For real numbers and , suppose and . What is ?

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SAT Nonlinear Equations & Systems Question 203 | Free SAT Prep | Aniko

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Question 203·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

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For real numbers $x$ and $y$ , suppose $x + y = 10$ and $x^2 + y^2 = 58$ . What is $|x - y|$ ?

Your answer

(Express the answer as an integer)

When you see a system involving $x+y$ and $x^2 + y^2$ , avoid solving for $x$ and $y$ individually. Instead, use the identities $(x+y)^2 = x^2 + 2xy + y^2$ and $(x-y)^2 = x^2 - 2xy + y^2$ to express everything in terms of $x^2 + y^2$ , $xy$ , and $x+y$ . First solve for $xy$ using the given values, then plug into the formula for $(x-y)^2$ , and finally take the positive square root to match the absolute value. This approach is much faster and cleaner than solving the full system of equations.

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Connect $x+y$ with $x^2 + y^2$

Try squaring $x+y$ . What identity do you get for $(x+y)^2$ in terms of $x^2$ , $y^2$ , and $xy$ ?

Find $xy$

After writing $(x+y)^2 = x^2 + 2xy + y^2$ , plug in the given values for $x+y$ and $x^2 + y^2$ and solve for $xy$ .

Express $(x-y)^2$ using known values

Recall the formula for $(x-y)^2$ . Rewrite it so it uses $x^2 + y^2$ and $xy$ , which you now know.

Relate $(x-y)^2$ and $|x-y|$

Once you have $(x-y)^2$ , how do you get $|x-y|$ ? Remember that absolute value must be nonnegative, even if $x-y$ itself could be positive or negative.

Desmos Guide

Compute $|x-y|$ directly from a combined formula

Use the identity $(x-y)^2 = 2(x^2 + y^2) - (x+y)^2$ . In a Desmos expression line, type sqrt(2*58 - 10^2). The value that Desmos displays is the value of $|x-y|$ for this problem.

Step-by-step Explanation

Relate the given expressions using $(x+y)^2$

Use the algebraic identity

(x+y)^2 = x^2 + 2xy + y^2.

You are given $x+y=10$ , so

(x+y)^2 = 10^2 = 100.

Substitute into the identity:

100 = x^2 + 2xy + y^2.

You are also given $x^2 + y^2 = 58$ , so plug that in:

100 = 58 + 2xy.

Solve for $xy$ :

2xy = 100 - 58 = 42 \Rightarrow xy = 21.

Write $(x-y)^2$ in terms of known quantities

Use the identity

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

Group the terms as $(x^2 + y^2) - 2xy$ :

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

You now know both $x^2 + y^2$ and $xy$ from the previous step.

Compute $(x-y)^2$ numerically

Substitute $x^2 + y^2 = 58$ and $xy = 21$ into

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

This gives

(x-y)^2 = 58 - 2(21) = 58 - 42 = 16.

So you have found the value of $(x-y)^2$ .

Use $(x-y)^2$ to find $|x-y|$

Since $(x-y)^2 = 16$ , take the square root of both sides:

|x-y| = \big|x-y\big| = \sqrt{16} = 4.

Because absolute value is always nonnegative, you choose the positive square root, so the final answer is $4$ .

Strategy

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Desmos Guide

Step-by-step Explanation

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Step-by-step Explanation