If real numbers and satisfy What is ?

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SAT Nonlinear Equations & Systems Question 188 | Free SAT Prep | Aniko

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Question 188·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

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If real numbers $x$ and $y$ satisfy

\begin{aligned} x + y &= 10 \\ x^2 + y^2 &= 58 \end{aligned}

What is $|x - y|$ ?

Your answer

(Express the answer as an integer)

When you see a system involving $x+y$ and $x^2 + y^2$ but the question asks for something like $|x-y|$ , avoid solving for $x$ and $y$ directly. Instead, use identities: write $(x+y)^2 = x^2 + 2xy + y^2$ to solve quickly for $xy$ , then use $(x-y)^2 = x^2 - 2xy + y^2$ (or an equivalent formula) to get $(x-y)^2$ and take the square root. This approach saves time, reduces algebra mistakes, and is especially useful on SAT questions that only ask for combinations of variables rather than the variables themselves.

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Use identities involving squares

You are given $x+y$ and $x^2+y^2$ . Think about the algebra identities for $(x+y)^2$ and $(x-y)^2$ . How can they connect these quantities?

Find $xy$ first

Write $(x+y)^2$ as $x^2 + 2xy + y^2$ . Plug in the values you know for $x+y$ and for $x^2 + y^2$ , and solve the resulting equation for $xy$ .

Relate $(x-y)^2$ to what you know

Once you know both $x^2 + y^2$ and $xy$ , use the identity $(x-y)^2 = x^2 - 2xy + y^2$ . Substitute your known values to find $(x-y)^2$ , then take the square root and remember the absolute value.

Desmos Guide

Graph the first equation

Enter the first equation as y = 10 - x. This is a straight line representing all pairs $(x,y)$ with sum 10.

Graph the second equation

Enter the second equation exactly as x^2 + y^2 = 58. This is a circle centered at the origin with radius $\sqrt{58}$ .

Find intersection points

Look for the intersection points of the line and the circle. Click on an intersection to see its coordinates $(x_1, y_1)$ .

Compute $|x-y|$ from an intersection

Using one intersection point $(x_1, y_1)$ , either type |x1 - y1| in a new Desmos line (using vertical bars for absolute value) or subtract the coordinates by hand and take the absolute value. The resulting number is the value of $|x - y|$ .

Step-by-step Explanation

Relate the given equations to $(x+y)^2$

We are told that $x+y=10$ and $x^2+y^2=58$ .

Use the algebra identity

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

Since $x+y=10$ , we know

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

Substitute $x^2+y^2=58$ into the identity:

100 = 58 + 2xy.

Solve for the product $xy$

From

100 = 58 + 2xy,

subtract $58$ from both sides:

\begin{aligned} 100 - 58 &= 2xy \\ 42 &= 2xy \end{aligned}

Divide both sides by $2$ :

xy = 21.

Now you know $x+y$ and $xy$ .

Express $(x-y)^2$ using $x^2+y^2$ and $xy$

Use the identity

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

We know $x^2 + y^2 = 58$ and $xy = 21$ , so

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

Compute the subtraction:

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

Find $|x-y|$ from $(x-y)^2$

We found that

(x-y)^2 = 16.

So $x-y$ must be a number whose square is $16$ , meaning $x-y = 4$ or $x-y = -4$ .

The question asks for $|x-y|$ , the absolute value, which is the nonnegative value of the difference, so

|x-y| = 4.

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation