Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College solving systems of equations by substitution and elimination by substitution. We're going to look at the two equations and figure out if there's a single variable we could solve for easily. In this first equation, we can see if I took the Y to the other side, I could get X by itself. So I get X equal y + 6. Then what we're going to do is we're going to substitute it into the second equation. So we're going to have two X + 3 Y equals 7, but instead of the X, we're going to put in what the X equal, which was y + 6. And what this does is it gives us a new equation that is only in one variable. So if we distribute two y + 12 + 3, Y equals 7 5Y. If I subtract the 12 to the other side, we get -5. So we can see Y is -1. If Y is -1, I can then substitute it in to figure out what my X is. So X is going to be 5. So this is the .5 negative 1. The facts that there is a solution means it's consistent and they're intersecting lines so they're in dependent. Now if we wanted to we could always check does 5 - -1 = 6 and the answer is yes. Now remember we have to check it in both lines. So 2 * 5 -, + 3 * -1 does that equal 710 -? 3 does equal 7. The reason we have to check it in both equations is it might work in one, but it might not work in the other. So it might be a point on one of the two lines, not on the other. In this case it is the intersecting point elimination. What we do is we want to try to get the coefficients on one or the other of the variables to be additive inverses. I personally look for if there's one that's a minus and one that's a plus. And then I think about what do I have to do to the coefficients to get them the same. Well, that top equation, I'm going to multiply 3 by three. If I multiply by three, I get three X -, 3 Y equal 18. I'm going to leave the second equation the way it is. 2X plus three y = 7 and I'm going to add. When I add these two equations, I get 5X. The YS are going to cancel because they're additive inverses 18 and sevens 25. If I divide by 5, I get X equal 5. Now go back and do this again, getting rid of the X's. I'm going to take this top equation and multiply it this time by a -2 so -2 X +2 Y equal -12. The second equation stays the same, 2X plus three y = 7. When we add here, we can see 5 Y equal -5 Y equal -1. Elimination is going to be much more advantageous as our problems get more difficult, especially once we start doing 3 variables. So let's look at another example. I'm going to do it both ways and this one. The first one is already solved for Y, so we're going to have -4 X -2 * -2 X plus 4 equaling -8 negative four X + 4 X -8 equal -8 negative 8 equal -8. What's it mean if the two sides are exactly the same and there are no variables? Well, it means it's a true statement, so the two lines must be the same line. So in this case we can say X, Y such that Y equal negative two X + 4. They're the same line, so they're consistent and they're dependent. If we wanted to do the same problem with elimination, the first step would be to get all the XS and YS on one side. So I'm going to add the 2X to the other side. Now I need them to be additive inverses. We can multiply the top one by a two, but we could also divide the second one by a 2. So let's do an example of showing a division. So I'm going to take that second one and divide by 2. Remember I can multiply or divide, it doesn't matter. So the top one I'm going to leave the same the bottom and I'm going to divide by 2. So I get negative two X -, y equaling -4. When I add those two together, the X's canceled, the Y's canceled, the four's cancel. 0 = 0 true statement, parallel lines. Once again, remember to write the solution correctly. Parenthesis X, Y such that Y equal negative two X + 4 consistent and dependent. Let's look at one more. If we have three X -, y equal 4, we want to do this by substitution. I'm going to choose this Y to solve for because it's the only one with A1 coefficient. So I'm going to have negative Y equal negative three X + 4 and divide by a one negative 1. So Y equal three X - 4. So when I put it into the other equation, we get negative six X + 2 * 3 X -4 equaling 5, negative six X + 6 X -8 equal 5. With this case we get -8 equal 5. Is that a true statement? No, So this is going to be parallel lines. No solution. So they're in dependent and they're inconsistent looking over here doing it. The same problem with elimination. I'm going to multiply this top one by A2. SO six X -, 2 Y equal 8. Leave the bottom one alone -6 X +2 Y equal 5. When I add the X's and the Y's cancel, so we get 0 equaling 13, zero equaling 13's. Not a true statement. So they're parallel lines. No solution, independent and consistent. You try the next one. Thank you and have a wonderful day.