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### Solving Simultaneous Equations and Factoring

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Date: 3/18/96 at 21:52:36
From: Mr. Michael J. Fitzpatrick
Subject: Math Problem

How would you use factoring to solve these problems?  It's 9th

Vanessa built a rectangular pen for her dogs.  She used an outside
wall of the garage for one of the sides of the pen.  She had to
buy 20 m of fencing in order to build the other sides of the pen.
Find the dimensions of the pen if its area is 48 m^2.

The Parkhursts used 160 yd of fencing to enclose a rectangular
corral and to divide it into two parts by a fence parallel to one
of the shorter sides.  Find the dimensions of the corral if its
area is 1000 yd^2.

Each edge of one cube is 2 cm longer than each edge of another
cube.  The volumes of the cubes differ by 98cm^3.  Find the
lengths of the edges of each cube.

Thank you.
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Date: 3/19/96 at 11:41:52
From: Doctor Elise
Subject: Re: Math Problem

Hi!

>Vanessa built a rectangular pen for her dogs...

To start out with, we're going to need 2 variables, one for the
length of the pen, and one for the width. Let's call the length L,
and say that it's the length of the garage wall and the side
opposite it. Then let's call the width W (real original, huh?) for
the other two walls.  Okay, so now we have a rectangle, with sides
L and W. We know that the total perimeter is the length of fence
bought, 20m, plus the length of the garage wall, L. And we know
how to calculate the perimeter of a rectangle, so our first
equation is:

L + W + L + W = 20 + L
2L + 2W = 20 + L
L + 2W = 20
L = 20 - 2W

Okay? Now we use the other bit of information, that the area is
48m^2. Well, the area of a rectangle is length times width, right?
So now we have

L*W = 48 And we can substitute for L from the first equation...

(20 - 2W)* W = 48

20W - 2W^2 = 48 I hate having a negative squared term,
(it's harder to factor) so I'm going to multiply everything by -1:

-20W + 2W^2 = -48

Now let's get everything together on one side of the
= sign so that we can start factoring:

2W^2 - 20W + 48 = 0 Now let's divide by 2

W^2 - 10W + 24 = 0      Can you factor that? Let's see.
If I factor 24, I get either 3 * 8 or 4 * 6, and I have to be able
to add or subtract factors to get 10. So it has to be 6 and
4. And they both have to be the same sign to get positive 24,
but the 10 is negative so they have to be negative.

(W - 6)(W - 4) = 0      Looks like this is our answer. The width
is either 6 or 4 meters. Now let's solve for the length and
make sure both of these numbers work in the real world:

L = 20 - 2W
L = 20 - 12 = 8 so if the width is 6, the length is 8

or

L = 20 - 8 = 12 or if the width is 4 then the length is 12. Both
of these combinations give us 20 meters of fencing and an area of
48 square meters.

Let's look at the Parkhursts.

>The Parkhursts used 160 yd of fencing to enclose a rectangular
>corral and to divide it into two parts by a fence parallel to one of
>the shorter sides. Find the dimensions of the corral if its area is
>1000 yd^2.

Let's use our old friends L and W again, and let's have W be the
short side. If you draw the corral, you'll see that instead of the
perimeter, we have the perimeter plus one extra short side's worth
of fencing adding up to 160.

L + W + L + W + W = 160

2L + 3W = 260
L = 1/2(160 - 3W)

And we know the total area, our old friend L * W, is 1000.

L * W = 1000 And we substitute for L again,

1/2(160 - 320 W) * W = 1000 Let's get rid of the 1/2, it won't
help us factor!

(160 - 3W) * W = 2000

3W^2 - 160W + 2000 = 0

Okay, we need to factor this mess. Both factors have
to be subtracted, because the 2000 is positive, (so
they're the same - plus times plus = plus, or neg times
neg = plus) and the 260 is negative, so they're both
negative. And we have a 3 thrown in:

(3W - ?)(W - ??)

Okay, 160. 3 times something, plus something else,
Does 40 * 40 = 2000? Nope, let's try again.

How about 60 and 100? Yes! 60/3 = 20, 20 * 100 is
2000!

(3W - 100)(W - 20) = 0 So our fence has a width of either 20
or 100/3.  If we plug W back into our first equation, we
get L = 50 if W = 20, or L = 30 if W = 100/3.

Phew! I had to do that one twice because I wrote the numbers
wrong!

Finally, Cubes.

>Each edge of one cube is 2 cm longer than each edge of another
>cube. The volumes of the cubes differ by 98cm^3. Find the lengths
>of the edges of each cube.

Okay, we have one cube that is L centimeters long on a side, and
one that is L + 2 cm on a side. The volume of the first one is L^3,
and the volume of the second one is 98 + L^3, right?

so (L + 2)^3 = 98 + L^3

(L^2 + 4L + 4)(L + 2) = 98 + L^3

L^3 + 4L^2 + 4L + 2L^2 + 8L + 8 = 98 + L^3
And the L^3 terms cancel out!

6L^2 + 12L + 8 = 98

6L^2 + 12L - 90 = 0

L^2 + 2L - 15 = 0

(L + ?)(L - ??) = 0 15 = 3 * 5, 5 > 3,

(L + 5)(L - 3) = 0 The possible values for L are L = -5 or L = 3.

In the real world, we aren't going to have a cube with a side of
length -5, so we throw that solution out as mathematically
possible, but not physically possible. So L = 3.

One cube is 3 cm on a side, and one cube is L + 2 or 5 cm on a
side.

I hope this helps! Word problems can be tricky to set up
mathematical equations for, but I think they're the real life
problems math was invented to solve in the first place!

-Doctor Elise,  The Math Forum

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Associated Topics:
High School Basic Algebra

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