Solving Linear-Quadratic Systems Algebraically

Solve linear-quadratic systems using substitution and elimination. Math 20-1 Alberta.

Lesson 2.3 of Linear Functions & Systems of Equations in Math 20-1 — Alberta curriculum lessons.

What you'll learn in this lesson

• The Method. Learn the two algebraic strategies — substitution and elimination — for solving linear-quadratic systems, and use the discriminant to predict how many solutions exist.
• Substitution Method. Solve a linear-quadratic system by isolating y in the linear equation, substituting into the quadratic, and finding both ordered-pair solutions.
• Elimination Method. Eliminate y by multiplying and adding the equations, then solve the resulting single-variable quadratic — including a system with no real solutions.
• Word Problems and Special Cases. Translate word problems into systems, handle context restrictions, work with right-triangle geometry, and find a parameter value that forces exactly one solution.

Inside the lesson — a free preview

The Big Idea

What is a Linear-Quadratic System?. A linear-quadratic system pairs one linear equation (a line) with one quadratic equation (a parabola). Solving algebraically means finding the coordinates of every point where they intersect.

There are two algebraic strategies:

Strategy 1 — Substitution. Solve the linear equation for one variable (usually y), then substitute that expression into the quadratic equation. The result is a single-variable quadratic you can solve.

Strategy 2 — Elimination. Add or subtract a multiple of one equation from the other to eliminate y. You must eliminate y — not x — because the quadratic contains an x^2 term that cannot be easily removed.

Why not eliminate x? Eliminating x would leave a messy expression because the quadratic's x^2 term has no matching term in the linear equation.

Steps for Solving Algebraically

Follow these six steps for any linear-quadratic system solved algebraically.

Possible Outcomes — Using the Discriminant

How Many Solutions?. After substituting or eliminating, you are left with a quadratic equation ax^2 + bx + c = 0. The number of real solutions depends on the discriminant D = b^2 - 4ac.

Worked example: Verify if a Point is a Solution

Part a) Is (3, 1) a solution to y = -x^2 + 10 and x - y = 2?. A point is a solution only if it satisfies both equations.

Test in equation 1: y = -x^2 + 10

1 = -(3)^2 + 10

1 = -9 + 10

1 = 1 ✓

Test in equation 2: x - y = 2

3 - 1 = 2

2 = 2 ✓

Yes — (3, 1) is a solution.

Part b) Is (3, -6) a solution to x - 2y = 15 and y = x^2 - x - 6?. Test in equation 1: x - 2y = 15

3 - 2(-6) = 15

3 + 12 = 15

15 = 15 ✓

Test in equation 2: y = x^2 - x - 6

-6 = (3)^2 - 3 - 6

-6 = 9 - 3 - 6

-6 = 0 ✗

No — (3, -6) is NOT a solution. It satisfies equation 1 but fails equation 2. Both must be true.

Key Takeaway — Both Equations Must Check Out. Even if a point works in one equation, it is not a solution to the system unless it satisfies both equations. Always verify in every equation.

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