Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: Translating prob into algebraic form
Replies: 3   Last Post: Jan 27, 2006 9:04 PM

 Messages: [ Previous | Next ]
 ticbol Posts: 116 Registered: 1/25/05
Re: Translating prob into algebraic form
Posted: Jan 27, 2006 5:19 AM

Protoman2050@gmail.com wrote:
> Could you translate this problem into algebraic form:
>
> The dimensions of a certain computer monitor screen are such that the
> legnth is 3 more than the width. If the lenght is increased by one,
> with all else remaining equal, the area will be 10 inches more than
> before. What are the dimensions?
>
> I got this far: x(x+3)=x(x+4).
> I don't even know if this is right. Could you help?

--------------------------
Sure, I can.
I love to blurt it out for you, one way to do it, so that you could
learn from it, or you could know the answers and see if you could get
them in another way.

In solving any word problem, if you know what to find or what the
problem wants you to find, then you "solved" it by 30 percent already.
Knowing what to find, you use all the formulas you know to relate those
what-to-find with the other data given in the problem. Make equation
per relation, jumble those, and when the smoke clears, you may have the

Here you are to find the original dimensions of the rectangular monitor
screen.
That's 30%, so we solve for the remaining 70% to get the answers.

Let w be the original width, in inches, and L be the original length,
in inches.
The problem mentioned "area" also, so relate that to the w and L,
Original area, A = w*L ------------(1).

The L is 3 inhes more than the w, so,
L = w+3 ---------------(2)

If the L is increased by 1 inch, and "all else remaining equal"
---?----, meaning "w" remains the same, then the resulting new area
will be 10 square inches more than the original A.
So,
A +10 = w*(L+1) -------------(3)

There, 3 equations, 3 unknowns. Solvable.

The "w" is present in all 3 equations, so let us solve for "w" first.
We "eliminate" the L and the A.
(L is present in all 3 equations also. That's another way. We use here
this way via the "w".)

Substitute the L from (2) into (1),
A = w*(w+3)
Sustitute that, and the L from (2), into (3),
w(w+3) +10 = w(w+3 +1)
w^2 +3w +10 = w^2 +4w
Bring them all into the lefthand side,
w^2 +3w +10 -w^2 -4w = 0
10 -w = 0
So, w = 10 inches -----------***
and L = w+3 = 13 inches ---------------***

Are those really correct?
Let us check.
" If the L is increased by 1 inch, and "all else remaining equal"
---?----, meaning "w" remains the same, then the resulting new area
will be 10 square inches more than the original A."
-------new length is 13+1 = 14 inches.
-------width remains the same, = 10 inches
------so new area = 10*14 = 140 sq.in.
-------original area, A = w*L = 10*13 = 130 sq.in.

Is the new area 10 sq.in more than before?
Yes.
Therefore, the original dimensions are 10-inch width by 13-inch