#### Help please

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X squared * Y squared =5

Y-X=3

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

Thanks!

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X squared * Y squared =5

Y-X=3

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

Thanks!

- Joined
- Dec 30, 2014

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X

X squared * Y squared =5

Y-X=3

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

Thanks!

- Joined
- Nov 12, 2017

- Messages
- 3,097

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.)What is the method for working out quadratic simultaneous equations?

X squared * Y squared =5

Y-X=3

After this, you can use substitution.

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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 problemDid 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?

In any case, I wouldn't call it

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Sorry it was a plus not a times.

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X squared * Y squared =5

Y-X=3

I was wondering if you might have done that, though it didn't seem likely!Sorry it was a plus not a times.

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.

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

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

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I like that method. Maybe there is hope for you(??)Ya made zee typo Halls...

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...

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

X squared * Y squared =5

Y-X=3

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

Thanks!

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

\(\displaystyle x + y + z = 7\).

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

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

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

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

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

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

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

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

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

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