Calculating solutions to a linear equation
About this video
Sal finds the missing y-value in a solution to y = 2x - 5 when x = 1, first by reading the graph and then by substituting into the equation. He shows that points on the line satisfy the equation and points off the line do not.
You'll be able to
- substitute an x-value into a linear equation to find the matching y-value.
- read a solution of a linear equation from its graph.
- explain that every point on the graph is a solution and every solution is on the graph.
Key words
- solution
- An (x, y) pair that makes the equation true when you substitute it in.
- substitute
- Replace a variable with a number to see what the expression equals.
- graph of an equation
- The set of all points whose (x, y) pairs make the equation true.
Transcript
- [Instructor] We're told consider the equation Y is equal to 2X minus five. Complete the missing value in the solution to the equation. So pause this video, have a go at that. So there's a couple of ways you could do this. You could just substitute X equals one into this original equation and figure out what you would get for Y.
And actually I'll do that in a second. That's actually probably the way that I would approach it. But they also gave us this graph, so maybe we should use this. Maybe that's what we're here for. Try to think about how to look at it from a graphical point of view.
So when X is equal to one, we can see that assuming this is the graph of Y is equal to two X minus five, what does it look like Y is equal to? It looks like Y is equal to negative three. But let's validate that using the technique I just said. So if we say Y is equal to two times one minus five, that is equal to two minus five, which is indeed equal to negative three.
So this checks out. You could also validate that something that's not on the graph right over here is not going to satisfy that equation. For example, if I picked this point right over here, one, negative one, if I had X equals one, two times one minus five, that is two, that is once again negative three, that's not going to be negative one.
Or maybe I could pick a point like this that's not going to be on the graph. Zero, zero not on the graph. So anything or not, and also does not satisfy the equation. So maybe the idea behind this question is just for you to appreciate that this graph is going to have any value, any value on this graph will satisfy this equation, and any value, any X,Y pair that satisfies this equation is going to sit on this graph.
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