Question 92·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

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A system of equations is given by

\begin{aligned} x + y &= 10 \\[1.4em] x^2 + y &= 40 \end{aligned}

If $(x,y)$ is a solution to the system, what is one possible value of $y$ ?

A nonlinear system has two equations sharing the same variables, with a squared term making one graph a curve. Type both equations into Desmos as written; a point where the graphs cross satisfies both equations. A line and a curve may cross more than once, so for one possible value of $y$ , read the second coordinate of either crossing, not the first.

Hints

Hint 1
A solution to a system makes both equations true. On a graph, that happens where the two graphs cross. You can type each equation into Desmos exactly as written, without rearranging either one.
Hint 2
An ordered pair $(x,y)$ lists $x$ first and $y$ second. The question asks for one possible value of $y$ , so which number should you read from a point where the graphs cross?

Step-by-step

Approach 1: Graph the system in Desmos
Step 1
Find the shared points
Type $x+y=10$ and $x^2+y=40$ on separate Desmos lines. Desmos shows a line crossing a parabola, the curved graph made by the squared term, at two points. Each crossing satisfies both equations, so either one can give a possible value of $y$ .
Step 2
Read the requested coordinate
Click the crossing on the right. Desmos labels it $(6,4)$ . In the ordered pair $(x,y)$ , $y$ is the second number, so $4$ is one possible value of $y$ . The $6$ is $x$ , not what the question asks for. Choice D.

Approach 2: Eliminate the matching terms
Step 1
Cancel $y$ by subtraction
Both equations contain $+y$ , so subtract the first equation from the second to eliminate $y$ :
$(x^2+y)-(x+y)=40-10.$
Simplify each side:
$x^2-x=30.$
Now one equation in $x$ replaces the two-equation system.
Step 2
Put the quadratic equal to zero
Subtract $30$ from both sides so you can factor the equation:
$x^2-x-30=0.$
The goal is to write the left side as a product equal to zero.
Step 3
Find one possible $x$
The numbers $-6$ and $5$ multiply to $-30$ and add to $-1$ , so factor:
$(x-6)(x+5)=0.$
A product is zero when a factor is zero. Since the question asks for one possible value, use $x-6=0$ . Add $6$ to both sides:
$x=6.$
Step 4
Find $y$ from the simpler equation
Put $x=6$ into the first equation, $x+y=10$ :
$6+y=10.$
Subtract $6$ from both sides:
$y=4.$
So one possible value of $y$ is $4$ . Choice D.

Question 92·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

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