System of two generic equations

Hi,

I am working on this problem

Consider the following system of linear equations in x and y.

ax + by = e

cx + dy = f

Under what conditions will the system have exactly one solution?

I can come up with specific examples of when this happens but not a generic one just using the variables involved.

Re: System of two generic equations

The condition is $\displaystyle ad-bc\ne0.$

Suppose $\displaystyle ad-bc\ne0.$ Then $\displaystyle x=\frac{de-bf}{ad-bc},y=\frac{af-ce}{ad-bc}$ is a solution. If $\displaystyle x=x',y=y'$ is another solution, let $\displaystyle X=\frac{de-bf}{ad-bc}-x',Y=\frac{af-ce}{ad-bc}-y'.$ This gives $\displaystyle aX+bY=0=cX+dY,$ which yields $\displaystyle X=0=Y,$ showing that the solution is unique.

Now suppose $\displaystyle ad-bc=0.$ We have $\displaystyle 0=(ad-bc)x=de-bf$ and $\displaystyle 0=(ad-bc)y=ce-af.$ Then either (1) $\displaystyle de=bf$ and $\displaystyle ce=af,$ in which case $\displaystyle x$ and $\displaystyle y$ can take any arbitrary values and so there are infinitely many solutions, or (2) $\displaystyle de\ne bf$ or $\displaystyle ce\ne af,$ in which case there is no solution in $\displaystyle x$ and $\displaystyle y.$

Hence the system has exactly one solution if and only if $\displaystyle ad-bc\ne0.$

Re: System of two generic equations

I solved for x and y but didn't know of or think of the test for a second solution by creating the x' and y'. This is fantastically simple. I appreciate the help immensely.

Also, unless I'm mistaken I think your last statement meant to say ad - bc not equal to 0, correct?

Re: System of two generic equations

Quote:

Originally Posted by

**bmoon123** Also, unless I'm mistaken I think your last statement meant to say ad - bc not equal to 0, correct?

Yes. Mistake corrected. (Blush)

Re: System of two generic equations

They are the equations of straight lines. Straight lines intersect at one point only unless they are parallel. So exactly one solution unless gradients equal, that is unless -b/a=-d/c, that is unless ad-bc=0

Re: System of two generic equations

Quote:

Originally Posted by

**biffboy** They are the equations of straight lines. Straight lines intersect at one point only unless they are parallel. So exactly one solution unless gradients equal, that is unless -b/a=-d/c, that is unless ad-bc=0

thanks, that makes sense. I think I tend to over think problems like this.

My final answer:

rewritten

y = -(a/b)x + (e/b)

y = -(c/d)x + (f/d)

if a/b = c/d then the slopes of the lines are the same and they are parallel or coincident

so a/b is not equal c/d is the condition needed for exactly one solution

Re: System of two generic equations

That's correct. We would usually finish it off by doing one extra step to say we require ad-bc not to be zero.