How many ordered pairs of real numbers satisfy the system of equations

Solution available with step-by-step explanation

SAT Nonlinear Equations & Systems Question 144 | Free SAT Prep | Aniko

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Question 144·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

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How many ordered pairs $(x, y)$ of real numbers satisfy the system of equations

\begin{cases} x^2 + y = 10 \\[3pt] y^2 + x = 10 \end{cases}

For nonlinear, symmetric systems like this, avoid messy substitution at first. Instead, subtract one equation from the other to eliminate constants, factor the result, and split into simple cases (such as $x = y$ and $x + y = \text{constant}$ ). For each case, substitute into one original equation to get a single-variable quadratic, use the discriminant to see how many real solutions it has, and then add the counts from all cases to get the total number of solution pairs—often you don’t even need the exact values of the solutions, just how many there are.

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Look for a way to combine the equations

Instead of trying to solve directly for $x$ and $y$ from one equation, think about what happens if you subtract one equation from the other. What cancels out?

Factor the result of the subtraction

After subtracting, you should get an expression involving $x^2 - y^2$ and some first-degree terms. Remember that $x^2 - y^2$ can be factored as $(x - y)(x + y)$ . Can you factor the entire expression?

Split into cases and solve quadratics

Once you factor, you should get a product of two expressions equal to 0. Treat each factor equal to 0 as a separate case, substitute into one of the original equations, and solve the resulting quadratic. How many real solutions does each quadratic have?

Desmos Guide

Graph the first equation

In Desmos, enter the first equation as a function: type y = 10 - x^2. This is a downward-opening parabola.

Graph the second equation in terms of x

Rewrite the second equation $y^2 + x = 10$ as $x = 10 - y^2$ , then in Desmos type x = 10 - y^2. This is a sideways-opening parabola.

Find and count the intersection points

Look at where the two curves intersect. Use Desmos’s intersection tool (or tap on each intersection) to see all intersection points. Each intersection represents one ordered pair $(x, y)$ that satisfies the system. Count how many intersection points there are and select the answer choice matching that count.

Step-by-step Explanation

Use symmetry and subtract the equations

Start with the system:

x^2 + y = 10

y^2 + x = 10

Subtract the second equation from the first:

(x^2 + y) - (y^2 + x) = 10 - 10

This simplifies to $x^2 - y^2 + y - x = 0$. Factor it as $(x - y)(x + y) - (x - y) = 0$, which is $(x - y)(x + y - 1) = 0$. So either $x - y = 0$ or $x + y - 1 = 0$, giving two separate cases: $x = y$ or $x + y = 1$. You will solve each case separately.

Case 1: Solve when x = y

If $x = y$ , substitute $y = x$ into one of the original equations, for example $x^2 + y = 10$ .

Then $x^2 + x = 10$ , so $x^2 + x - 10 = 0$ .

Use the quadratic formula on $x^2 + x - 10 = 0$ :

x = \dfrac{-1 \pm \sqrt{1^2 - 4(1)(-10)}}{2(1)} = \dfrac{-1 \pm \sqrt{1 + 40}}{2} = \dfrac{-1 \pm \sqrt{41}}{2}

There are two distinct real values of $x$ . For each, $y$ equals the same value (since $x = y$ ), giving two ordered pairs from this case.

Case 2: Solve when x + y = 1

If $x + y = 1$ , rewrite as $y = 1 - x$ .

Substitute $y = 1 - x$ into $x^2 + y = 10$ :

x^2 + (1 - x) = 10

Simplify: $x^2 - x + 1 = 10$ , so $x^2 - x - 9 = 0$ .

Use the quadratic formula on $x^2 - x - 9 = 0$ :

x = \dfrac{1 \pm \sqrt{(-1)^2 - 4(1)(-9)}}{2(1)} = \dfrac{1 \pm \sqrt{1 + 36}}{2} = \dfrac{1 \pm \sqrt{37}}{2}

Again, there are two distinct real values of $x$ . For each $x$ , compute $y = 1 - x$ , giving two ordered pairs from this case.

Combine the solutions from both cases

From Case 1 ( $x = y$ ), you found 2 ordered pairs. From Case 2 ( $x + y = 1$ ), you found 2 more ordered pairs. None of these pairs are the same, because in Case 1 the coordinates are equal ( $x = y$ ), while in Case 2 they add to 1 (so $x$ and $y$ must be different real numbers).

Therefore, there are $2 + 2 = 4$ ordered pairs that satisfy the system, so the correct answer choice is 4.

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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