### Video Transcript

The given graph shows the lines π¦ equals three π₯ minus one and π¦ equals four π₯ minus seven. Determine the point whose coordinates are solutions to both equations simultaneously.

So we have on our graph two lines. So weβve got π¦ equals three π₯ minus one and π¦ equals four π₯ minus seven. And we know that the blue line is π¦ equals three π₯ minus one and the orange line is π¦ equals four π₯ minus seven. And we know this because actually both of our equations are in the form π¦ equals ππ₯ plus π, so the general form for a straight line. And what we have is that the π β so the coefficient of π₯ β is our slope. But the last number β so our π β is actually the π¦-intercept. And we can actually see that the blue line crosses the π¦-axis at negative one and the orange line crosses the π¦-axis at negative seven. So this is how we actually identified them.

Okay, so now, what we want to do is actually want to find the point whose coordinates are solutions to both equations. So the way weβre gonna do this is actually use two methods: the first one is actually graphically and the second method is actually to do it algebraically.

So for the first method, what we do is we look at the graph. And we actually see right do the two lines actually meet? So yes, they do. They actually cross at the point Iβve circled here in pink. And then, what we do is we actually read off the values for π₯ and π¦ for where the point is. So we have an π₯-value of six and a π¦-value of 17. So therefore, the point whose coordinates are solutions to both equations simultaneously is six, 17 cause like we said the π₯-coordinate was six and the π¦-coordinate was 17.

Okay, great, so weβve actually solved it graphically. What I can also do is actually show you how to do it algebraically. Well, if weβre gonna solve it algebraically, what we can actually do is say that well we know that one line is π¦ equals three π₯ minus one and the other line is π¦ equals four π₯ minus seven. So therefore, where they meet theyβre actually gonna be equal to each other. So therefore, we can actually form the equation three π₯ minus one equals four π₯ minus seven. And when we do this, it means we can actually now solve this to find our π₯-value.

So as we actually have the greater number of π₯ is on the right-hand side, so our coefficient of π₯ is greater on the right-hand side, what weβre gonna do is actually subtract three π₯ from both sides of the equation. And when we do that, we get negative one is equal to π₯ minus seven. And then what I need to do is actually add seven to each side of the equation. And when we do that, we get six is equal to π₯. So we got an π₯-value of six.

Okay, great, so now, what we need to do is actually find out what our π¦-value is going to be. And the way we do this is we actually substitute π₯ equals six into one of our equations. And what Iβm gonna do is actually substitute it into π¦ equals three π₯ minus one. It doesnβt matter which one you substitute it into. It should work for either of them.

So when we do this, weβre gonna get π¦ is equal to three multiplied by six minus one. And thatβs because weβve substituted six in for π₯. So therefore, weβre gonna say that π¦ is equal to 18 minus one, which is 17. So this again gives us the coordinates of a point which are six, 17, where π₯-value is six, π¦-value is 17.

So great, this is actually exactly the same as what we found using the graphical method. So therefore, we can say that if the graph shows the lines π¦ equals three π₯ minus one and π¦ equals four π₯ minus seven, the point whose coordinates are solutions to both equations simultaneously is six, 17.