Quadratic Simultaneous Equations (Cambridge (CIE) IGCSE International Maths)

Revision Note

Author

Mark Curtis

Expertise

Maths

Quadratic Simultaneous Equations

What are quadratic simultaneous equations?

When there are two unknowns (e.g. x and y) in a problem, we need two equations to be able to find them both; these are called simultaneous equations
If there is an x² or y² or xy in one of the equations then they are quadratic (or non-linear) simultaneous equations

How do I solve quadratic simultaneous equations?

Use substitution
- Substitute the linear equation, y = ... (or x = ...), into the quadratic equation
  - Do not try to substitute the quadratic equation into the linear equation
E.g. To solve $x^{2} + y^{2} = 25$ and $y - 2 x = 5$
- Rearrange the linear equation into $y = 2 x + 5$
- Substitute this into the quadratic equation, replacing all y's with $(2 x + 5)$
  - $x^{2} + {(2 x + 5)}^{2} = 25$
Expand and solve this quadratic equation
- $\begin{array}{rcl} x^{2} + 4 x^{2} + 20 x + 25 & = & 25 \end{array}$
- $\begin{array}{rcl} 5 x^{2} + 20 x & = & 0 \end{array}$
- $\begin{array}{rcl} 5 x (x + 4) & = & 0 \end{array}$
- $x = 0$ and $x = - 4$
Substitute each value of x into the linear equation, $y = 2 x + 5$ , to find the corresponding y values
- $y = 2 (0) + 5 = 5$
- $y = 2 (- 4) + 5 = - 3$
Present your solutions in a way that makes it obvious which x belongs to which y
- x = 0, y = 5 or x = -4, y = -3

Check your final solutions satisfy both equations

What if the quadratic has repeated roots or no roots?

If the resulting quadratic after substituting has a repeated root,
- then the line is a tangent to the curve
  - i.e. the curve and the line intersect in one place only
- There is only one solution for x and y
If the resulting quadratic to be solved has no roots,
- then the line does not intersect with the curve
- There are no solutions to the simultaneous equations
- If this happens it may be an indicator that your working is wrong!

What if I can't substitute one equation into the other straight away?

If the linear equation is not in the form y = ... or x = ...
- You will need to rearrange it first, so that it can be substituted into the quadratic equation
Consider solving $x y = 3$ and $x + y = 4$
Either:
- Rearrange the second equation to $y = 4 - x$ and substitute into $x y = 3$
  - $x (4 - x) = 3$
  - Expanding produces a quadratic that can be solved for $x$
  - $4 x - x^{2} = 3$
- Or rearrange the first equation to $y = \frac{3}{x}$ and substitute into $x + y = 4$
  - $x + \frac{3}{x} = 4$
  - Multiplying both sides by $x$ produces a quadratic that can be solved for $x$
  - $x^{2} + 3 = 4 x$

How do I use a graph to solve quadratic simultaneous equations?

Plot both equations on the same set of axes
- To do this, you can use a table of values
- Or for straight lines it can help to rearrange into y = mx + c
Find the point where the lines intersect
- The x and y solutions to the simultaneous equations are the x and y coordinates of the point of intersection
E.g. To solve y = x² + 3x + 1 and y = 2x + 1 simultaneously
- First plot them both (see graph below)
- Then find the points of intersection, (-1, -1) and (0, 1)
- So the solutions are x = -1 and y = -1 or x = 0 and y = 1

Graphs of x^2 + 3x + 1 and y=2x+1 on the same graph

Exam Tip

When giving your final answer, make sure you indicate which x and y values go together

Worked Example

Solve the equations

$\begin{array}{rcl} x^{2} + y^{2} & = & 36 \\ x - 2 y & = & 6 \end{array}$

Number the equations

$\begin{array}{rcl} x^{2} + y^{2} & = & 36 \\ x - 2 y & = & 6 \end{array}$ $1 2$

There is one quadratic equation and one linear equation so this must be done by substitution

Equation 2 can be rearranged to make $x$ the subject, which can then be substituted into equation 1

You could rearrange to make $y$ the subject instead, but this results in a fraction which can be more tricky to deal with

Rearranging equation 2

$x = 2 y + 6$

Substituting into equation 1

${(2 y + 6)}^{2} + y^{2} = 36$

Expand the brackets
Remember that a bracket squared should be treated the same as double brackets

$\begin{array}{rcl} (2 y + 6) (2 y + 6) + y^{2} & = & 36 \\ 4 y^{2} + 6 (2 y) + 6 (2 y) + 6^{2} + y^{2} & = & 36 \end{array}$

Simplify

$\begin{array}{rcl} 4 y^{2} + 12 y + 12 y + 36 + y^{2} & = & 36 \\ 5 y^{2} + 24 y + 36 & = & 36 \end{array}$

Rearrange to form a quadratic equation that is equal to zero
Do this by subtracting 36 from both sides

$\begin{array}{rcl} 5 y^{2} + 24 y & = & 0 \end{array}$

Take out the common factor of $\begin{array}{rcl} y \end{array}$

$\begin{array}{rcl} y (5 y + 24) & = & 0 \end{array}$

Solve to find the values of $y$ by equating each factor to zero

$y = 0$ or $5 y + 24 = 0$

Solve the linear equation above

$y = - \frac{24}{5}$

So the two $y$ values are

$\begin{array}{rcl} y_{1} & = & 0 \\ y_{2} & = & - \frac{24}{5} \end{array}$

Substitute the values of $y$ into one of the equations (the linear equation is easiest) to find the values of $x$

$x = 2 y + 6$

$x_{1} = 2 (0) + 6 = 6 x_{2} = 2 (- \frac{24}{5}) + 6 = - \frac{18}{5}$

Write the final solutions in clear pairs

$x_{1} = 6, y_{1} = 0$
$x_{2} = - \frac{18}{5}, y_{2} = - \frac{24}{5}$

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