What if the resulting quadratic equation yields only one solution?

This indicates that the line is tangent to the quadratic curve, intersecting it at exactly one coordinate point.

What if the quadratic equation yields no real solutions?

If the discriminant $b^2 - 4ac$ is negative, the line and the curve do not intersect in the real Cartesian plane.

Can I use the elimination method instead?

Elimination is generally inefficient for these systems unless the non-squared variables align perfectly. Substitution is the standard and universally reliable method.

How to solve a system with one linear and one quadratic equation

Solving a system with one linear and one quadratic equation requires the substitution method. This procedure applies when finding the geometric intersection points of a line and a conic section such as a parabola or circle.

The setup

Identify the linear equation and the quadratic equation in the system. The linear equation contains variables with a maximum degree of 1, whereas the quadratic equation contains at least one squared term.

The steps

Isolate one variable (either $x$ or $y$ ) in the linear equation. 2. Substitute this isolated expression into the quadratic equation in place of the corresponding variable. 3. Expand and rearrange the resulting equation into standard quadratic form $ax^2 + bx + c = 0$ . 4. Solve for the remaining variable using factoring or the quadratic formula. 5. Substitute these values back into the isolated linear equation to compute the corresponding values of the other variable. 6. State the final solutions as coordinate pairs $(x, y)$ .

Checking the result

Substitute each coordinate pair $(x, y)$ into both original equations. Both equations must evaluate to true statements for the pair to be a confirmed solution.

Common errors

Failing to expand binomials correctly, such as writing $(x+a)^2 = x^2 + a^2$ instead of $x^2 + 2ax + a^2$ . Substituting the first variable's solutions back into the quadratic equation instead of the linear one, which can generate extraneous solutions.

Worked example

Solve the system $y = x^2 - 4x + 3$ and $y = 2x - 5$ .

The linear equation is already isolated for $y$ : $y = 2x - 5$ . Substitute $2x - 5$ for $y$ in the quadratic equation: $2x - 5 = x^2 - 4x + 3$ . Subtract $2x$ and add $5$ to both sides to equate to zero: $0 = x^2 - 6x + 8$ . Factor the quadratic equation: $0 = (x - 2)(x - 4)$ . Solve for $x$ : $x = 2$ or $x = 4$ . Substitute $x = 2$ into the linear equation: $y = 2(2) - 5 = -1$ . First solution is $(2, -1)$ . Substitute $x = 4$ into the linear equation: $y = 2(4) - 5 = 3$ . Second solution is $(4, 3)$ . The solutions to the system are $(2, -1)$ and $(4, 3)$ .

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References: OpenStax College Algebra Section 7.1 · Khan Academy Unit: Nonlinear systems of equations