Simultaneous equations with a quadratic · Grade 9
In Simultaneous equations with a quadratic (Grade 9) you practise 3 skills: A parabola and a line · Two parabolas · Solving from a graph. When one or both of the equations are quadratic, a solution of the simultaneous equations is a pair of numbers that satisfies both equations. On a graph, each solution is a point the two graphs share: where the parabola and the line, or the two parabolas, meet.
To solve, set the two expressions equal, gather everything on one side and solve the quadratic equation you get. Each solution gives the x coordinate of a point; substitute it into one of the rules to find the y coordinate. A quadratic equation can have two solutions, one or none. So a parabola and a line can share two points, one point (when the line touches the parabola) or none.
Tip: Check each solution by substituting it into both equations: a point that fits only one of them is not a solution.
What you learn
A parabola and a line
For example: Here are the simultaneous equations: y = x² − 3x + 4 y = −6x + 8 Find the values of x where the parabola and the line cross.
Answer and solution
The answer: x = −4 or x = 1
Set the two expressions for y equal: x² − 3x + 4 = −6x + 8 Gather everything on one side: x² + 3x − 4 = 0 The expression under the square root in the quadratic formula: 3² − 4 × 1 × (−4) = 25 √25 = 5 x = (−3 + 5) ÷ 2 = 1 x = (−3 − 5) ÷ 2 = −4
Two parabolas
For example: Here are the simultaneous equations: y = x² + 2 y = −x² + 4 Find the values of x where the two parabolas meet.
Answer and solution
The answer: x = −1 or x = 1
Set the two expressions for y equal: x² + 2 = −x² + 4 Gather everything on one side: 2x² − 2 = 0 2x² = 2 x² = 1 so x = 1 or x = −1
Solving from a graph
For example:
The picture shows the graphs of the simultaneous equations: y = x² + 4x + 3 y = 3x + 3 Read from the picture the point on the left where the parabola and the line cross.Answer and solution
The answer: (−1, 0)
The parabola and the line meet at (−1, 0) and (0, 3) . The one on the left is (−1, 0) . Check by substituting: y = (−1)² + 4 × (−1) + 3 = 0 and y = 3 × (−1) + 3 = 0
Not on the site yet
Parts of this topic that are not built yet: Simultaneous equations with a circle · Drawing the graphs.
The exercises are made by rules and the answers are worked out in code, so there are new exercises every time. The topic follows the Israeli Ministry of Education's grade 9 maths curriculum (2026–27).