Simultaneous Equations with One Quadratic
Solve a linear and a quadratic equation together by substituting the linear one into the quadratic, and read the answers as the points where a line meets a curve.
What a learner can do afterwards
- Substitute the linear equation into the quadratic and collect terms to one side
- Give the answers as coordinate pairs, not just x values
- Say what a single repeated solution means about the line and the curve
1 · Read
You have sketched parabolas. A nonvertical line and a quadratic graph meet where both equations hold. Each solution is a coordinate pair (x, y), not an x-value alone. Substitution first finds intersection x-coordinates; substitute each into either equation to find y.
If y = ax² + bx + c and y = mx + n, equate their right sides and subtract mx + n: ax² + (b - m)x + c - n = 0. Check both subtractions before solving.
For y = x² - 14 and y = 2x + 1, equate the right sides: x² - 14 = 2x + 1. Collect to get x² - 2x - 15 = 0. Find two numbers with product -15 and sum -2: -5 and 3. Factorise as (x - 5)(x + 3) = 0. Set either factor to zero, giving x = 5 or x = -3. These are the intersection x-coordinates.
The discriminant of ax² + bx + c = 0 is D = b² - 4ac. For x² + 6x + 9 = 0, a = 1, b = 6 and c = 9. Thus D = 36 - 36 = 0.
A negative discriminant gives no real roots. For x² + 4x + 7 = 0, D = 16 - 28 = -12. No real x makes the two substituted expressions equal, so there are no real intersections.
For a known intersection x-coordinate, use the line to find y. If y = 3x + 2 and x = -4, then y = -12 + 2 = -10. Keep the values together as (-4, -10).
Both equations give the same y at a true intersection. Use whichever has less arithmetic for the particular value. For a line already written as y = mx + n, direct substitution is often convenient, but it is not a universal rule.
After substitution, roots describe all possible intersection x-coordinates. If D < 0, there are none. This conclusion concerns real intersections and does not require attempting factorisation.
When the collected equation is monic, look for two numbers whose product is its constant and whose sum is its x coefficient. x² - 7x + 10 = (x - 2)(x - 5), giving roots 2 and 5.
Solve y = x² - 11 and y = 4x + 1. Equating gives x² - 4x - 12 = 0. Numbers -6 and 2 have product -12 and sum -4, so factorise as (x - 6)(x + 2) = 0. Thus x = 6 or -2. The line gives y = 25 or -7. The intersection points are (6, 25) and (-2, -7).
Count distinct real roots of the substituted quadratic to count intersections: D > 0 gives two, D = 0 gives one and D < 0 gives none. A repeated root is one x-coordinate, not two distinct points.
For y = x² and y = 6x - 9, substitution gives (x - 3)² = 0. There is one root, x = 3, and y = 9. The nonvertical line is tangent to the quadratic at (3, 9).
To reconstruct a line through two points, calculate m = (y₂ - y₁)/(x₂ - x₁), then find c in y = mx + c. Through (0, 2) and (2, 8), m = 6/2 = 3 and c = 2, so y = 3x + 2.
To find a sum of intersection x-coordinates, first obtain both roots. For y = x² - 19 and y = 2x + 5, equating and collecting gives x² - 2x - 24 = 0. Numbers -6 and 4 have product -24 and sum -2, so (x - 6)(x + 4) = 0. The roots are 6 and -4. Add them with their signs: 6 + (-4) = 2. Do not add y-coordinates or absolute values.
When both equations contain an x term, collect that term carefully. Equating x² + bx + c with mx + n changes the x coefficient to b - m and the constant to c - n. Then factorise or use the formula.
For a nonvertical tangent to a quadratic-function graph, substitution gives a quadratic with a repeated root. A quadratic has no further distinct root, so this line meets that parabola only at the tangent point. This claim does not apply to every kind of curve.
For ax² + bx + c = 0, use x = (-b ± √(b² - 4ac))/(2a). For x² - 4x - 2 = 0, x = (4 ± √24)/2 = 2 ± √6. Evaluate the requested sign before rounding.
You can establish that a line misses a quadratic by substitution. Keep the signs when calculating D. For x² - 2x + 5 = 0, D = 4 - 20 = -16, proving that there are no real intersections.
Check a proposed intersection by substituting both coordinates into both original equations. Even with a correct x-value, the y-value must also be correct. If a point fails either equation, replace or reject that proposed pair.
A downward quadratic can be solved with the same method. After equating it with the line, multiplying the entire equation by -1 may make factorisation easier. This changes neither its roots nor the corresponding intersection points.
For y = x² and y = 3x + k, substitution gives x² - 3x - k = 0. Tangency requires D = 9 + 4k = 0, so k = -9/4. Find the parameter before finding the tangent point.
A line with gradient m and y-intercept c has equation y = mx + c. Build that equation from the given information before substituting into the quadratic. A point (0, c) gives the intercept directly.
For y = x + c and y = x² - 5x + 10, substitution gives x² - 6x + 10 - c = 0. No real intersections require D = 36 - 4(10 - c) < 0. Hence c < 1; the largest allowed integer is 0.
Find the tangent to y = x² - 10x + 21 at (3, 0). Write y = m(x - 3). Substitution gives x² - (10 + m)x + 21 + 3m = 0. A repeated root at 3 means this equals (x - 3)². Compare coefficients: 10 + m = 6, so m = -4; also 21 + 3m = 9. The line is y = -4x + 12.
You can reconstruct a system from its collected equation. If the equation is Q(x) - L(x) = 0, it comes from y = Q(x) and y = L(x). Expand and collect to check a proposed pair of original rules.
To test supplied candidate points, substitute each into both equations rather than solving again. Keep only pairs for which both equalities hold. A point on the line alone is not enough.
Substitution can eliminate x instead of y. If x = y - 2 and x = y² - 8, equating gives y² - y - 6 = 0. Solve for y, then substitute into x = y - 2. The final pair still uses the order (x, y).
For a horizontal line y = k and a quadratic y = (x - h)² + v, substitution gives (x - h)² = k - v. A positive right side gives two intersections, zero gives tangency, and a negative right side gives none.
Substitute, collect terms and solve; then find y for each x. Two distinct real roots give two points, a repeated root gives tangency, and a negative discriminant gives no real meeting points.
The rewritten reading was written for this dataset, with solving notes and additional assessment questions written for Marble.
2 · Watch
3 · Play
Slide the line across the curve: two crossings, one touch, or a clean miss.
Take it off screen
Where it sits
Where this leads
Jobs that lean on this skill. Follow one to see everything it is built on.
29 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.