form: simplify_cartoon%28+4w%2B60=900+%29
    For tutors: simplify_cartoon( 4w+60=900 )
  • If you have a website, here's a link to this solution.

DETAILED EXPLANATION

Look at 4%2Aw%2B60=highlight_red%28+900+%29.
Moved these terms to the left highlight_green%28+-900+%29
It becomes 4%2Aw%2B60-highlight_green%28+900+%29=0.
Look at 4%2Aw%2Bhighlight_red%28+60+%29-highlight_red%28+900+%29=0.
Added fractions or integers together
It becomes 4%2Aw%2Bhighlight_green%28+-840+%29=0.
Look at 4%2Aw%2Bhighlight_red%28+-840+%29=0.
Removed extra sign in front of -840
It becomes 4%2Aw-highlight_green%28+840+%29=0.
Look at highlight_red%28+4%2Aw-840+%29=0.
Solved linear equation highlight_red%28+4%2Aw-840=0+%29 equivalent to 4*w-840 =0
It becomes highlight_green%28+0+%29=0.
Result: 0=0
This is an equation! Solutions: w=210.

Universal Simplifier and Solver


Done!



With w=210 and w%2B30=240. And P=2%28210%2B240%29=900 which is a true statement

Step 5. ANSWER: The width is 210 yards and the length is 240 yards.

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