Quadratic simultaneous equations: X squared * Y squared =5, Y-X=3

New member
What is the method for working out quadratic simultaneous equations?

X squared * Y squared =5
Y-X=3

I only know how to work these out through trial and error

Thanks!

Jomo

Elite Member
What is the method for working out quadratic simultaneous equations?

X squared * Y squared =5
Y-X=3

I only know how to work these out through trial and error

Thanks!
X2Y2 =5 and Y-X=3. Why not solve for one variable and plug it into the other equation? For us to help you, you need to show us where you are stuck, else we have no idea where you might need help.

Dr.Peterson

Elite Member
What is the method for working out quadratic simultaneous equations?

X squared * Y squared =5
Y-X=3
One way to simplify this is to first take the square root of both sides of the first equation. But be careful! You need to use plus-or-minus, because there are two square roots. (This caution will likely be needed sooner or later no matter what method you use.)

After this, you can use substitution.

Dr.Peterson

Elite Member
Did you make that up?
If not, is it the original problem IN FULL?
What do you mean with "simultaneous quadratic equations"?
After substituting (as per Jomo's post), you'll end up
with a 4th degree equation, having 4 solutions...are you at that level yet?
It's not all that hard to solve, at least if you do it the way I suggested, and if you know the quadratic formula; but the answer is not pretty, so it is quite possible that the problem was copied wrong or invented as an example.

In any case, I wouldn't call it beginning algebra; but it is common to mix that up, so the OP may know more than we would assume.

New member
Sorry it was a plus not a times.

Dr.Peterson

Elite Member
X squared * Y squared =5
Y-X=3
Sorry it was a plus not a times.
I was wondering if you might have done that, though it didn't seem likely!

So it's

x^2 + y^2 = 5
y - x = 3

You can still do exactly what was suggested: solve one equation for one variable, and substitute in the other. I might solve the second for y, and put the result in place of y in the first.

Have you tried that? Please show us your work, so we can see if you are doing it correctly. You should end up with a single quadratic equation that can be solved easily by factoring.

HallsofIvy

Elite Member
Except that now you get a nice quadratic equation rather than the nasty fourth degree equation from before!

Since x- y= 3, y= x- 3. Then $$\displaystyle x^2+ y^2= x^2+ (x- 3)^2= x^2+ x^2- 6x+ 9= 5$$.

$$\displaystyle 2x^3- 6x+ 4= 0$$, $$\displaystyle x^3- 3x+ 2=(x- 2)(x- 1)= 0$$.

Jomo

Elite Member

aNUTter way:

x^2 + y^2 = 5 [1]
y - x = 3 [2]

[2]^2 : x^2 - 2xy + y^2 = 9 [3]

Subtract [1] from [3] : xy = -2

Continue...
I like that method. Maybe there is hope for you(??)

JeffM

Elite Member
What is the method for working out quadratic simultaneous equations?

X squared * Y squared =5
Y-X=3

I only know how to work these out through trial and error

Thanks!
You can attempt to solve any system of simultaneous equations by the method of substitution. Example.

$$\displaystyle x^2 + y - z = 41$$.

$$\displaystyle x + y + z = 7$$.

$$\displaystyle 5x + 3y + 4z = 18.$$

$$\displaystyle z = \dfrac{1}{4} * (18 - 5x - 3y) \ \because \ 5x + 3y + 4z = 18.$$

$$\displaystyle \therefore x + y + z = 7 \implies 28 = 4x + 4y + 18 - 5x - 3y \implies y = 10 + x.$$

$$\displaystyle \text {And } x^2 + y - z = 41 \implies 4x^2 + 4y - (18 - 5x - 3y) = 164 \implies$$

$$\displaystyle 4x^2 + 7y + 5x - 18 = 4x^2 + 7(10 + x) + 5x - 18 = 164 \implies$$

$$\displaystyle 4x^2 + 12x - 112 = 0 \implies x^2 + 3x - 28 = 0 \implies (x + 7)(x - 4) = 0 \implies$$

$$\displaystyle x = 4 \text { or } x = -\ 7.$$

$$\displaystyle x = 4 \implies y = 14 \implies z = \dfrac{1}{4} * (18 - 20 - 42) = -\ 11.$$

$$\displaystyle x = -\ 7 \implies y = 3 \implies z = \dfrac{1}{4} * (18 + 35 - 9) = 11.$$

It must be admitted, however, the systems of non-linear equations can get very ugly even with substitution.