3 equations, 3 unknowns, how do i solve ?

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In summary, the author is looking for help solving three equations. They start by multiplying both sides of the equations by the denominator of the fraction. They then define d=bc and solve for a, b, and c. They are then able to solve for c using one of the original equations.
  • #1
bjr_jyd15
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3 equations, 3 unknowns, how do i solve!?

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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  • #2
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.
 
  • #3
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
 
  • #4
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.
 
  • #5
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.
 
  • #6
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.
 
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  • #7
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
 
  • #8
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.
 
  • #9
bjr_jyd15 said:
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? :smile:
 
  • #10
thank you all for your help--it's appreciated! :!)
 
  • #11
StatusX said:
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.
 

1. How does the Gauss-Jordan elimination method work?

The Gauss-Jordan elimination method is a systematic process for solving a system of linear equations with three unknowns. It involves using a series of operations, such as adding or subtracting equations, to eliminate one of the variables in each equation until only one variable remains. Then, by substituting the value of this variable into the other equations, the values of the remaining variables can be determined.

2. Can the substitution method be used to solve a system of three equations with three unknowns?

Yes, the substitution method can be used to solve a system of three equations with three unknowns. This method involves solving one of the equations for one of the variables and then substituting the resulting expression into the other equations. By repeating this process for each variable, the values of all three variables can be found.

3. What is the difference between consistent and inconsistent systems of equations?

A consistent system of equations has at least one solution, while an inconsistent system has no solutions. In other words, a consistent system has a set of values for the variables that satisfy all of the equations, while an inconsistent system has no such set of values.

4. Are there any special cases when solving a system of three equations with three unknowns?

Yes, there are a few special cases that can occur when solving a system of three equations with three unknowns. One special case is when one of the equations is a multiple of another equation, resulting in an infinite number of solutions. Another special case is when two of the equations are equivalent, resulting in an infinite number of solutions. Lastly, when all three equations are parallel, there is no solution to the system.

5. How can I check if my solution to a system of three equations with three unknowns is correct?

To check if your solution is correct, you can substitute the values of the variables into each of the original equations and see if they all hold true. If they do, then your solution is correct. You can also use a graphing calculator to plot the equations and see if they intersect at the values you found for the variables.

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