Consider the system of equations What is the sum of all distinct real values of that satisfy the system?

Solution available with step-by-step explanation

SAT Nonlinear Equations & Systems Question 90 | Free SAT Prep | Aniko

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Question 90·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

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Consider the system of equations

\begin{cases} x^2 + y = 13 \\[4pt] y^2 + x = 13 \end{cases}

What is the sum of all distinct real values of $x$ that satisfy the system?

For symmetric systems like this, look for a quick algebraic shortcut instead of solving by substitution directly. Subtract one equation from the other to eliminate constants and expose a factor like $(x - y)$ , which splits the problem into two simple cases. Then, when quadratics appear, use factoring or the quadratic formula—but rely on the sum-of-roots formula $-b/a$ whenever you only need the sum of solutions. This minimizes messy arithmetic and saves time on test day.

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Look at how the two equations are related

Compare $x^2 + y = 13$ and $y^2 + x = 13$ . What happens if you subtract one of these equations from the other?

Factor the expression you get

After subtracting, you should get an expression involving $x^2 - y^2$ and linear terms in $x$ and $y$ . Try to factor it using the difference of squares and then factor out a common term.

Solve both resulting cases

Your factored expression should split into two factors, giving two simpler relationships between $x$ and $y$ . Solve the system in each case, list all distinct $x$ -values, and then add them together.

Desmos Guide

Graph both equations as implicit curves

In Desmos, type x^2 + y = 13 on one line and y^2 + x = 13 on another line. Desmos will draw both curves in the coordinate plane.

Find all intersection points

Use the intersection tool (tap/click on the points where the curves cross) to see the coordinates of every intersection. Record the $x$ -coordinates of all intersection points; these are the $x$ -values that satisfy the system.

Compute the sum of the x-values

In a new Desmos line, type the sum of the distinct $x$ -coordinates you found (for example, something like x1 + x2 + x3 + x4 or directly enter the numbers). The resulting value is the sum of all distinct real solutions for $x$ .

Step-by-step Explanation

Use the symmetry of the system

The system is

\begin{cases} x^2 + y = 13 \\ y^2 + x = 13 \end{cases}

Because the equations look very similar, a good first move is to subtract one from the other to eliminate the constant 13 and connect $x$ and $y$ directly.

Subtract the second equation from the first:

(x^2 + y) - (y^2 + x) = 13 - 13.

This simplifies to

x^2 - y^2 + y - x = 0.

Factor the result to get two cases

Factor $x^2 - y^2$ as a difference of squares and then factor the whole expression:

\begin{aligned} x^2 - y^2 + y - x &= (x - y)(x + y) + (y - x) \\ &= (x - y)(x + y) - (x - y) \\ &= (x - y)(x + y - 1). \end{aligned}

So we have

(x - y)(x + y - 1) = 0.

This gives two possible cases:

Case 1: $x - y = 0$ , so $x = y$ .
Case 2: $x + y - 1 = 0$ , so $x + y = 1$ .

Case 1: Solve when x = y

If $x = y$ , plug into one of the original equations, for example $x^2 + y = 13$ .

Since $y = x$ :

x^2 + x = 13.

Rewriting:

x^2 + x - 13 = 0.

This is a quadratic in $x$ . It has two real solutions. By the sum-of-roots formula for $ax^2 + bx + c = 0$ , the sum of the two $x$ -values from this case is

-\frac{b}{a} = -\frac{1}{1} = -1.

(You do not need the exact individual roots here, only their sum.)

Case 2: Solve when x + y = 1

If $x + y = 1$ , then $y = 1 - x$ .

Substitute into $x^2 + y = 13$ :

\begin{aligned} x^2 + (1 - x) &= 13 \\ x^2 - x + 1 &= 13 \\ x^2 - x - 12 &= 0. \end{aligned}

Now factor the quadratic:

(x - 4)(x + 3) = 0.

So the $x$ -values from this case are

x = 4 \quad \text{or} \quad x = -3.

Combine all distinct x-values and find their sum

From both cases, the distinct real $x$ -values that satisfy the system are:

From Case 1: two solutions whose sum is $-1$ .
From Case 2: $x = 4$ and $x = -3$ , whose sum is $4 + (-3) = 1$ .

Now add the sums from both cases:

(-1) + 1 = 0.

So the sum of all distinct real values of $x$ that satisfy the system is $0$ .

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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