### Video Transcript

Consider the linear equation π¦ equals five minus two π₯. We can draw a straight line to represent this equation. Complete the table to find the coordinates of points on the line.

In order to work out the coordinates, we need to substitute the π₯-values into the equation π¦ equals five minus two π₯. When π₯ is equal to negative two, π¦ is equal to five minus two multiplied by negative two. Negative two multiplied by negative two is positive four. Therefore, π¦ is equal to nine.

When π₯ equals negative one, π¦ is equal to five minus two multiplied by negative one. This is equal to seven as five plus two equals seven.

When π₯ is equal to zero, π¦ is equal to five minus two multiplied by zero. This gives us an answer of five when π₯ equals zero.

substituting in π₯ equals one into the equation five minus two π₯ gives us a π¦-value of three. And finally, when π₯ is equal to two, π¦ is equal to one.

This gives us a set of five coordinates that lie on the line π¦ equals five minus two π₯. They are negative two, nine; negative one, seven; zero, five; one, three; and two, one.

The second part of the question asked us which of the following is the graph of the equation. All four of the graphs have a π¦-intercept of five. The second and fourth graph have a positive gradient. So we can rule these out as our equation has a gradient or slope of negative two. The third graph has a gradient or slope of negative one, whereas the first graph has a slope of negative two. Therefore, the graph that shows the equation π¦ equals five minus two π₯ is the first graph.

The final part of the question asked, which of these points is not on the line? Is it A six, negative six; B four, negative three; C negative three, 11; D three, negative one; or E negative four, 13.

The π¦-coordinate of all the points that lie on the line π¦ equals five minus two π₯ must be odd. This means that point A six, negative six is not on the line. We can check this by substituting π₯ equals six into the equation five minus two multiplied by six. This gives us an answer of negative seven and not negative six. The point six, negative seven lies on the line, but the point six, negative six does not. All four other points four, negative three; negative three, 11; three, negative one; negative four, 13 lie on the line π¦ equals five minus two π₯.