### Video Transcript

The lines π₯ equals three π‘ one minus two, π¦ equals three π‘ one plus two, π§ equals nine π‘ one minus two and π₯ equals ππ‘ two minus two, π¦ equals π‘ two plus one, π§ equals ππ‘ two minus two are parallel. What is π plus π?

Our first line is represented by these three equations. And our second line is represented by these three equations. Now when you first see this, you might be tempted to try to do some kind of algebra, some kind of substitution. And you do have enough equations to do that. You could solve their repetitive substitution. But itβs going to take a very long time.

And thereβs a better way to do this. This format is called the parametric form of the equation line. And when weβre given this format and in this format the coefficients of the π‘-variable make up a vector that is parallel to this line. In line one, for the π₯ equals, there is a three as the coefficient to π‘ one. In the π¦-equation, there is a three as a coefficient to π‘ one. And in the π§-equation, thereβs a nine as the coefficient to the π‘ term.

Letβs find the vector for line two. For the π₯ terms, the coefficient of the π‘ two variable is π. If we look carefully at the π¦-equation, there is nothing by the π‘ two variable. And that means the coefficient is one. And finally the coefficient for the π§-equation is π.

So how does this help us? We know that these two lines are parallel. And that means these two vectors should be proportional to each other. If we look carefully at the third coefficient, we have a three in line one and a one in line two. To go from three to one, we divide by three. And that means the π-variable will be equal to nine divided by three. And the π-variable will be equal to three divided by three.

π equals one. And π equals three. Our goal is to solve π plus π. One plus three equals four.