stuck on Algebra question for TEAS!

Howdy,
I have been studying the manual and am done with the algebra portion. I missed 5 (4 because I read the question wrong). This one question, I am totally stumped how to get the answer.
In my manual, it is on page 87, question number 43.
There is an X/Y graph to the right with X marked at 1.5 and Y marked at 3.
The problem states;
Using the graph, determine the value of Y when X = 4.
The answer is 5.
I have looked through my basic and intermediate algebra books along with info in the TEAS study guide as well and just can't figure out how they got 5 for an answer!
I have been going bonkers over this for 2 days. Could someone please help me out on this????????
Thanks!
Diana 

May 16, '07DianaJH,
Does the graph on the page have a line or just a point? From the description you posted we only know one point on the graph, without having a slope or yintercept the line could be going in any direction, so without more information I'm not aware of how this problem could be solved definitively.
However, if the graph shows a line you can calculate the yintercept or the slope and then using basic algebra with y=mx+b you should be able to find y for any given x.
EXAMPLE:
3 = m(1.5) + b & 5 = m(4) + b
Calculate Slope: rise / run = (5  (3)) / (4  1.5) = 8 / 2.5 = 3.2 = m
Find yintercept: y  y1 = m(x  x1)
y  (3) = 3.2(x  1.5)
y + 3 = 3.2x  4.8
y = 3.2x  7.8
Check answer:
5 = 3.2(4)  7.8
5 = 12.8  7.8
5 = 5 // SUCCESS!
Hope that helps. Best of luck.Last edit by aHolisticStudent on May 16, '07 
May 16, '07Hi, thanks for the reply!
It does have the line in the graph, but do you know what? I have forgotten this stuff! I aced both algebra classes and can't remember this for anything. I am not even understanding what you typed except that you did get the answer of 5! I remember the phrases of slope, rise, run and y intercept. I even remember the formula of y = mx + b, but this is looking totally foreign to me. I took the classes in 2004. At least now I can go back to the algebra book and review all of the terms and know where to go from there!
Thank you SO much. Now I have some learning to do! (again)
Diana 
May 16, '07if it makes you feel better i did not encounter any of these type of questions on the teas....

May 16, '07Quote from DianaJHHappy to helpHi, thanks for the reply!
Quote from DianaJHOK, let's see if I can clarify what I posted.It does have the line in the graph, but do you know what? I have forgotten this stuff! I aced both algebra classes and can't remember this for anything. I am not even understanding what you typed except that you did get the answer of 5!
The original information you posted gave us a point (x, y) of (1.5, 3) which I plugged into the equation y = mx + b to get:
3 = m(1.5) + b.
You also provided the other point on the line. X = 4 from the original question and Y = 5 from the answer in your book which I once again plugged into the equation to get:
5 = m(4) + b
At this point we know neither the slope of the line (m), nor the yintercept (b). So the first thing I did was determine the slope of the line (m) by calculating the rise of line over the run of the line defined as:
rise / run = (Change in Y) / (Change in X) = (Y2  Y1) / (X2  X1) =
(5  (3)) / (4  1.5) = 8 / 2.5 = 3.2
So if we plug in the newly calculated slope of the line (m) to our two equations earlier we get:
3 = 3.2(1.5) + b
5 = 3.2(4) + b
The only thing left to determine is the yintercept (b). So using the following formula: y  y1 = m(x  x1) with the first point provided from the problem (1.5, 3) we end up with:
y  (3) = 3.2(x  1.5)
Then we solve and reduce:
y + 3 = 3.2x  4.8
y = 3.2x  7.8
So to find the yintercept we set x = 0 which gives us:
y = 3.2(0)  7.8
which reduces to:
y = 7.8
So now we know:
y = 3.2x  7.8 will provide us with y for any given x value.
By rearranging the equation we can find x for any given y by using:
x = (y + 7.8) / 3.2
Given all the previous steps, by plugging in the supplied value where x = 4, we can solve for y.
y = 3.2(4)  7.8
y = 12.8  7.8
y = 5
And from your book, you know that the correct answer was indeed 5.
Hope that helps to bring it all back to you.
Best of luck on the exam and getting into your preferred nursing program. 
May 16, '07Cajun Moma, it DOES make me feel better knowing you didn't see this problem on the test....especially after I read the explanation from Aholisticstudent!
OHMYGOSH!
Aholisticstudent, Thank you SO much for taking the time to type all of that out for me. It could not have been fun! I DO see the actions now and putting it all together. I went back to my algebra book and the reason it was foreign to me was because that type of problem started in chapter 3 section 6 and guess what... we stopped at ch 3 section 5 and skipped the rest all the way to chapter 5! I printed off all of your message and followed it along and was amazed how it worked out. A big thank you from me!
Diana