Quadratic Simultaneous Equations (Edexcel IGCSE Maths)

Spec point — 2.7D

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 $( 2 x + 5 )$

    • $x^{2} + ( 2 x + 5 )^{2} = 25$

• Expand and solve this quadratic equation

  • $x^{2} + ( 2 x + 5 ) ( 2 x + 5 ) = 25$
  • $\begin{gathered} x^{2} + 4 x^{2} + 20 x + 25 & = 25 \end{gathered}$
  • $\begin{gathered} 5 x^{2} + 20 x & = 0 \end{gathered}$
  • $\begin{gathered} 5 x ( x + 4 ) & = 0 \end{gathered}$
  • $x=0$ and $x=-4$

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

  • $y = 2 ( 0 ) + 5 = 5$
  • $y = 2 ( - 4 ) + 5 = - 3$

• 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 $xy=3$ and $x+y=4$

• Either:

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

    • $x ( 4 - x ) = 3$
    • 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 $x$
    • $x^{2}+3=4x$

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

Worked Example Solve the equations

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

Answer:

Number the equations

$\begin{gathered} x^{2} + y^{2} & = 36 \\ x - 2 y & = 6 \end{gathered}\quad \text circled{1}$ $> \text circled{2}$* *

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

$x=2y+6$

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

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

Simplify

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

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

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

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

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

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

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

Solve the linear equation above

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

So the two $y$ values are

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

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

$x=2y+6$

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

Write the final solutions in clear pairs

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