Simultaneous Quadratic Equations are a more complicated version of simultaneous linear equations. In this new quadratic type, one of the two equations will have an \(x^2\)
and/or \(y^2\) term. In the familiar linear type, both equations only have terms like \(3x, -6x, 3y, or -2y\).
Remember in linear equations, the highest power of \(x\) or \(y\) is 1.
In quadratic equations it is 2.
So \(5x+3y=7\) is linear.
\(5x^2+3y=7\) is quadratic, as is \(5x+3y^2=7\).
\(5x^2+3y^3\) is neither linear or quadratic. Although there is an \(x^2\) term, the highest power is \(3\), in \(3y^3\).
Solving them is not much different to the substitution method for simultaneous linear equations. The only difference is that you end up getting a quadratic equation to solve, and hence get up to two sets of \(x,y\) values instead of just one.
For example the two sets of values could be \(x=3, y=4\), or \(x=-1,y=2\).
In graphical terms, these represent the points where two curves intersect. While linear equations give straight lines with at most one intersection point, quadratic equations can create curves that intersect at up to two points, giving us up to two solutions.
What You Need to Know
Higher content only:
- Solve simultaneous equations including linear/quadratic pairs
- How to interpret the solutions graphically
🔑 Key fact
These type of questions are usually worth more marks than the average question, so if you get them right it gives your score in the exam a really good boost.
Steps For Solving Simultaneous Quadratic Equations
The easiest approach for solving a simultaneous quadratic-linear equation pair is the substitution method:
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Rearrange the linear equation to express one variable in terms of the other
So something like \(y=3x+5\), or \(x= 3 – 2y\)
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Substitute this expression into the quadratic equation
You get a new quadratic equation with just \(x\) or just \(y\) in it
- Solve this new quadratic equation to find possible values for one variable
- Substitute these values back into the linear equation to find the corresponding values of the other variable
- Write the solutions as pairs of coordinates \((x, y)\)
Let’s work through examples to see how this works.
Example 1
Solve these simultaneous equations:
\(y = x^2 – 2x + 1\) (quadratic equation)
\(y+3 = 3x\) (linear equation)
Step 1: Rearrange the linear equation to express \(y\) in terms of \(x\).
\(y + 3 = 3x\)
\(y = 3x – 3\)
Step 2: Substitute this expression for \(y\) into the quadratic equation.
\(y = x^2 – 2x + 1\)
\(3x – 3 = x^2 – 2x + 1\)
Step 3: Rearrange to get a standard form quadratic equation with zero on one side.
\(3x – 3 = x^2 – 2x + 1\)
\(3x – 3 – x^2 + 2x – 1 = 0\)
\(-x^2 + 5x – 4 = 0\)
\(x^2 – 5x + 4 = 0\) (multiplying everything by -1)
Step 4: Solve the quadratic equation. We can factorise this as:
\((x – 1)(x – 4) = 0\)
\(x = 1\) or \(x = 4\)
Step 5: Find the corresponding \(y$-values using the linear equation.
For $x = 1\): \(y = 3(1) – 3 = 0\)
For \(x = 4\): \(y = 3(4) – 3 = 9\)
So the solutions are \((1, 0)\) and \((4, 9)\)
Example 2
Solve these simultaneous equations: \(y = x^2 – 2x + 1\) (quadratic equation) \(y + 3 = 3x\) (linear equation)
Step 1: Rearrange the linear equation to express \(y\) in terms of \(x\).
\(y + 3 = 3x\)
\(y = 3x – 3\)
Step 2: Substitute this expression for \(y\) into the quadratic equation.
\(y = x^2 – 2x + 1\)
\(\therefore 3x – 3 = x^2 – 2x + 1\)
Step 3: Rearrange to get a standard form quadratic equation with zero on one side.
\(3x – 3 = x^2 – 2x + 1\)
\(\therefore 3x – 3 – x^2 + 2x – 1 = 0\)
\(\therefore -x^2 + 5x – 4 = 0\)
\(\therefore x^2 – 5x + 4 = 0\) (multiplying everything by -1)
Step 4:Solve the quadratic equation. We can factorise this as:
\((x – 1)(x – 4) = 0\)
So \(x = 1\) or \(x = 4\)
Step 5: Find the corresponding \(y$-values using the linear equation.
For $x = 1\): \(y = 3(1) – 3 = 0\)
For \(x = 4\): \(y = 3(4) – 3 = 9\)
So the solutions are \((1, 0)\) and \((4, 9)\)
Example 3
Solve these simultaneous equations: \(x^2 + y^2 = 25\) (quadratic equation) \(y = x + 1\) (linear equation)
Step 1: The quadratic equation has both \(x^2\) and \(y^2\) terms, so we need to substitute the expression for \(y\) from the linear equation, \(y = x + 1\).
Step 2: Substitute this into the quadratic equation.
Put \(y = x + 1\) into \(x^2 + y^2 = 25\)
\(\therefore x^2 + (x + 1)^2 = 25\)
\(\therefore x^2 + (x^2 + 2x + 1) = 25\)
Step 3: Rearrange to get a standard form quadratic equation with zero on one side.
\(x^2 + (x^2 + 2x + 1) = 25\)
\(\therefore 2x^2 + 2x + 1 = 25\)
\(\therefore 2x^2 + 2x – 24 = 0\)
\(\therefore x^2 + x – 12 = 0\) (dividing by 2)
Step 4: Solve the quadratic equation. We can factorise this as:
\(x^2 + x – 12 = 0\)
\(\therefore (x + 4)(x – 3) = 0\)
So \(x = -4\) or \(x = 3\)
Step 5: Find the corresponding \(y$-values using the linear equation.
For $x = -4\): \(y = -4 + 1 = -3\)
For \(x = 3\): \(y = 3 + 1 = 4\)
So the solutions are \((-4, -3)\) and \((3, 4)\)
Example 4
This example is hard, so don’t feel obliged to complete it. It is very unlikely to be this hard in the exam.
Occasionally, the solutions will not have nice values.
Solve these simultaneous equations: \(y = 2x^2 – 5x\) and \(y = 3 – 7x\)
Step 1: Set the two expressions for \(y\) equal to each other.
\(2x^2 – 5x = 3 – 7x\)
Step 2: Rearrange into the standard quadratic format.
\(2x^2 – 5x = 3 – 7x\)
\(2x^2 – 5x + 7x – 3 = 0\)
\(2x^2 + 2x – 3 = 0\)
Step 3: Solve the quadratic equation. This doesn’t factorise nicely, so we’ll use the quadratic formula:
\(x = \dfrac{-b \pm \sqrt{b^2 – 4ac}}{2a}\)
With \(a = 2\), \(b = 2\), and \(c = -3\):
\(x = \dfrac{-2 \pm \sqrt{4 + 24}}{4}\)
\(x = \dfrac{-2 \pm \sqrt{28}}{4}\)
\(x = \dfrac{-2 \pm 2\sqrt{7}}{4}\)
\(x = \dfrac{-1 \pm \sqrt{7}}{2}\)
So \(x = \dfrac{-1 + \sqrt{7}}{2} \approx 0.82\)
or \(x = \dfrac{-1 – \sqrt{7}}{2} \approx -1.82\)
Step 4: Find the corresponding \(y$-values.
For $x \approx 0.82\): \(y = 3 – 7(0.82) \approx 3 – 5.74 \approx -2.74\)
For \(x \approx -1.82\): \(y = 3 – 7(-1.82) \approx 3 + 12.74 \approx 15.74\)
The solutions are approximately \((0.82, -2.74)\) and \((-1.82, 15.74)\)
(You can verify these by substituting into the original equations)
Graphical Approach to Solving Simultaneous Equations
🔑 Key Fact
The solutions to simultaneous equations correspond to the coordinates of the points where the two curves intersect.
When we have two equations:
- A linear equation creates a straight line on a graph.
- A quadratic equation typically creates a parabola (a U-shaped curve).
- Draw the straight line and the parabola on the same graph.
- The solutions to the simultaneous equations are the points where these curves intersect.
Questions
The first three questions have integer solutions for both x and y, while the last two questions have fractional solutions.
Solve the following systems of equations:
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\(x² + 2y = 10\)
\(x – y = -1\)
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\(2x² + 3y = 30\)
\(2x – 5 = y – 3\)
-
\(2x² + y = 7\)
\(3x – 2y = -13\)
-
\(x² + 8y = 10\)
\(3x + 4y = 9\)
-
\(9x² + 4y = 5\)
\(6x – y = \dfrac{15}{4}\)
Solutions
Question 1
Solve the system of equations: \(x^2 + 2y = 10\) and \(x – y = -1\)
Step 1: Rearrange the linear equation to make \(y\) the subject.
\(x – y = -1\)
\(y = x + 1\)
Step 2: Substitute this expression for \(y\) into the quadratic equation.
\(x^2 + 2y = 10\)
\(x^2 + 2(x + 1) = 10\)
\(x^2 + 2x + 2 = 10\)
\(x^2 + 2x – 8 = 0\)
Step 3: Solve the resulting quadratic equation.
\(x^2 + 2x – 8 = 0\)
\((x + 4)(x – 2) = 0\)
\(x = -4\) or \(x = 2\)
Step 4: Find the corresponding \(y\) values.
For \(x = -4\): \(y = -4 + 1 = -3\)
For \(x = 2\): \(y = 2 + 1 = 3\)
Step 5: Check the solutions in both original equations.
When \(x = -4, y = -3\):
\(x^2 + 2y = (-4)^2 + 2(-3) = 16 – 6 = 10\) ✓
\(x – y = -4 – (-3) = -4 + 3 = -1\) ✓
When \(x = 2, y = 3\):
\(x^2 + 2y = 2^2 + 2(3) = 4 + 6 = 10\) ✓
\(x – y = 2 – 3 = -1\) ✓
Therefore, the solutions are \((x, y) = (-4, -3)\) or \((x, y) = (2, 3)\).
Question 2
Solve the system of equations: \(2x^2 + 3y = 30\) and \(2x – 5 = y – 3\)
Step 1: Rearrange the linear equation to make \(y\) the subject.
\(2x – 5 = y – 3\)
\(2x – 5 + 3 = y\)
\(y = 2x – 2\)
Step 2: Substitute this expression for \(y\) into the quadratic equation.
\(2x^2 + 3y = 30\)
\(2x^2 + 3(2x – 2) = 30\)
\(2x^2 + 6x – 6 = 30\)
\(2x^2 + 6x – 36 = 0\)
\(x^2 + 3x – 18 = 0\)
Step 3: Solve the resulting quadratic equation.
\(x^2 + 3x – 18 = 0\)
\((x + 6)(x – 3) = 0\)
\(x = -6\) or \(x = 3\)
Step 4: Find the corresponding \(y\) values.
For \(x = -6\): \(y = 2(-6) – 2 = -12 – 2 = -14\)
For \(x = 3\): \(y = 2(3) – 2 = 6 – 2 = 4\)
Step 5: Check the solutions in both original equations.
When \(x = -6, y = -14\):
\(2x^2 + 3y = 2(-6)^2 + 3(-14) = 2(36) – 42 = 72 – 42 = 30\) ✓
\(2x – 5 = y – 3\) → \(2(-6) – 5 = -14 – 3\) → \(-12 – 5 = -17\) → \(-17 = -17\) ✓
When \(x = 3, y = 4\):
\(2x^2 + 3y = 2(3)^2 + 3(4) = 2(9) + 12 = 18 + 12 = 30\) ✓
\(2x – 5 = y – 3\) → \(2(3) – 5 = 4 – 3\) → \(6 – 5 = 1\) → \(1 = 1\) ✓
Therefore, the solutions are \((x, y) = (-6, -14)\) or \((x, y) = (3, 4)\).
Question 3
Solve the system of equations: \(2x^2 + y = 7\) and \(3x – 2y = -13\)
Step 1: Rearrange the linear equation to make \(y\) the subject.
\(3x – 2y = -13\)
\(-2y = -13 – 3x\)
\(y = \dfrac{13 + 3x}{2}\)
Step 2: Substitute this expression for \(y\) into the quadratic equation.
\(2x^2 + y = 7\)
\(2x^2 + \dfrac{13 + 3x}{2} = 7\)
\(4x^2 + 13 + 3x = 14\)
\(4x^2 + 3x – 1 = 0\)
Step 3: Solve the resulting quadratic equation.
\(4x^2 + 3x – 1 = 0\)
\((4x – 1)(x + 1) = 0\)
\(x = \dfrac{1}{4}\) or \(x = -1\)
Step 4: Find the corresponding \(y\) values.
For \(x = \dfrac{1}{4}\): \(y = \dfrac{13 + 3(\dfrac{1}{4})}{2} = \dfrac{13 + \dfrac{3}{4}}{2} = \dfrac{13\cdot 4 + 3}{8} = \dfrac{52 + 3}{8} = \dfrac{55}{8}\)
For \(x = -1\): \(y = \dfrac{13 + 3(-1)}{2} = \dfrac{13 – 3}{2} = \dfrac{10}{2} = 5\)
Step 5: Check the solutions in both original equations.
When \(x = \dfrac{1}{4}, y = \dfrac{55}{8}\):
\(2x^2 + y = 2(\dfrac{1}{4})^2 + \dfrac{55}{8} = 2(\dfrac{1}{16}) + \dfrac{55}{8} = \dfrac{2}{16} + \dfrac{55}{8} = \dfrac{1}{8} + \dfrac{55}{8} = \dfrac{56}{8} = 7\) ✓
\(3x – 2y = 3(\dfrac{1}{4}) – 2(\dfrac{55}{8}) = \dfrac{3}{4} – \dfrac{110}{8} = \dfrac{6}{8} – \dfrac{110}{8} = -\dfrac{104}{8} = -13\) ✓
When \(x = -1, y = 5\):
\(2x^2 + y = 2(-1)^2 + 5 = 2(1) + 5 = 2 + 5 = 7\) ✓
\(3x – 2y = 3(-1) – 2(5) = -3 – 10 = -13\) ✓
Therefore, the solutions are \((x, y) = (\dfrac{1}{4}, \dfrac{55}{8})\) or \((x, y) = (-1, 5)\).
Question 4
Solve the system of equations: \(x^2 + 8y = 10\) and \(3x + 4y = 9\)
Step 1: Rearrange the linear equation to make \(y\) the subject.
\(3x + 4y = 9\)
\(4y = 9 – 3x\)
\(y = \dfrac{9 – 3x}{4}\)
Step 2: Substitute this expression for \(y\) into the quadratic equation.
\(x^2 + 8y = 10\)
\(x^2 + 8(\dfrac{9 – 3x}{4}) = 10\)
\(x^2 + \dfrac{8(9 – 3x)}{4} = 10\)
\(x^2 + \dfrac{72 – 24x}{4} = 10\)
\(x^2 + 18 – 6x = 10\)
\(x^2 – 6x + 8 = 0\)
Step 3: Solve the resulting quadratic equation.
\(x^2 – 6x + 8 = 0\)
\((x – 4)(x – 2) = 0\)
\(x = 4\) or \(x = 2\)
Step 4: Find the corresponding \(y\) values.
For \(x = 4\): \(y = \dfrac{9 – 3(4)}{4} = \dfrac{9 – 12}{4} = \dfrac{-3}{4} = -\dfrac{3}{4}\)
For \(x = 2\): \(y = \dfrac{9 – 3(2)}{4} = \dfrac{9 – 6}{4} = \dfrac{3}{4}\)
Step 5: Check the solutions in both original equations.
When \(x = 4, y = -\dfrac{3}{4}\):
\(x^2 + 8y = 4^2 + 8(-\dfrac{3}{4}) = 16 – 6 = 10\) ✓
\(3x + 4y = 3(4) + 4(-\dfrac{3}{4}) = 12 – 3 = 9\) ✓
When \(x = 2, y = \dfrac{3}{4}\):
\(x^2 + 8y = 2^2 + 8(\dfrac{3}{4}) = 4 + 6 = 10\) ✓
\(3x + 4y = 3(2) + 4(\dfrac{3}{4}) = 6 + 3 = 9\) ✓
Therefore, the solutions are \((x, y) = (4, -\dfrac{3}{4})\) or \((x, y) = (2, \dfrac{3}{4})\).
Question 5
Solve the system of equations: \(9x^2 + 4y = 5\) and \(6x – y = \dfrac{15}{4}\)
Step 1: Rearrange the linear equation to make \(y\) the subject.
\(6x – y = \dfrac{15}{4}\)
\(-y = \dfrac{15}{4} – 6x\)
\(y = 6x – \dfrac{15}{4}\)
Step 2: Substitute this expression for \(y\) into the quadratic equation.
\(9x^2 + 4y = 5\)
\(9x^2 + 4(6x – \dfrac{15}{4}) = 5\)
\(9x^2 + 24x – 15 = 5\)
\(9x^2 + 24x – 20 = 0\)
Step 3: Solve the resulting quadratic equation.
\(9x^2 + 24x – 20 = 0\)
\(9x^2 + 30x – 6x – 20 = 0\)
\(3x(3x + 10) – 2(3x + 10) = 0\)
\((3x + 10)(3x – 2) = 0\)
\(x = -\dfrac{10}{3}\) or \(x = \dfrac{2}{3}\)
Step 4: Find the corresponding \(y\) values.
For \(x = -\dfrac{10}{3}\): \(y = 6(-\dfrac{10}{3}) – \dfrac{15}{4} = -20 – \dfrac{15}{4} = -\dfrac{80}{4} – \dfrac{15}{4} = -\dfrac{95}{4}\)
For \(x = \dfrac{2}{3}\): \(y = 6(\dfrac{2}{3}) – \dfrac{15}{4} = 4 – \dfrac{15}{4} = \dfrac{16}{4} – \dfrac{15}{4} = \dfrac{1}{4}\)
Step 5: Check the solutions in both original equations.
When \(x = -\dfrac{10}{3}, y = -\dfrac{95}{4}\):
\(9x^2 + 4y = 9(-\dfrac{10}{3})^2 + 4(-\dfrac{95}{4}) = 9(\dfrac{100}{9}) – 95 = 100 – 95 = 5\) ✓
\(6x – y = 6(-\dfrac{10}{3}) – (-\dfrac{95}{4}) = -20 + \dfrac{95}{4} = -\dfrac{80}{4} + \dfrac{95}{4} = \dfrac{15}{4}\) ✓
When \(x = \dfrac{2}{3}, y = \dfrac{1}{4}\):
\(9x^2 + 4y = 9(\dfrac{2}{3})^2 + 4(\dfrac{1}{4}) = 9(\dfrac{4}{9}) + 1 = 4 + 1 = 5\) ✓
\(6x – y = 6(\dfrac{2}{3}) – \dfrac{1}{4} = 4 – \dfrac{1}{4} = \dfrac{16}{4} – \dfrac{1}{4} = \dfrac{15}{4}\) ✓
Therefore, the solutions are \((x, y) = (-\dfrac{10}{3}, -\dfrac{95}{4})\) or \((x, y) = (\dfrac{2}{3}, \dfrac{1}{4})\).
By Atomic