Systems of equations with elimination: -3y+4x=11 & y+2x=13

Solve for x and y. And we have two equations with two unknowns. Negative 3y plus 4x is equal to 11, and y plus 2x is equal to 13. What we could do is try to solve this using elimination. Maybe we can add these two equations together to get some variables to cancel out. But if we just add y and negative 3y, those won't cancel out. And if we just add a 2x and 4x, those won't cancel out. But maybe we can scale one of these equations, so that we will get some cancellation. If this y could somehow become a 3y, then 3y and negative 3y would cancel out. And so the easiest way to turn this y into a 3y is to just multiply this entire equation, so we don't actually change the information in it, by 3. So let's multiply both sides by 3. If we multiply the left-hand side by 3, we get 3y plus 6x. We have to multiply every term by 3. 3y plus 6x is equal to 3 times 13 is 39. So this equation and this equation are the exact same equation. I've just multiplied both sides by 3 to go from here to here. So I multiplied by 3. And of course, we have this original equation up here. That negative 3y plus 4x is equal to 11. And now, if we add the left-hand side and the right-hand sides of this equation, something interesting happens. The negative 3y and the positive 3y cancel out. That was the whole point behind multiplying this second guy over here by 3. So let's add them. So if we add, these guys cancel out. We have 4x plus 6x is 10x. And that is equal to 11 plus 39, which is 50. And then we can divide both sides by 10. We get x is equal to 50 divided by 10 is 5. And then we could go substitute back to solve for y. If x is 5, we could use the second equation right over here. We get y plus 2 times 5. I'll do it in that same color. 2 times 5 is equal to 13. Or y plus 10 is equal to 13. We can subtract 10 from both sides. And we will get y. These cancel out. y is equal to 3. So our solution that we get for x and y is that x is equal to 5. And y is equal to 3. And we can verify that it works in both equations. In this top equation-- let me do it in a new color-- negative 3 times 3 plus 4 times 5 is-- this is negative 9 plus 20, which is, indeed, equal to 11. So both of these satisfy the first equation. And then if we take the second equation, 3 plus 2 times 5, that's 3 plus 10, that does, indeed, equal 13. So it definitely satisfies both. And I want to be clear. You could have solved this with substitution. Or you could have changed which equation you scale so that we get different cancellations when we add or different eliminations. So another way that we could have done this, we could've kept the first equation the same. Negative 3y plus 4x is equal to 11. But maybe we want this 2x to cancel with this 4x. And the only way we could do that is if this 2x could become a negative 4x. And the way to get from a 2x to a negative 4x is to multiply it by negative 2. So let's multiply this entire equation by negative 2. So it would become negative 2y minus 4x is equal to negative 26. And if we then add the two equations, the x's will cancel out. You have negative 3y minus 2y, which is negative 5y is equal to 11 minus 26. That's negative 15. Now, you can divide both sides by negative 5. And you get-- the negatives cancel out-- y is equal to 3. So in this situation, we eliminated the x's first to solve for y. And then you could go back, substitute, and solve for x. You'll get x is equal to 5. The first time we did it, we eliminated the y's. We scaled this equation right here so that the y's would cancel out. And so we solved for the x first and then substituted back. Either way works. You get x is equal to 5. y is equal to 3.