Heart of algebra
Solving systems of linear equations — Harder example
- [Instructor] Consider the system of linear equations above. Which of the following choices of a will result in a system of equations with no solutions? No solutions. So a system has no solutions is if both lines and these are both linear equations, they actually tell us these are linear equations, is if you have two lines that are parallel, then you have no solutions. They are never going to intersect. There's not xy pair that satisfies both of them. So if they are parallel, if they are parallel, then you're going to have no solutions. So what makes two lines parallel? Well if they have the same slope but they have different y-intercepts, they have different y-intercepts, you're going to be parallel. So essentially we need to pick the a here that makes the second line parallel to the first. And to do that I'm actually write both of them in kind of the slope intercept form where y is equal to the slope times x plus the intercept. So let's do that. Let's do that for both of these. So first I'll do this one up here. So you have 9x minus 14y is equal to -3. See if I subtract 9x from both sides I would get -14y is equal to -9x minus three. See if I divide both sides by negative 14 I'm going to get y is equal to negative 9 divided by negative 14 is positive nine over 14x and then plus three over 14. So that's this line written in slope intercept form. Now let me write the second line in slope intercept form. So I have 2x minus ay is equal to -6 and let's see if I subtract two from both sides, I get -ay is equal to -2x. I'm subtracting 2x from both sides so I get minus 2x minus six and then I can divide both sides by negative a and I get y is equal to -2 divided by negative a is positive two over a x, and then plus six over six over a. Alright so we need to set up a situation, we need to set up a situation where two over a is equal to nine over 14. These two things have to have the same slope. And then when we're able to figure out that a we have to verify that they have different y-intercepts 'cause if they have the same slope and the same y-intercept instead of having no solutions, they would have an infinite number of solutions 'cause then it would be the same line. But let's solve for a. So we know that nine over 14 we know that nine over 14 needs to be equal to two over a. Two over a, or another way of thinking about this, there's a bunch of sometimes people say cross multiply and all of that. I like to just do logical algebraic operations. But we can take the reciprocal of both sides so we could have, and actually let me swap both sides. So we could say a over two is equal to 14 over nine. Is equal to 14 over nine. And then multiply both sides by two. Multiply both sides by two, and you're gonna get a is equal to 28 over nine. A is going to be this right over here. This is 28 over nine. So if I'm under time pressure I already see a choice that's looking pretty good. But if we really want to care that we've, if a is 28 over nine, these two things are going to have the same slope. But let's make sure they don't have the same y-intercept and I wouldn't do this if I was under time pressure. But just for to feel good about it, if this y-intercept is six over a. So it's going to be six over 28 over nine, which is six divided by a fraction is the same thing dividing by a fraction is the same thing as multiplying by its reciprocal, so it's gonna be the same thing as six times nine over 28. And this is going to be equal to six times nine is 54 over 28, which is clearly different than three over 14 so if a is 28 over nine, same slope, different y-intercepts and you're dealing with parallel lines. They will not intersect and you're going to have no solutions.