Is it possible to solve a system of three equations with two unknowns?

[nobahar]

Hello!
How do I determine if there is a solution to the following simultaneous equation?:

2a+3b=7
5a+7b=19
9a+5b=32

I have given a specific example but I was interested in the general case. I am confused because I have seen examples with two equations and two unkowns where it has been argued that, if, for example, I have:

2a+4b=x₁
5a+6b=x₂
Then:
$$a=\frac{1}{2}(x_{2}-\frac{3}{2}x_{1})$$
$$b=\frac{1}{4}({\frac{5}{2}x_{1}-x_{2})$$

and I can plug in any x₁ and any x₂ and get an answer, as there is no division by zero, etc, and so there is no reason why I can't get an output for a and c.
I can represent my original three equations in a similar format; presenting a as a combination of the outputs and b as a combination of the outputs. But when I put them into the original equation it doesn't work!

Has this something to do with an assumption being made? That with two equations and two unknowns, I can always find a solution for any x₁ and x₂ (provided that they're not parallel), but for three equations and two unknowns, there is not always going to be a solution? If this is the case, how do I ascertain that there is not a solution (without graphing the functions)?

As always, any help appreciated.

 

[debsuvra]

The simultaneous equations you gave here have conflicting values of a and b. Solving the first two equations, we can have a=8 and b=-3. On the other hand putting a=8 on 3rd equation doesn't satisfy b=-3.

Since the equations are of first order, there will be single discreet solution for a and b. The third equation is conflicting with this.

 

[Borek]

If there are three equations in two unknowns there are basically two possibilities - either they are inconsistent, or one is redundant. Solve any two equations and put results into third - if it is satsified, equation was redundant, if not - it was inconsistent.

 

[HallsofIvy]

Graphically, a single equation in two unknowns can be represented as a line in the plane. Two such equations, of course, represent two lines and, except in the unusual case that the lines are parallel, have a single point of intersection, giving a single solution to the two equations.

Three equations in two unknowns represent three lines in the plane and now the "standard" situation is that two of the lines intersect in a single point while the third line intersects those two lines at point different from the original point- there is no point in common to the three lines and so no x, y that satisfy all three equations.

 

[nobahar]

Thanks Debsuvra, Halls and Borek.

The simultaneous equations you gave here have conflicting values of a and b. Solving the first two equations, we can have a=8 and b=-3. On the other hand putting a=8 on 3rd equation doesn't satisfy b=-3.

Since the equations are of first order, there will be single discreet solution for a and b. The third equation is conflicting with this.

Graphically, a single equation in two unknowns can be represented as a line in the plane. Two such equations, of course, represent two lines and, except in the unusual case that the lines are parallel, have a single point of intersection, giving a single solution to the two equations.

Three equations in two unknowns represent three lines in the plane and now the "standard" situation is that two of the lines intersect in a single point while the third line intersects those two lines at point different from the original point- there is no point in common to the three lines and so no x, y that satisfy all three equations.

Using these two pieces of information, it is possible that I can compute the unknowns using equations 1) and 2); 1) and 3); and 2) and 3) and arrive at appropriate values that satisfy two of the three equations, depending on which two of the three equations I use, but will not necessarily work for all three.

If there are three equations in two unknowns there are basically two possibilities - either they are inconsistent, or one is redundant. Solve any two equations and put results into third - if it is satsified, equation was redundant, if not - it was inconsistent.

Is this the only method of determining whether there is an answer. Solve for two and see if it works for the third? If I get a gobbeldygook answer such as 8=12, then there is no solution for all three? I found some information on Reduced Row Echelon Form, is this valid here?

Many thanks for the responses.

 

[HallsofIvy]

Yes, you can use "Row Echelon form". If one of the equations is a linear combination of the other two, your last row will be all "0"s and the solution is given by the first two rows. If not you will get something like "0 0 1" in the last row which corresponds to "0x+ 0y= 1" and is impossible.

 

[nobahar]

Thanks again.

This gets a little more confusing when introducing a fourth variable. I shall not attempt an explanation as they end up being long and often do not elicit a response! My question is, if I have three variables in three equations, therefore three planes, and which do not all intersect at a single point or line, if I then introduce a fouth variable, does it make it possible to solve a simultaneous equation where there is a solution to the three equations? It's impossible to imagine and I do not know where to begin to prove it algebraicly.
Any responses would be appreciated.
Many thanks.