In the previous section we saw how to use graphs to solve simultaneous equations. In this section we will look at ways to solve simultaneous equations algebraically.

To solve a set of simultaneous equations you need to have as many equations as you have variables. For example, to find the point of intersection of two straight lines you need the equation of each of the lines. There are many ways to solve sets of equations but each method relies on creating an equation with one variable, solving that equation then back substituting to find the other variable(s). In Example 5.1 we have $y=$ one function of $x$ and $y=$ a different function of $x$. At the point of intersection $y$ is the same in both equations so we can set then equal to each other which eliminates the $y$ term.

**Sanity Check:** Once we have values for $x$ and $y$ we need to check whether our answer is correct. We do this by inserting the value for $x$ into the other equation. If the two values for $y$ are the same then you can be confident the answers are correct.

**Example 5.1**: Find the point of intersection of $y = 2x - 3$ and $y = -x + 3$

We have $y$ | $=$ | $2x-3$ | (1) | |

and $y$ | $=$ | $-x+3$ | (2) | |

so we can write $2x-3$ | $=$ | $-x+3$ | ||

Rearranging we get $2x+x$ | $=$ | $3+3$ | ||

simplifying $3x$ | $=$ | $6$ | ||

so $x$ | $=$ | $2$ | ||

Substituting $x=2$ into equation (1) we get $y=1$ | ||||

Sanity Check: | ||||

Substituting $x=2$ into equation (2) we get $y=1$ so we can be confident the point of intersection is at (2, 1) |

Finding the point of intersection of two straight lines is straight forward. You set the lines equal to each other and solve for $x$ and $y$. If the lines are parallel they do not intersect otherwise you can calculate the point of intersection.

In principle finding points of intersection for higher order curves is the same as for straight lines; set the lines equal to each other then solve for $x$ and $y$. Naturally, the solution can be more involved than for the case of straight lines.

**Example 5.2**: Find the points of intersection between $y = x^2+2x - 3$ and $y = 3x-1$

We have $y$ | $=$ | $x^2+2x-3$ | (1) |

and $y$ | $=$ | $3x-1$ | (2) |

so $x^2+2x-3$ | $=$ | $3x-1$ | |

Rearranging $x^2-x-2$ | $=$ | $0$ | |

$=$ | $(x+1)(x-2)$ | ||

which means the lines intersect at $x=-1$ and $x=2$. Using equation (1) we can see that when $x=-1\ \ y=-4$ and when $x=2\ \ y=5$ so the points of intersection are at $(-1, -4)$ and $(2, 5)$ | |||

Sanity Check: | |||

Using equation (2) we can see that when $x=-1\ \ y=-4$ and when $x=2\ \ y=5$ so we can be confident that the points of intersection are at $(-1, -4)$ and $(2, 5)$ |