Systems of Equations

New Zealand

Level 7 - NCEA Level 2

Lesson

We've already looked at how to solve a system of simultaneous equations using the substitution method and the elimination method. The coordinates that we found using these algebraic methods were the point of intersection of two straight lines.

We can use similar processes to find the point of intersection between a straight line and a curve.

Remember!

There may be 1 or 2 points of intersection. In fact there may be heaps depending on the type of curve that the line is intersecting with.

To find a point of intersection, we can graph the curve and the straight line to see where the two lines cross. Remember, there may be more than one point of intersection.

Consider the equations $y=4x-x^2$`y`=4`x`−`x`2 and $y=-2x+5$`y`=−2`x`+5.

Complete the table of values for the two equations.

$x$ `x`$1$1 $2$2 $3$3 $y=4x-x^2$ `y`=4`x`−`x`2$\editable{}$ $\editable{}$ $\editable{}$ $y=-2x+5$ `y`=−2`x`+5$\editable{}$ $\editable{}$ $\editable{}$ On the same set of axes, graph $y=4x-x^2$

`y`=4`x`−`x`2 and $y=-2x+5$`y`=−2`x`+5.Loading Graph...Hence state the coordinates of the points of intersection in the form $\left(x,y\right)$(

`x`,`y`) on the same line separated by a comma.

Consider the circle with equation $x^2+y^2=45$`x`2+`y`2=45 and the line with equation $y=-2x$`y`=−2`x`.

Form an equation to solve for the $x$

`x`-coordinate(s) of their point(s) of intersection.Hence state the coordinates of their points of intersection in the form $\left(x,y\right)$(

`x`,`y`). Write the coordinates of each point**on a separate line**.

Finding the intersection algebraically requires us to find where the two equations are equal. That is, we make one equation equal to the other and then solve.

Here is an example of finding the intersection of a straight line and quadratic algebraically.

The straight line $y=5x-6$`y`=5`x`−6, intersects the quadratic $y=x^2-9x+18$`y`=`x`2−9`x`+18 in $2$2 places. Find the coordinate of the point with the value of $x<4.5$`x`<4.5

So we start by setting the $2$2 equations to be equal, and then we solve.

$5x-6$5x−6 |
$=$= | $x^2-9x+18$x2−9x+18 |

$0$0 | $=$= | $x^2-14x+24$x2−14x+24 |

$0$0 | $=$= | $(x-12)(x-2)$(x−12)(x−2) |

By applying the null factor theorem here, I can see that either $x=12,orx=2$`x`=12,`o``r``x`=2 are solutions. These are the two x-coordinates of the $2$2 points of intersections. I only need the coordinate for the $x=2$`x`=2 one though.

So substituting back into either equation the value of $x=2$`x`=2, will give us the y-coordinate.

$y=5x-6$`y`=5`x`−6, so at $x=2,y=4$`x`=2,`y`=4. Hence the coordinate of the point of intersection is (2,4).

This is also confirmed using the graph.

Consider the circle with equation $x^2+y^2=22$`x`2+`y`2=22 and the line with equation $y=2$`y`=2.

Form an equation to solve for the $x$

`x`-coordinate(s) of their point(s) of intersection, leaving your answer in exact surd form.Hence state the coordinates of their points of intersection in the form $\left(x,y\right)$(

`x`,`y`). Write the coordinates of each point**on a separate line**.