We are given the system of equations:

1. $x^2 + xy = 60$
2. $y^2 + xy = 20$

To solve for $x$ and $y$, let's subtract the second equation from the first:

$(x^2 + xy) - (y^2 + xy) = 60 - 20$

Simplifying the left side:

$x^2 - y^2 = 40$

This is a difference of squares, so we can factor it as:

$(x - y)(x + y) = 40$

Now, we have a new equation:

$(x - y)(x + y) = 40 \quad \text{(Equation 3)}$

Next, we can try to manipulate the system further by adding the two original equations. Adding equations 1 and 2:

$(x^2 + xy) + (y^2 + xy) = 60 + 20$

This simplifies to:

$x^2 + y^2 + 2xy = 80$

Now, we can express $x^2 + y^2$ in terms of $(x + y)^2$. Recall the identity:

$x^2 + y^2 = (x + y)^2 - 2xy$

Substitute this into the equation:

$(x + y)^2 - 2xy + 2xy = 80$

Simplifying:

$(x + y)^2 = 80$

Now we know:

$x + y = \sqrt{80} = 4\sqrt{5} \quad \text{(Equation 4)}$

We now have two key equations:

• $(x - y)(x + y) = 40$
• $x + y = 4\sqrt{5}$

Substitute $x + y = 4\sqrt{5}$ into Equation 3:

$(x - y)(4\sqrt{5}) = 40$

Solving for $x - y$:

$x - y = \frac{40}{4\sqrt{5}} = \frac{10}{\sqrt{5}} = 2\sqrt{5}$

Now we have:

• $x + y = 4\sqrt{5}$
• $x - y = 2\sqrt{5}$

To solve for $x$ and $y$, we can add and subtract these two equations:

1. Adding them:

$(x + y) + (x - y) = 4\sqrt{5} + 2\sqrt{5} = 6\sqrt{5}$

So:

$2x = 6\sqrt{5} \quad \Rightarrow \quad x = 3\sqrt{5}$

2. Subtracting them:

$(x + y) - (x - y) = 4\sqrt{5} - 2\sqrt{5} = 2\sqrt{5}$

So:

$2y = 2\sqrt{5} \quad \Rightarrow \quad y = \sqrt{5}$

Thus, the solutions are:

$x = 3\sqrt{5}, \quad y = \sqrt{5}$

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Would you like further explanation or more details on this process?

Here are some related questions:

1. How do you solve quadratic systems of equations with variables multiplied together?
2. What does factoring a difference of squares imply for solving systems of equations?
3. How can you use substitution or elimination methods in quadratic systems?
4. What role does the identity $x^2 + y^2 = (x + y)^2 - 2xy$ play in solving systems?
5. How do you check the correctness of your solutions once you find them?

Tip: When dealing with quadratic systems, it's often useful to manipulate the equations to isolate terms or eliminate one variable using substitution or combination strategies like adding or subtracting.