A linear-nonlinear system pairs a line with a curve; a squared variable is the cue. Graph both equations as written in Desmos and click every intersection, since each shared point solves both equations. If the question asks for the sum of possible -values, add the first coordinate of each point. Subtracting the coordinates finds the gap between them instead.
Hints
- Hint 1
A system asks for values that make both equations true at once. Graph each equation on its own Desmos line. Where do the line and the curved graph meet?
- Hint 2
An intersection gives a point , with the -value first. Find every point where the graphs meet, since each one may give a possible value of .
- Hint 3
A sum means add the -coordinates, keeping their signs. Subtracting one from the other would answer a different question: how far apart are the values?
Step-by-step
Approach 1: Graph both equations in Desmos
Step 1Find every intersection
Type and on separate lines, then click both intersections. Desmos shows and . An intersection lies on both graphs, so its first coordinate is an -value that satisfies the system.
- Step 2
Add the possible x-values
Add the first coordinates, keeping the negative sign: . Use every possible -value in the sum, not the -coordinates or the distance between the points. The sum of all possible values of is . Choice B.
Approach 2: Use the sum of quadratic roots
Step 1Eliminate y
No Desmos needed. Adding the equations removes , and the resulting quadratic gives its sum of roots directly. Add the left sides and the right sides; and cancel:
- Step 2
Put the quadratic in standard form
Subtract to get zero on one side: . A root is an -value that makes this equation true. Each root also gives a value of through either original equation, so these roots are exactly the possible -values.
- Step 3
Read the sum from the coefficients
For a quadratic , the sum of its roots is . Here and , so: . The sum of all possible values of is . Choice B.