A linear-nonlinear system pairs a line with a curve. A squared variable in one equation is the cue. Graph both equations as written in Desmos and find where they meet. A line can touch a parabola without crossing it, so check for a touching point before reading the coordinate the question asks for.
Hints
- Hint 1
A solution to a system is one point that makes both equations true. On a graph, where would that point lie?
- Hint 2
A line can meet a parabola, the curve formed by a squared variable, at one point without crossing it. Look for a touching point as well as a crossing.
- Hint 3
An ordered pair lists first and second. Once Desmos labels the shared point, which coordinate does the question ask for?
Step-by-step
Approach 1: Find the shared point in Desmos
Step 1Identify what a solution looks like
The first equation has , so its graph is a parabola; the second equation graphs as a line. A solution must make both equations true, so look for a point the graphs share. A touching point counts as an intersection, too.
- Step 2
Read the requested coordinate
Type and on separate Desmos lines. Click where the graphs touch; Desmos labels the point . In the ordered pair , comes first, so the requested value is , not . Choice C.
Approach 2: See why there is only one value
Step 1Remove y by subtraction
The matching terms in let you eliminate , meaning remove it without finding its value. Subtract the second equation from the first: . Cancel the terms: .
- Step 2
Make a square
The middle term fits . Make a perfect square, an expression with a repeated factor, by adding to both sides: . Factor the left side: .
- Step 3
Find the only possible x-value
A square is only when the expression inside it is . Set the repeated factor to zero: . Add to both sides: . So the system has only one possible value of : . Choice C.