Question

Find all orders pairs (x, y) that satisfy the system:

Answer 1

\[x ^{3}-x ^{2}y+5xy ^{2}-y ^{3}=124\]

\[x ^{3}-5x ^{2}y+xy ^{2}-y ^{3}=4\]

Answer 2

I'm tempted to subtract the second from the first.

Don't forget to notice the symmetry on x = y when you do that.

Answer 3

So would \[x=\sqrt{30/(y ^{2}}+y)\]

Answer 4

Here's something interesting.
Add them...\[2x ^{3}-6x ^{2}y+6xy ^{2}-y ^{3}=128\]Divide both sides by two...\[x ^{3}-3x ^{2}y+3xy ^{2}-y ^{3}=64\]Factor...\[(x-y)^{3}=64\]Solve...\[x-y=4\]\[y=x-4\]I'm not sure how to interpret this though.

Answer 5

I'm not sure if you can add and divide system of equations like that.

Answer 6

Why not? If a = b and c = d, can we not say that a+c = b+d? That's all we're doing.

Except for Euler's typo on y^3 (s/b 2y^3), that is a wonderful solution.

Answer 7

System of equations are lines on a graph. When you subtract one equation from the other you have to keep the other one the same. I'm not sure what adding them together then dividing it by 2 will do to the line. If it distorts it, you're no longer solving for where the original two lines intersect.

Answer 8

Not a thing. go ahead and explore a little.

y = 3x - 5

Is this a different line?

2y = 6x - 10

No.

Is this a different line?

\(\sqrt{2}y = 3\sqrt{2}x - 5\sqrt{2}\)

No.

Answer 9

But if you add y = -3x + 5 to that equation, y=0 is not the solution.

Answer 10

I got busy on another problem... thank you for catching my typo @tkhunny.
We are definitely allowed to add equations when trying to solve. It is commonly used for the addition/elimination method for solving linear systems. This, however, is not a linear system. The graphs of these equations will not be lines because they are multiplied and raised to various powers. If anything, I would say that they are 2 surfaces that appear to intersect at a line.

Answer 11

@tkhunny was merely demonstrating the balanced equations concept (what you do to one side of an equation, you do to the other, and the relationship is not changed).
These methods (adding/ subtracting equations and multiplying/ dividing by constants) are your main tools when you first start solving systems of equations and their successors... matrices.

Answer 12

2x+y=4
2x-y=8
how would you solve this system? would you subtract it? would you sub it?
The reason you're able to subtract them is because x and y must satisfy both eqs at the same time. if you can't subtract them, it's because x and y in the first eq is not the same as the x and y in the second one.
i.e.
\(2x_1 +y_1 =4\)
\(2x_2-y_2=8\) then you cannot subtract.

\(2x +y =4\)
\(2x-y=8\) then you can subtract.
you follow?