gotta math problem, can't quite remember how to solve it anymore :-/?

Question

I am assuming the unknown equation to be of the form:
Y = (X - a) * (Q - b) - c

where:
X & Q are independent variables
a, b, & c are constants

Given:
values of X, Y, Q are known in the following 3 instances:
|X|Y|Q|
|508|22,623|48.48|
|515|22,086|46.76|
|1,570|72,871|47.95|

I'm trying to find the values of: a, b, c

any ideas anyone?

Answer 1

by putting the values in the initial equation u'll have 3 equations which will find your 3 unknowns.
Here is how to do:
your equation is:
Y = (X - a) * (Q - b) - c
now substitute the values of x,y&q into your initial equation and get 3 new equations:
1_) 22,623 = (508 - a) * (48.48 - b) - c
2_) 22,086 = (515 - a) * (46.76 - b) - c
3_) 72,871 = (1570 - a) * (47.95 - b) - c

So now you have 3 equations which will give you the answers of your 3 unknowns. you can use a computer program like MATLAB to solve it but if you gonna solve it manually here is how:
First you need to expand the equations.Expand means make it as basic as possible. like no parenthesis or stuff:
1_) 22623 = 24624.84 - 508b - 48.48a + ab - c
2_) 22086 = 24081.4 - 515b - 46.76a + ab - c
3_) 72871 = 75281.5 - 1570b - 47.95a + ab - c
now multiply the fisrt equation by (-1)
{Extra explanation...why we do that? cause in both equations, we have ab and -c. so if we multiply one equation by -1 and then add it to the other, ab and c would be eliminated and thus our job becomes easier}
{22623 = 24624.84 - 508b - 48.48a + ab - c } * (-1) =
-22623 = -24624.84 + 508b + 48.48a - ab + c
now add it to the other two equations
step 1:
-22623 = -24624.84 + 508b + 48.48a - ab + c
+ 22086 = 24081.4 - 515b - 46.76a + ab - c
----------------------------------------------------------------------
-537 = -543.44 - 7b + 1.72a Eq. (I)
{ As you see ab and c are gone}
step 2:
-22623 = -24624.84 + 508b + 48.48a - ab + c
+ 72871 = 75281.5 - 1570b - 47.95a + ab - c
-----------------------------------------------------------------------
50248 = 50656.66 -1062b + 0.53a Eq.(II)

This way you've reduced the number of your equations to 2:
-537 = -543.44 - 7b + 1.72a Eq. (I)
50248 = 50656.66 -1062b + 0.53a Eq.(II)

now you should multiply the Eq (I) by -0.53/1.72 which is
-0.308.
{Extra explanation: why we multiply by -0.53/1.72? cause we want to eliminate another variable and make our job easier. Multiplying the above fraction into the Eq(1) will change the number at the side of variable "a" from 1.72 into -0.53 and then by addind the two equations together, variable "a" would be eliminated too}
-0.308 * {-537 = -543.44 - 7b + 1.72a}
= 165.396 = 167.380 + 2.156b - 0.53a
now add this equation to Eq(II)
youll have:
165.396 = 167.380 + 2.156b - 0.53a
+ 50248 = 50656.66 -1062b + 0.53a
----------------------------------------------------------
50413.396 = 50824.04 -1059.844b
{As you see "a" is gone and we have an equation only containing "b"}

from this quation, b= 0.387
substitute this value in the Eq(I) and you'll have an equation just containing variabla "a".
-537 = -543.44 - 7(0.387) + 1.72a
=> a = 5.319
Now substitute these two values in the equation 1_)
22623 = 24624.84 - 508(0.378) - 48.48(5.319) + (0.378)(5.319) - c
This is an equation only containing the variable "c" and yields c = 1553.961
So your answers are:
a = 5.319 , b= 0.387 , c = 1553.961

Answer 2

Y = (X-a)(Q-b) - c
= XQ - Qa - Xb + ab - c
So we have
(Q)a + (X)b + (c-ab) = XQ - Y.
If we think of our three variables as a, b and (c-ab) then each data point gives us a linear equation. So we get
48.48 a + 508 b + (c-ab) = 2004.84
46.76 a + 515 b + (c-ab) = 1995.4
47.95 a + 1570 b + (c-ab) = 2410.5

You can express this as a matrix and solve; equivalently you can eliminate variables one at a time as below. (This is really all we're doing with matrices anyway.)

By subtracting the second equation from the other two we get
1.72 a - 7 b = 9.44
1.19 a + 1055 b = 415.1

By eliminating a (1.72 times the second equation minus 1.19 times the first equation) we get
1822.93 b = 702.7384
and hence b = 0.3855 to 4 d.p.

Then 1.72 a - 7 b = 9.44 gives us a = 12.1385 / 1.72 = 7.0573 to 4 d.p.
and from 48.48 a + 508 b + (c-ab) = 2004.84
we get c-ab = 1466.87011 and hence c = 1469.59 to 6 s.f.

So a = 7.057, b = 0.3855 and c = 1470 all to 4 s.f.

Answer 3

Hey, I think it'll be pretty simple.
If you take the first equation and expand it you get something like: Y = XQ -Xb -Qa +ab -c

Just substitute the 3 combinations of X,Y and Q, that'll give you three more equations which have only a,b and c. Take two of these and eliminate c from the view.

Like,for example. you have something of the sort 3a + 4b -c = 8 and 7a - 3b +c =1
Adding these two, you'll get 10a -b =9 ----> This was an example

similarly take the next two a,b,c combos and you'll get another a,b equation. And with these two a,b equations you can easily get the values for a and b. Then it's just the matter of substituting all the values to get c.

Hope that made some sense and you can work it out,all the best!!

Answer 4

from the values of X, Y and Q you can set three equations
22,623 = (508-a)(48.48-b) - c ...........(1)
22,086 = (515-a)(46.76-b) - c ...........(2)
72,871 = (1570-a)(47.95-b) - c...........(3)
exchanging the positions equation (1) becomes c = (508-a)(48.48-b) - 22,623
substituting c in equation (2)
22,086 = (515-a)(46.76-b) - ((508-a)(48.48-b) - 22,623)..... (4)
exchanging the positions equation (4)
b = continue like this

Answer 5

Y = (X - a) * (Q - b) - c
Expanding,
XQ - bX - aQ + ab - c = Y
(508)(48.48) - 508b - 48.48a + ab - c = 22,632
(515)(46.76) - 515b - 46.76a + ab - c = 22,086
(1570)(47.95) - 1570b - 47.95a + ab - c = 72,871
24,627.84 - 508b - 48.48a + ab - c = 22,623
24,081.4 - 515b - 46.76a + ab - c = 22,086
75,281.5 - 1570b - 47.95a + ab - c = 72,871
Eliminating (ab - c)
7b - 1.72a = - 9.44
1055b + 1.19a = 415.1
- 1822.93a = - 12864.9
a = 7.057265
7b - 1.72(7.057265) = - 9.44
b = 0.3854994
24,627.84 - 508(0.3854994) - 48.48(7.057265) + 2.720571 - c = 22,623
c = 1469.591

Answer 6

Substitute the values of x , t and q

take two equations, add and subtract them by solving simultaneously and then express one constant in terms of the other two in the third equation and any one equation and solve simultaneously again. Just remember that you have to first eliminate one of the constants.