# Solving Equations.

• Sep 15th 2011, 05:12 PM
JibblyJabbly
Solving Equations.
I have two questions which I have attempted to do with no success, can you thoroughly explain what to do?

The length of a swimming pool is 2.5m greater than its width. The pool's perimeter is 29m. Find the lenghth of the pool.

So, for this one. I knew that the perimeter is 2l + 2w

so I ended up with this 29 =2(2.5l(w))

then I don't know what to do after that.

The second question is

$\frac{x}{5}+1\frac{1}{2}x-\frac{11}{20}=\frac{5x-1}{4}$

Do I turn the mixed fraction into a improper one first then move 5x to the left hand side? then move 11/20 to right hand side so it would be:

$\frac{x}{5}+\frac{3}{2}x+5x=-\frac{11}{20}+\frac{-1}{4}$

Am I right?
• Sep 15th 2011, 05:22 PM
skeeter
Re: Solving Equations.
Quote:

Originally Posted by JibblyJabbly
I have two questions which I have attempted to do with no success, can you thoroughly explain what to do?

The length of a swimming pool is 2.5m greater than its width. The pool's perimeter is 29m. Find the lenghth of the pool.

So, for this one. I knew that the perimeter is 2l + 2w

so I ended up with this 29 =2(2.5l(w))

then I don't know what to do after that.

correction ...

P = 2(L + W)

P = 2[(2.5+W) + W]

P = 2(2.5 + 2W)

P = 5 + 4W

The second question is

$\frac{x}{5}+1\frac{1}{2}x-\frac{11}{20}=\frac{5x-1}{4}$

Do I turn the mixed fraction into a improper one first then move 5x to the left hand side? then move 11/20 to right hand side so it would be:

$\frac{x}{5}+\frac{3}{2}x+5x=-\frac{11}{20}+\frac{-1}{4}$

$\frac{x}{5}+1\frac{1}{2}x-\frac{11}{20}=\frac{5x-1}{4}$

multiply every term by the common denominator to clear the fractions ...

$4x + 30x - 11 = 5(5x-1)$

easier now?
• Sep 16th 2011, 12:52 AM
JibblyJabbly
Re: Solving Equations.
Thank you very much, Skeeter! I almost understand it but:

P = 2(L + W)

P = 2[(2.5+W) + W]

P = 2(2.5 + 2W)

P = 5 + 4W

Where did you get the + W from?
• Sep 16th 2011, 03:02 AM
HallsofIvy
Re: Solving Equations.
It's not so much "+W" as it is "+ 2.5". The problem says "The length of a swimming pool is 2.5m greater than its width." If write W for the width then the length is 2.5 meters greater: W+ 2.5, assuming that width and length are measured in meters.
• Sep 16th 2011, 03:55 AM
skeeter
Re: Solving Equations.
Quote:

Originally Posted by jibblyjabbly
where did you get the + w from?

p = 2(l + w)

p = 2[(2.5 + w) + w]