### Video Transcript

Find the slope of the line negative two π₯ plus three π¦ minus two equals zero and the π¦-intercept of this line.

The slope or gradient of a straight line is the steepness of that line. It tells us for every one unit the line moves to the right, how many units up or down the line moves. Lines with a positive slope slope upwards from left to right. And, lines of a negative slope slope downwards from left to right. The π¦-intercept of a line is the π¦-value at which the line intercepts the π¦-axis.

We donβt need to draw a graph of a straight line in order to work out its slope and π¦-intercept. We can actually determine each of these values if the equation of our straight line is in a particular form, slope-intercept form, π¦ equals ππ₯ plus π. If the equation of a straight line is in this particular form, then the value of π, the coefficient of π₯, gives the slope of the line. And, the value of π, the constant term, gives the π¦-intercept.

The equation of the line weβve been given is not in this form. But by rearranging the equation, we can bring it into the form π¦ equals ππ₯ plus π. We need to perform a series of steps which will leave π¦ on its own on the left-hand side of this equation. We know that, to begin with, π₯ and π¦ are both on the left-hand side of the equation. So, we start by adding two π₯ to each side. This will leave three π¦ minus two on the left-hand side. And on the right-hand side, we now have two π₯.

Next, we add two to both sides of our equation, leaving just three π¦ on the left-hand side. And on the right-hand side, we now have two π₯ plus two. We could have performed these last two steps in either order. We could have added two first and then added two π₯, if we wish, or indeed added both terms in the same step. This is starting to look a little bit more like π¦ equals ππ₯ plus π. But, instead of having π¦ on its own on the left-hand side, we have three π¦.

So, our next step is going to be to divide both sides of this equation by three. Doing so leaves just π¦ on its own on the left-hand side because three π¦ divided by three is just π¦. And on the right-hand side, we have two π₯ plus two all over three. Now, we can separate this fraction into the sum of two separate fractions with the same denominator of three. Two π₯ plus two all over three is equal to two π₯ over three plus two over three. And, we can write two π₯ over three as just two over three multiplied by π₯ or two-thirds of π₯.

Now, compare this to the general form, π¦ equals ππ₯ plus π, that we wrote down to begin with. We can identify that the slope of our line is the coefficient of π₯, which is two-thirds. We must make sure we include both parts of this fraction. The slope isnβt just two. π₯ is being multiplied by two over three, two-thirds. The π¦-intercept of our line is the constant term. Thatβs positive two-thirds.

Now, we donβt actually need to include that positive sign as this is a positive value. But certainly, if it was a negative value, we would need to include the negative sign. By rearranging the equation of this straight line into the slope-intercept form, π¦ equals ππ₯ plus π. Weβve found then that both the slope and the π¦-intercept of this line are two-thirds.