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 \\[1.4em] x^2 + y^2 &= 58 \end{aligned}

What is $|x - y|$ ?

Your answer

A nonlinear system has equations for the same unknowns, with at least one squared variable. Graph both equations in Desmos and click every intersection, since more than one pair may satisfy them. Then use a pair in the expression the question asks for. With an absolute difference, reversing the pair changes the sign, not the distance.

Hints

Hint 1
An intersection is a point shared by two graphs, so its coordinates make both equations true. Enter the equations as written in Desmos and check every crossing, not only the first one.
Hint 2
The absolute value of a number is its distance from $0$ , so it can't be negative. If two solution pairs reverse $x$ and $y$ , what happens to the size of their difference?

Step-by-step

Approach 1: Graph the system in Desmos
Step 1
Find the pairs that satisfy both equations
Type $x+y=10$ and $x^2+y^2=58$ on separate Desmos lines. Click both intersections: $(3,7)$ and $(7,3)$ . An intersection lies on both graphs, so each pair satisfies both given equations.
Step 2
Find the absolute difference
The absolute value of a number is its distance from $0$ . Swapping $x$ and $y$ changes the sign of $x-y$ , but not $|x-y|$ . Use either pair: type $|7-3|$ below the equations. Desmos displays $4$ , so the requested absolute difference is $4$ . Grid in 4.

Approach 2: Find the difference without finding x and y
Step 1
Expand the sum
An identity is an equation true for any $x$ and $y$ . Here, squaring the sum and the difference creates opposite $xy$ terms. Expand the sum:
$(x+y)^2=x^2+2xy+y^2.$
Step 2
Expand the difference
Expand the difference:
$(x-y)^2=x^2-2xy+y^2.$
Step 3
Cancel the xy terms
Add the equations so the $xy$ terms cancel:
$(x+y)^2+(x-y)^2=2(x^2+y^2).$
Step 4
Isolate the difference squared
The question asks for $|x-y|$ , so first isolate $(x-y)^2$ . Subtract $(x+y)^2$ from both sides:
$(x-y)^2=2(x^2+y^2)-(x+y)^2.$
Step 5
Use the two given values
The system gives $x^2+y^2=58$ and $x+y=10$ , so substitute those whole expressions:
$(x-y)^2=2(58)-10^2.$
Step 6
Evaluate the expression
Type $2(58)-10^2$ in Desmos to evaluate the multiplication, square, and subtraction. Desmos displays $16$ , so $(x-y)^2=16$ .
Step 7
Take the nonnegative square root
A difference whose square is $16$ can be positive or negative, but absolute value asks for its nonnegative size. Take the square root:
$|x-y|=\sqrt{16}.$
Evaluate it:
$|x-y|=4.$
So the absolute difference is $4$ . Grid in 4.

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

Strategy

Hints

Step-by-step

Lessons that teach this

Study Hints

Strategy

Hints

Step-by-step

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