### Video Transcript

By drawing the straight lines five π¦ equals π₯ plus one and π¦ equals negative three π₯ plus five on a grid, find their point of intersection.

In this problem, we are given these two equations, and we need to find their point of intersection. Thatβs the point where the lines meet or cross. So, there are a number of different ways in which we can determine what the graphs of these two equations will look like. However, the most useful way to create the graph of any type of equation is by creating a table of values. A table of values will allow us to create ordered pairs of the input or π₯-values and the output or π¦-values associated with those. So, letβs take the first equation five π¦ equals π₯ plus one.

We might choose to rearrange this so that we have π¦ on its own as the subject of this equation. By dividing through by five, we can say that this equation is π¦ equals π₯ plus one over five. We now need to choose some input values. We can choose any values, but keeping some simple values and a negative and a positive value is always good practice. So, letβs pick π₯ is equal to negative one, zero, and one. We can then take this equation π¦ equals π₯ plus one over five and substitute in the value π₯ is equal to negative one. This gives us π¦ is equal to negative one plus one over five. And the numerator negative one plus one simplifies to zero, and zero over five is also zero. So, the output value π¦ is zero when π₯ is equal to negative one.

We can then take the second input value π₯ is equal to zero and substitute this into the equation, which gives us a π¦-value of one-fifth. To find the third π¦-value in the table, we substitute in π₯ is equal to one, which gives us π¦ is equal to two-fifths. We now know that there are three coordinates that lie on the line five π¦ equals π₯ plus one. They are negative one, zero; zero, one-fifth; and one, two-fifths. As an aside, we really only need two coordinates to create a straight line. However, if we work out three coordinates, then we have a checking system. When we draw these three points, if one of them doesnβt lie on the line, then we know that we must have made a mistake somewhere.

But before we plot these three coordinates on a grid, letβs get three coordinates for the second equation. That means weβll know exactly how large the graph should be. Weβll take the same three π₯-values of negative one, zero, and one. Only this time, we substitute the π₯-values into the equation π¦ equals negative three π₯ plus five. Substituting in π₯ is equal to negative one, we have π¦ is equal to negative three times negative one plus five, which is equal to eight. The first coordinate we have found for this line then is negative one, eight.

Substituting in π₯ is equal to zero gives us π¦ is equal to five, and substituting in π₯ is equal to one gives us π¦ is equal to two. We have therefore found the two remaining coordinates zero, five and one, two. We can then choose some grid paper and draw the grid. When we start plotting the coordinates for the line of five π¦ equals π₯ plus one, we might realize that itβs quite difficult to plot these small π¦-values of one-fifth and two-fifths, in which case we might choose to go back to the table of values and pick some points which might make plotting a little easier.

Remembering that this equation is equivalent to π¦ equals π₯ plus one over five, we might choose to input the π₯-value of four. Doing so would give us a π¦-value equal to one. So, we know that the coordinate four, one lies on this line. Adding this coordinate to the grid allows us to more easily draw this equation five π¦ equals π₯ plus one. We can then complete this graph drawing with the second graph of π¦ equals negative three π₯ plus five. And so, to answer this question, we need to find the point of intersection. We can see that these two lines meet at the point which has an π₯-value of 1.5 and a π¦-value of 0.5. We can therefore say that the point of intersection is 1.5, 0.5.