Solve a system of a linear equation and a quadratic equation in two variables algebraically and graphically

Learning goal: Solve a simple system consisting of a linear equation and a quadratic equation in two variables, algebraically by substitution and graphically as the points of intersection of a line with a parabola or a circle, and interpret the number of solutions from the geometry; for example, find the points of intersection between the line y = −3x and the circle x² + y² = 3.

The two kinds of quadratic equations in two variables in this lesson. A quadratic equation in two variables is one in which the highest total degree of a term is 2. Two shapes matter here. The graph of y = ax² + bx + c with a ≠ 0 is a parabola. The graph of x² + y² = r² with r > 0 is a circle of radius r centered at the origin, because x² + y² is the square of the distance from (x, y) to (0, 0). A linear equation in two variables has a line as its graph. A system made of one linear and one quadratic equation asks for every point that is on both graphs, so the solutions are the points of intersection of the line with the parabola or the circle.

How many solutions there can be. A non-vertical line y = mx + b can cross a parabola y = ax² + bx + c in two points, touch it at one point (a tangent), or miss it entirely. (A vertical line such as x = 1 meets that parabola exactly once without being tangent, and substitution then gives no quadratic at all; this lesson's lines are non-vertical.) The same three possibilities hold for a line and a circle. So a linear–quadratic system has two, one or zero real solutions. Algebraically, substituting the linear equation into the quadratic one produces a quadratic equation in one variable, and its discriminant sorts the three cases: positive means two intersection points, zero means one point of tangency, negative means the graphs do not meet. A negative discriminant produces complex values of x, which are legitimate algebraic solutions of the one-variable equation but do not correspond to points in the coordinate plane; when the question is about intersection points, the answer in that case is that there are none.

The method: substitute the linear equation into the quadratic one. Solve the linear equation for one variable (it is already solved when it reads y = mx + b), substitute into the quadratic equation, solve the resulting one-variable quadratic by whatever method fits its form, and then find the matching value of the other variable from the linear equation. Finally check every point in both original equations. Using the linear equation to recover the second coordinate is the safe choice; using the quadratic equation can produce extra values that are not on the line.

A fully worked model: the line y = −3x and the circle x² + y² = 3. Substitute y = −3x into x² + y² = 3: x² + (−3x)² = 3, so x² + 9x² = 3, 10x² = 3, x² = 3/10. Take square roots: x = ±√(3/10). Rationalize: √(3/10) = √30/10. Then y = −3x, so when x = √30/10, y = −3√30/10, and when x = −√30/10, y = 3√30/10. The two points of intersection are (√30/10, −3√30/10) and (−√30/10, 3√30/10), approximately (0.5477, −1.6432) and (−0.5477, 1.6432). Check the first point in the circle equation exactly: x² = 30/100 = 3/10 and y² = 9(30)/100 = 27/10, and 3/10 + 27/10 = 30/10 = 3. Check it in the line: −3(√30/10) = −3√30/10. Both equations hold. Graphically, the circle has radius √3 ≈ 1.732 and the line through the origin with slope −3 is steep, falling from upper left to lower right; it must cross the circle twice because it passes through the center, and the two crossings are symmetric about the origin, which the ± in the algebra reflects.

A line and a parabola with two solutions. Solve y = 2x − 7 and y = x² − 4x + 1. Set the expressions for y equal: x² − 4x + 1 = 2x − 7, so x² − 6x + 8 = 0, which factors as (x − 2)(x − 4) = 0. So x = 2 or x = 4. From the line, y = 2(2) − 7 = −3 and y = 2(4) − 7 = 1. Points: (2, −3) and (4, 1). Check in the parabola: 4 − 8 + 1 = −3 and 16 − 16 + 1 = 1. Both points are on both graphs.

One solution: a tangent line. Solve y = 2x − 1 and y = x². Then x² = 2x − 1, so x² − 2x + 1 = 0, which is (x − 1)² = 0, a repeated solution x = 1, with y = 1. The line touches the parabola at the single point (1, 1) without crossing it. Check: 1 = 2 − 1 and 1 = 1². The zero discriminant is the algebraic signature of tangency.

No solution. Solve y = x − 1 and y = x² + 2. Then x² + 2 = x − 1, so x² − x + 3 = 0, with discriminant 1 − 12 = −11 < 0. No real x satisfies it, so the line and parabola do not meet. Geometrically the parabola never goes below y = 2 and the line at x = 3 is at y = 2 while the parabola there is at 11; more precisely, the vertical gap between them is (x² + 2) − (x − 1) = x² − x + 3 = (x − 1/2)² + 11/4, which is never less than 11/4, so they never touch.

Exact and approximate solutions. Solve y = x + 1 and y = x² − 3. Then x² − 3 = x + 1, so x² − x − 4 = 0, and the formula gives x = (1 ± √17)/2, approximately 2.5616 and −1.5616. From the line, y = x + 1 = (3 ± √17)/2, approximately 3.5616 and −0.5616. Check the first point approximately in the parabola: 2.5616² − 3 ≈ 6.5618 − 3 = 3.5618, which matches 3.5616 to the rounding used. The exact coordinates are the solution; the decimals are what a graph would let you estimate, and reading (2.6, 3.6) off a drawing is a good estimate that is not the exact answer.

The misconception and why it is wrong. Students sometimes find the x-values and stop, or find the y-values by substituting x into the quadratic equation and get a value that does not lie on the line. A solution of a system in two variables is a point, so both coordinates are required, and the point must satisfy both equations. Using the linear equation to find y, then checking the point in the quadratic equation, avoids the error. A second misconception is that a line and a circle always meet twice; a line can miss a circle entirely, as x + y = 10 misses x² + y² = 3, because the closest point of the line to the origin is farther than the radius.

Worked line and circle not through the center: Solve x + y = 7 and x² + y² = 25. From the line, y = 7 − x. Substitute: x² + (7 − x)² = 25, so x² + 49 − 14x + x² = 25, 2x² − 14x + 24 = 0, x² − 7x + 12 = 0, (x − 3)(x − 4) = 0. So x = 3, y = 4 or x = 4, y = 3. Check (3, 4): 3 + 4 = 7 and 9 + 16 = 25. Check (4, 3): 7 and 16 + 9 = 25. The two points of intersection are (3, 4) and (4, 3), symmetric about the line y = x, as the symmetric equations suggest.

Worked line and parabola with factoring: Solve y = 3x − 9 and y = x² − 2x − 3. Then x² − 2x − 3 = 3x − 9, so x² − 5x + 6 = 0, (x − 2)(x − 3) = 0, x = 2 or x = 3. From the line: y = −3 and y = 0. Points (2, −3) and (3, 0). Check in the parabola: 4 − 4 − 3 = −3 and 9 − 6 − 3 = 0. Graphically, the parabola's lowest point is at x = 1, so both crossings are on its rising side: at (2, −3) and at the parabola's x-intercept (3, 0).

Worked circle and line through the origin with integer answers: Solve y = 2x and x² + y² = 20. Substitute: x² + 4x² = 20, 5x² = 20, x² = 4, x = ±2. Then y = 4 or y = −4. Points (2, 4) and (−2, −4). Check: 4 + 16 = 20 for both. The line passes through the center, so the two points are reflections of each other through the origin.

Worked irrational intersection with approximation: Solve y = x + 3 and y = x². Then x² − x − 3 = 0, x = (1 ± √13)/2 ≈ 2.3028 and −1.3028; y = x + 3 = (7 ± √13)/2 ≈ 5.3028 and 1.6972. Check (2.3028, 5.3028) in y = x²: 2.3028² ≈ 5.3029, agreeing to the rounding. Exact points: ((1 + √13)/2, (7 + √13)/2) and ((1 − √13)/2, (7 − √13)/2). A sketch shows the line cutting the parabola once left of the y-axis and once to the right, near x = 2.3, which is the approximation the graph supplies.

Intersection record: Draw, on separate axes, the circle x² + y² = 3 with the line y = −3x, the parabola y = x² − 4x + 1 with the line y = 2x − 7, and the parabola y = x² with the line y = 2x − 1. Mark and label every intersection point with its exact coordinates and, where irrational, a four-decimal approximation. Beside each drawing write the one-variable quadratic produced by substitution, its discriminant, and how the sign of the discriminant matches the number of points you drew.

Independent performance: Write three systems of your own, each consisting of one linear and one quadratic equation in two variables: one line-and-circle system with two solutions, one line-and-parabola system with exactly one solution (choose the line so the discriminant is zero), and one system with no real solution. Solve each algebraically by substitution, give exact coordinates for every point of intersection and decimal approximations where the coordinates are irrational, and check every point in both original equations. Sketch each system and explain in a sentence how the sketch confirms the number of solutions. Then write a short explanation of why substituting the linear equation into the quadratic one is the right move, and why the y-coordinate should be recovered from the linear equation. Success criteria: every solution is a point with both coordinates, every point is checked in both equations, the discriminant is used to predict the number of intersection points before they are found, and the no-solution system is explained both algebraically and geometrically.