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3 equations, 3 unknowns, how do i solve ?

by bjr_jyd15
Tags: equations, solve, unknowns
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bjr_jyd15
#1
Jan27-06, 07:37 PM
P: 75
hi everyone, i have 3 equations and 3 unknowns:

x = (a/b) + c
y = (a/(b+1)) + c
z = (a/(b+7)) + c

where x, y, z are given in the problem (real #'s), and we need to find the values of a, b, and c.

can anyone tell me how to go about this problem? i tried rigourous algebra of solving and substituting but it became too difficult to isolate one of the variables a, b, or c.

thank you in advance!
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StatusX
#2
Jan27-06, 07:42 PM
HW Helper
P: 2,566
Multiply both sides by the denominator of the fraction. Then, define d=bc. You'll have three linear equations for a,b,d, and at the end you can get c from d and b.
bjr_jyd15
#3
Jan27-06, 07:47 PM
P: 75
hmm i'm not sure i follow. so would i have xb = a + bc ? and then replace bc with d? how do i go from there then when i have xb = a + d?

thanks

StatusX
#4
Jan27-06, 07:54 PM
HW Helper
P: 2,566
3 equations, 3 unknowns, how do i solve ?

Right, do that for all three equations, and then you'll have linear equations (since x,y,z are constants). These should be easy to solve. If you know linear algebra, that helps, but if not, it's just straightforward eliminating variables.
bjr_jyd15
#5
Jan27-06, 08:04 PM
P: 75
okay so i have

bx = a + d
by + y = a + d + c
bz + 7z = a + d + 7c

where x,y,z are #'s and d=bc...
if i want to put this in a matrix, is it possible? because i see the b's on the left side of the equation that are unknown...

thank you for your patience.
StatusX
#6
Jan27-06, 08:14 PM
HW Helper
P: 2,566
Sorry, I kinda rushed through this before and made a mistake. You probably didn't need to introduce d. But it doesn't really matter, and it'll still work if you use the first equation to plug in bx for a+d in the other two equations. This will eliminate a in both of them, and that will leave you two linear equations for b and c. Just group all the b's and c's on one side.
bjr_jyd15
#7
Jan27-06, 08:41 PM
P: 75
so now i have two linear equations:

by - bx = c -y
bz - bx = 7c - 7z

which i'm not sure how set up from there as a linear eqn...should i factor a b out of the left side of the first equation then divide? please let me know!

thanks
jtbell
#8
Jan27-06, 08:58 PM
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P: 11,630
You now have two equations in two unknowns (b and c). You've done that sort of thing before, right?

For example, you could solve one equation for c and substitute it for c in the other one. Then you have a single equation for b only. Solve it. Then substitute your solution for b back into one of those two equations. Either one will work. Solve for c. Finally, take your solutions for b and c and substitute them back into one of your original equations. Any one of them will work. Solve for a.
VietDao29
#9
Jan27-06, 09:05 PM
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VietDao29's Avatar
P: 1,421
Quote Quote by bjr_jyd15
so now i have two linear equations:

by - bx = c -y
bz - bx = 7c - 7z

which i'm not sure how set up from there as a linear eqn...should i factor a b out of the left side of the first equation then divide? please let me know!

thanks
You know, x, y, z. Rearrange a bit gives:
[tex]\left\{ \begin{array}{l} (x - y)b + c = y \\ (x - z)b + 7c = 7z \end{array} \right.[/tex]
Now that you have your matrix. You can go from here, right?
bjr_jyd15
#10
Jan27-06, 10:23 PM
P: 75
thank you all for your help--it's appreciated!!!
abszero
#11
Jan27-06, 10:56 PM
P: 49
Quote Quote by StatusX
Multiply both sides by the denominator of the fraction. Then, define d=bc. You'll have three linear equations for a,b,d, and at the end you can get c from d and b.
Be double careful that you don't multiply by zero and stumble across a solution that actually does not exist.


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