Solve Linear-Quadratic Systems by Substitution
Solving linear-quadratic systems by substitution is a Grade 11 Algebra 1 algebraic method from enVision Chapter 9. The 6 steps: identify linear and quadratic equations, solve the linear equation for one variable, substitute into the quadratic, solve the resulting single-variable quadratic, substitute back to find the other variable, and verify as ordered pairs. For the system y = x - 1 and x^2 + y^2 = 13: substitute y = x-1 into the circle equation to get 2x^2 - 2x - 12 = 0, which factors to give solutions (3, 2) and (-2, -3).
Key Concepts
To solve a linear quadratic system by substitution:.
Step 1. Identify which equation is linear and which is quadratic. Step 2. Solve the linear equation for either variable. Step 3. Substitute the expression from Step 2 into the quadratic equation. Step 4. Solve the resulting quadratic equation. Step 5. Substitute each solution from Step 4 into the linear equation to find the other variable. Step 6. Write each solution as an ordered pair and check it in both original equations.
Common Questions
What are the steps to solve a linear-quadratic system by substitution?
Identify linear and quadratic equations, solve the linear for one variable, substitute into the quadratic, solve the resulting quadratic, substitute back to find the remaining variable, and write ordered pair solutions.
How do you solve the system y = x - 1 and x^2 + y^2 = 13?
Substitute y = x-1 into x^2 + y^2 = 13: x^2 + (x-1)^2 = 13. Expand to 2x^2 - 2x - 12 = 0. Factor: 2(x-3)(x+2) = 0. Solutions: x=3,y=2 and x=-2,y=-3.
Why is substitution useful for linear-quadratic systems?
The linear equation is easily solved for one variable, giving an expression that substitutes cleanly into the quadratic, reducing the system to a single-variable equation.
How many solutions can this method produce?
Zero, one, or two ordered pairs, depending on how many times the line intersects the quadratic curve.
Why must you substitute back into the linear equation (not the quadratic) to find y?
The linear equation is simpler and less prone to arithmetic errors. You can verify using the quadratic as a check.
What happens if the resulting quadratic has no real solutions?
The discriminant is negative, meaning the line and the quadratic curve do not intersect. The system has no real solutions.