Quadratic Simultaneous Equations (Edexcel IGCSE Maths B): Revision Note

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 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

How do I solve quadratic simultaneous equations using algebra?

• 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 $y-2x=5$

  • Rearrange the linear equation into $y=2x+5$

  • Substitute this into the quadratic equation, replacing all y's with $\left(2x+5\right)$

    •  $x^2+{\left(2x+5\right)}^2=25$

• Expand and solve this quadratic equation

  • $x^2+\left(2x+5\right)\left(2x+5\right)=25$

  • $\begin{array}{rcl}{x}^2+4x^2+20x+25& =& 25\end{array}$

  • $\begin{array}{rcl}5x^2+20x& =& 0\end{array}$

  • $\begin{array}{rcl}5x\left(x+4\right)& =& 0\end{array}$

  • $x=0$ and $x=-4$

• Substitute each value of x into the linear equation, $y=2x+5$, to find the corresponding y values

  • $y=2\left(0\right)+5=5$

  • $y=2\left(-4\right)+5=-3$

• 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 $xy=3$ and $x+y=4$

• Either:

  • Rearrange the second equation to $y=4-x$ and substitute into $xy=3$

    • $x\left(4-x\right)=3$

    • Expanding produces a quadratic that can be solved for $x$

    • $4x-x^2=3$

  • 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 $x$

    • $x^2+3=4x$