For positive integers and that satisfy the system of equations what is the value of ?

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SAT Nonlinear Equations & Systems Question 96 | Free SAT Prep | Aniko

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Question 96·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

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For positive integers $x$ and $y$ that satisfy the system of equations

\begin{cases} x^2 + y = 13,\\ y^2 + x = 19 \end{cases}

what is the value of $x + y$ ?

Your answer

(Express the answer as an integer)

For nonlinear systems with integer constraints, first try combining the equations (by adding or subtracting) to eliminate constants or create factorable expressions. Factor what you get, then use integer factor pairs and the positivity/integer conditions to narrow down possible solutions quickly, instead of attempting to solve the nonlinear equations directly or graphing from scratch. Always verify candidate pairs in the original equations before computing the requested quantity.

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Eliminate terms by combining equations

Instead of trying to solve each equation alone, try subtracting one equation from the other to combine like terms and simplify.

Look for factoring opportunities

After you subtract the equations, you will see an expression involving $y^2 - x^2$ and $x - y$ . Remember that $y^2 - x^2$ can be factored using the difference of squares.

Use factor pairs and positivity

Once you have a product of two integer expressions equal to $6$ , think about all integer factor pairs of $6$ and which ones could come from positive integers $x$ and $y$ that also satisfy the original equations.

Desmos Guide

Graph both equations as functions of x

In Desmos, enter the first equation as y = 13 - x^2. For the second equation, rewrite $y^2 + x = 19$ as $y^2 = 19 - x$ , so $y = \sqrt{19 - x}$ (only the positive square root, since $y$ is positive). Enter this as y = sqrt(19 - x).

Locate the intersection point in the first quadrant

Look for the intersection of the curves y = 13 - x^2 and y = sqrt(19 - x) where both $x$ and $y$ are positive. Tap or click that intersection point and note its coordinates $(x, y)$ .

Compute the required sum

Take the $x$ -coordinate and the $y$ -coordinate of the intersection you found and add them (you can type x_value + y_value into Desmos if you like). That sum is the value of $x + y$ for the system.

Step-by-step Explanation

Subtract the two equations

Start by subtracting the first equation from the second to eliminate the constant terms:

From

\begin{cases} x^2 + y = 13,\\ y^2 + x = 19 \end{cases}

subtract the first from the second:

(y^2 + x) - (x^2 + y) = 19 - 13.

This simplifies to

y^2 - x^2 + x - y = 6.

Factor the expression

Recognize $y^2 - x^2$ as a difference of squares and factor the left side:

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

Rewrite $x - y$ as $-(y - x)$ to factor out $(y - x)$ :

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

So the equation becomes

(y - x)(x + y - 1) = 6.

Use integer factor pairs of 6

Because $x$ and $y$ are positive integers, both $y - x$ and $x + y - 1$ must be integers, and their product is $6$ .

List the integer factor pairs of $6$ :

$1 \cdot 6$
$2 \cdot 3$
$3 \cdot 2$
$6 \cdot 1$
(and the same pairs with both factors negative)

For each pair, set

$y - x$ equal to one factor, and
$x + y - 1$ equal to the other factor,

then solve the resulting small system to find possible integer $(x, y)$ pairs. Keep only those with positive $x$ and $y$ that also satisfy the original equations.

Find the valid integer solution and compute the sum

Testing the factor pairs, you will find that the only pair that leads to positive integers $x$ and $y$ which satisfy both original equations is

x = 3, \quad y = 4.

Check:

$x^2 + y = 3^2 + 4 = 9 + 4 = 13$ ,
$y^2 + x = 4^2 + 3 = 16 + 3 = 19$ .

Thus,

x + y = 3 + 4 = 7.

So the value of $x + y$ is $7$ .

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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