# Homework Help: Find number of real solutions of system of equations

1. Aug 5, 2016

### songoku

1. The problem statement, all variables and given/known data
I got this question from my friend.

How many real solutions for this system of equations:
x2 - xy - 2 = 0
x2 + x + 2y = 0

2. Relevant equations
Don't know

3. The attempt at a solution
I have never learnt about this type of question. What chapter or topic should I learn in order to be able to answer this question?

Thanks

2. Aug 5, 2016

### haruspex

You can just treat them as simultaneous equations, though they are nonlinear.
With linear simultaneous equations, the standard approach is to use one equation to express one unknown in terms of the other(s) then use that to substitute for it in all the other equations. That looks more awkward here because it will give you cubic or worse terms. So instead you could use one equation to express a quadratic term as a sum of linear terms, then use that to eliminate that quadratic from the other equation. See where that leads.

3. Aug 5, 2016

### songoku

I am not sure I get your hint but I managed to do this:
After subtracting the two equations:
x + xy + 2y + 2 = 0
x(1 + y) + 2(y + 1) = 0
(x + 2)(y + 1) = 0
x = -2 , y = -1

Is this the only solution? How can we know or check whether there are other solutions?

Thanks

4. Aug 5, 2016

### ehild

No. If (x + 2)(y + 1) = 0, either x=-2 or y=-1, not necessarily both. Assuming x=-2, replace it back into the original equations and find y. Also, assume y=-1 and find the value(s) of x.

5. Aug 5, 2016

### songoku

Ah, my bad. I got (-2, -1) and (1, -1). So there are only 2 solutions? How can we know that there won't be any other solutions? Is there formula or method to know exactly number of possible solutions?

Thanks

6. Aug 5, 2016

### ehild

The system of equations is liner in y, but quadratic in x. For any value of x, both equations yield some single value for y. As this is a system of equations, the y values coming from the single equations must be the same.
Isolating y from the equations, you get that $y=\frac{-x^2-x}{2}$ and $y=\frac{x^2-2}{x}$, so $\frac{-x^2-x}{2}=\frac{x^2-2}{x}$
That is the same as $x^3+3x^2-4=0$ (for x≠0). This equation can have three real roots. You know two roots already, x=1 and x=-2. So (x-1)(x+2) divides the polynomial x3+3x2-4, and the synthetic division gives the third factor in the factorized equation x3+3x2-4=(x-1)(x+2)(x-a). What is that new root a?

7. Aug 5, 2016

### haruspex

Certainly plugging in x=-2 will give a unique value for y since the resulting equations are linear, but as ehild writes, plugging in y=-1 will yield a quadratic in x, so generally two solutions or none.
All that is pure deduction, so you know there can be no other solutions. The risk can be that somewhere along the way you made an algebraic step which introduced extra solutions, so to complete it you should check all the solutions you have satisfy the original equations.

8. Aug 5, 2016

### ehild

The system of equations is linear in y:
xy=x2-2
2y=-x2-x.
Pretend x is a parameter. A system of linear equations can have a single root, one root or no roots, depending on the value of the parameter(s).
The system of equations can be transformed to $y =\frac{-x^2-x}{2}$ and $\frac{x^2-2}{x}-\frac{-x^2-x}{2}=0$.
If $\frac{x^2-2}{x}-\frac{-x^2-x}{2}$ was zero for all values of x, you would have infinite many solution for y, one for all x values. But it is not zero for all x. If the expression is not zero, there is no solution for y. If the expression is zero for certain values of x, you have a $y =\frac{-x^2-x}{2}$ value for all these x values. In principle, you can have 3 solutions, but two of them coincide.