Simultaneous Equations

We can think of simultaneous equations as being two equations in two unknowns, say x and y. Note that the word simultaneous means ‘at the same time’. It follows that for the values of x and y found both equations must be true at the same time. Sometimes it is easy to inspect the equations and guess the answers. However, when one of the equations is quadratic this becomes less likely. The answers could be surds, in which case, this is very difficult to guess.

Note that a question may ask you to solve simultaneous equations explicitly. In others, it will be implied and you must deduce that it is simultaneous equations to solve. For example, the question could ask you  to find out which points two curves have in common. See Example 2. (What if there are 3 equations?)

Methods for Solving Simultaneous Equations

There are three methods for solving simultaneous equations:

  1. Elimination– this is where you multiply both equations through by different coefficient in order to eliminate one of the unknowns. This page will focus on substitution since it works for more complicated simultaneous equations. For example, when one of the equations is a quadratic. Click here to see an example using elimination.
  2. Substitution – one of the equations can be quadratic, in which case, substitution is the method to use. You will need to know how to solve quadratics. Make x or y the subject of one of the equations, then substitute into the other. See the Worked Example and Example 1.
  3. Graphical method – we can interpret the solution of simultaneous equations as the intersection of their graphs. This plot shows the graphs of y=2x-3 in red and 4x+5y=6 in blue. Their intersection lies on the x-axis and has coordinates (1.5,0). This is the solution when solving simultaneously. Also see Example 2.

Simultaneous Equations Worked Example

Solve the simultaneous equations




This example requires solution via substitution, i.e. make either x or y the subject of one equation and insert it into the other. The obvious choice would be to make x the subject of the second equation – it is the quickest, least complicated choice. The second equation tells us that x=5-2y. We can insert this into the first equation: (5-2y)^2+y^2=10. By multiplying out the brackets and simplifying we see that this is a quadratic equation in y:


Write out the brackets: (5-2y)(5-2y)+y^2=10

Expand the brackets: 25-10y-10y+4y^2+y^2=10

Simplify: 5y^2-20y+15=0

Divide both by sides by 5: y^2-4y+3=0

Factorise: (y-3)(y-1)=0

This tells us that y has to be either 3 or 1. If y=3, then x=5-2\times 3=-1 (from the second equation rearranged) and if y=1 then x=5-2\times 1=3.

We obtain the solutions (x_1,y_1)=(-1,3) and (x_2,y_2)=(3,1).


Solve the simultaneous equations:


Substitute the expression for y in the first equation into the second then simplified to give:

2x^2-x(x-4)=8\Rightarrow x^2+4x=8

Rearrange and complete the square to solve the resulting quadratic as follows:

x^2+4x-8=0\Rightarrow (x+2)^2-12=0\Rightarrow x=-2\pm\sqrt{12}

Simplify the surd to get x=-2\pm2\sqrt{3} with corresponding y solutions given by y=-6\pm2\sqrt{3}. Note that these y solutions are best found by substituting x into the most basic of the original equations.

Sketch the graphs of x^2+y^2=10 and x+2y=5 on the same plot. Determine the coordinates of the intersection points.


The first equation represents a circle centred at the origin with radius \sqrt{10}. The second equation is a straight line. Both can be seen in this figure.

The coordinates of the intersection points were calculated in the worked example. They are given by (-1,3) and (3,1).


Sketching a quadratic against a cubic and using algebra to find their intersection points.

Using simultaneous equations and discriminants to find an inequality in terms of unknowns satisfied when a straight line cuts a circle.

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