### Video Transcript

Find the equation of the straight line that passes through the point of intersection of the two straight lines π₯ minus eight π¦ equals two and negative six π₯ minus eight π¦ equals one and is parallel to the π¦-axis.

So weβre looking for the equation of this particular straight line. And the first thing we notice that this line is parallel to the π¦-axis. Now we can recall that lines which are parallel to the π¦-axis are vertical lines. And they therefore have equations of the form π₯ equals π, where π is some constant. And π is the value at which these lines cross the π₯-axis.

We also know that the straight line weβre looking for passes through the point of intersection of the two straight lines whose equations weβve been given. So what weβre going to need to do is solve these two equations simultaneously in order to find their point of intersection. Or at least find the π₯-coordinate of their point of intersection. So these are the two equations weβre looking to solve. Theyβre linear simultaneous equations in two variables.

There are a couple of different approaches we could take. Firstly, we could note that both equations have the same coefficient of π¦. They both have negative eight π¦. And so we could use the acronym SSS, standing for same sign subtract, in order to eliminate the π¦-variables. If we subtract equation two from equation one, then we have π₯ minus negative six π₯, which is π₯ plus six π₯ or seven π₯. We have negative eight π¦ minus negative eight π¦. Thatβs negative eight π¦ plus eight π¦. So the π¦s are eliminated. And on the right-hand side, we have two minus one, which is equal to one. We therefore have the equation seven π₯ equals one, and weβve eliminated the π¦-variable. To solve this equation, we just need to divide both sides by seven, giving π₯ equals one-seventh.

An alternative method would be to use the method of substitution. We could rearrange equation one to give negative eight π¦ is equal to two minus π₯. And we can then substitute this expression for negative eight π¦ into equation two. Doing so gives negative six π₯ plus two minus π₯ is equal to one. And now we have an equation in π₯ only.

We can then simplify to give negative seven π₯ plus two equals one. Subtract two from each side to give negative seven π₯ equals negative one, and then divide by negative seven. Giving π₯ equals negative one over negative seven, which is equal to one-seventh. In both cases then, we found that the π₯-coordinate of the point of intersection of these two straight lines is one-seventh.

Now we donβt need to go any further because we donβt need to know the π¦-coordinate of the point of intersection. Remember, we said straight lines parallel to the π¦-axis have equations of the form π₯ equals some constant. So if our line passes through the point where π₯ equals one-seventh, its equation must just be π₯ equals one-seventh.

So by first recalling the general equation of a straight line which is parallel to the π¦-axis and then partially solving the equations of the two straight lines to find the π₯-coordinate of their point of intersection. Weβve found that the equation of the straight line that passes through their point of intersection and is parallel to the π¦-axis is π₯ equals one-seventh.