Question 199·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

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If $(x, y)$ is a real solution to the system

\begin{aligned} x + y &= 5 \\[1.4em] x^2 - y &= 7 \end{aligned}

what is one possible value of $x$ ?

A nonlinear system has equations in the same variables, with at least one equation whose graph isn't a line. Graph both equations in Desmos and click their intersections: each crossing satisfies both equations, and a curve may cross a line more than once. If the question asks for $x$ , read the first coordinate of a crossing, not a number from one of the equations.

Hints

Hint 1
A solution to a system is a point that makes both equations true. On a graph, where would a point have to lie to satisfy both equations?
Hint 2
The $x^2$ makes one graph a curve, so there may be more than one intersection with the line. Type both equations as written and look for every place the graphs meet.
Hint 3
In a point written $(x,y)$ , $x$ is the first coordinate and $y$ is the second. Which first coordinate from an intersection matches an answer choice?

Step-by-step

Approach 1: Graph the system in Desmos
Step 1
Find where the graphs meet
Type $x+y=5$ and $x^2-y=7$ on separate lines. Desmos shows a line and a curve with two crossings. The $x^2$ makes the second graph a curve, so don't assume the system has only one solution.
Step 2
Read a possible x-value
Click both crossings. Desmos labels them $(-4,9)$ and $(3,2)$ . Each crossing satisfies both equations, and $x$ is its first coordinate. So $x$ could be $-4$ or $3$ . Among the choices, $3$ is one possible value of $x$ . Choice D.

Approach 2: Add the equations to eliminate y
Step 1
Cancel the y terms
The equations have opposite $y$ terms, so adding them removes $y$ and leaves an equation in $x$ . Add both sides:
$x^2+x=5+7.$
Add the numbers:
$x^2+x=12.$
Step 2
Put zero on one side
Subtract $12$ from both sides to get zero on one side, which lets you factor:
$x^2+x-12=0.$
Step 3
Factor the quadratic
To factor, find two numbers whose product is $-12$ and whose sum is $1$ . The numbers $4$ and $-3$ work, so write the factored equation:
$(x+4)(x-3)=0.$
Step 4
Find the possible x-values
A product is zero only if at least one factor is zero. Set each factor to zero:
$x+4=0\quad\text{or}\quad x-3=0.$
Solve each equation:
$x=-4\quad\text{or}\quad x=3.$
One possible value of $x$ among the choices is $3$ . Choice D.

Question 199·Medium·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

Lessons that teach this