Question 141·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

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Let $(x, y)$ be a pair of real numbers satisfying

(x-4)(y+3)=3 \\[1.4em] x^2 + y^2 = 25.

Which of the following is a possible value of $x$ ?

A nonlinear system asks for a point that satisfies two equations at once. A circle equation can limit the possible $x$ -coordinates, but being on the circle isn't enough. Graph both original equations in Desmos and click where they intersect. Then read the coordinate the question asks for, not the whole point.

Hints

Hint 1
The equation $x^2+y^2=25$ describes a circle centered at $(0,0)$ with radius $5$ . How far left or right can a point on that circle be?
Hint 2
An intersection is a point shared by two graphs, so its coordinates satisfy both equations. Type each equation into Desmos as written and look for a shared point.
Hint 3
In a point $(x,y)$ , $x$ is the first coordinate. After clicking an intersection, compare that first number with the offered values.

Step-by-step

Graph both equations and read the shared point
Step 1
Limit where the circle can reach
Since $25=5^2$ , $x^2+y^2=25$ is a circle centered at $(0,0)$ with radius $5$ . So its $x$ -coordinates run only from $-5$ to $5$ ; $6$ cannot work.
Step 2
Find a point that satisfies both equations
Type $(x-4)(y+3)=3$ and $x^2+y^2=25$ on separate Desmos lines. Click where the graphs meet at the circle's right edge. Desmos shows $(5,0)$ . An intersection satisfies both equations, so this point is a solution. Its first coordinate is $x=5$ . Choice C.

Question 141·Hard·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