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## Elimination Method

So far, we've seen how to solve a system of equations using graphs and substitution. Let's try and come up with one more method. Here's a system of two equations, and what I want you to do is I want you to add these two equations together. What would you get? You could write your answer in this box.

## Elimination Method

If I add the x terms together, I get 3x. And if I add positive y and negative y, those sum to 0, negative 11 plus 26 equals 15. If you have trouble thinking about that, try 26 minus 11, that would also b 15. Without even realizing it, you've actually already performed elimination. We've eliminated the y variables, they added to 0.

## Find the Point of Intersection

Now that we've eliminated the y variable, I want you to finish solving this system of equations. What's the x value? What's the y value?

## Find the Point of Intersection

To finish solving for x we divide both sides by 3, so x would equal 5. So we know the x coordinate is 5, but remember we still need to find y. I'm going to use this first equation to find y. I'm going to plug in or substitute the value of x, which is 5 into this equation. Negative 2 times 5 is negative 10 and then I add 10 to both sides of the equation to get y equals negative 1. So this is my solution to the system. And, of course, if we want to know if we're right, we can check our solution. We'll plug in x equals 5, and y equals negative 1 into both equations, and it should work. And when we check the first equation, we get negative 11 equals negative 11. And when we check the second equation, we get 26 is equal to 26. So, yes, this has to be our solution.

## Set Up for Elimination

In general, the elimination method is useful when one variable has the same number or the same coefficient in front of it as the other. But they need to be opposite in sign, just like these y's. We can eliminate the y variables in the last problem, because the y here had a positive 1 as a coefficient and this y had a negative 1 as its coefficient. We know, if we added these two terms together, we would get 0. So this is what we want to strive for. We want variables with the same coefficient, but with opposite sides.

## Multiplying by Constants

Here's a different system of equations. The coefficients to the x are different, and the coefficients for y are different. Hm, so what could we do? Well, we can actually use the properties of equality to change one of these equations or both of them. Here's what I mean. I can think about changing the second equation by multiplying it by a positive 2. When I do that I distribute this 2 to each of the terms, 2 times 3x is 6x, 2 times negative 4y is negative 8y and 2 times 10 is 20. Make sure that you distribute this number to each term. So I know this green equation is the same as this equation, it's been multiplied by 2. So I've rewritten my system of equation here, and notice the y variables have the same coefficient and opposite signs. This is great. Now using our elimination method, I want you to solve this system. You can put your answer here.

## Multiplying by Constants

If I add my equations together, I get 13x equals 26. Positive 8y and negative is equal to 2. I found x, so now, I need to find y. I'm going to use this first equation, and write it over here, so I can substitute the value of x, which is 2 into this equation. So, plugging in a value of 2 for x, I get 7 times 2, which is 14. The rest of my equation stays the same. I subtract 14 from both sides to get 8y equals negative 8. Remember, I have more negatives than positives here, so definitely negative 8. Then, we divide both sides by 8 to get y equals negative 1. We write a solution as a point. I'm not going to show the check for this equation but you can always double check your answer.

## Elimination Again

So let's test your knowledge. What number should you multiply the first equation by to perform elimination. Enter the number here.

## Elimination Again

We should multiply the equation by 3. If I multiply every term by 3, then I'll get 9x. 3 times negative 4y is negative 12y, and 3 times negative 7 is negative 21. So this equation turns into this equation, and they're actually the same equation. They just look different. And wow, look at that. We can actually eliminate the y terms. I have negative 12y here and positive 12y here. Because these terms are opposite in sign, we know that they sum to zero. I'm not going to solve this system of equations, but if you want some more practice, go for it. Remember to check your answer at the end, and then you can know if you're right.

## Multiply Both Equations

You might be wondering, why should we even learn elimination to begin with? Well, would substitution really work here? Would it be a good choice? Well, I could take this first equation and try and solve it for either x or y. Remember, for substitution we're trying to get one of the variables alone on one side of the equation. I'm going to solve this equation for x. I subtract 3y from both sides to get 2x is equal to negative 3y minus 15. I just list these terms, because they're not like. Then we divide both sides by 2, so x is equal to negative 3 halves y minus 15 halves. Notice that this equation is a little complicated. We have two negative signs, and we also have fractions. So when we're working with substitution, this could make our work more difficult. Keep in mind though, that when we perform substitution, we don't have to just solve for x. We could have solved this equation for y. I could subtract 2x from both sides, to get 3y equals negative 2x plus 15. Then I would divide each term by 3. So, 1y would equal negative 2 3rds x plus 5. If I wanted to use this for substitution, it's still pretty complicated. We have a fraction and we have a negative sign. So let's try instead using elimination. When solving this system using elimination, I'm going to try and eliminate these x variables. I could choose to try and eliminate the y variables. It doesn't actually matter. I'm going to multiply this first equation by 5. So I get 5 times 2x, which is 10x, 5 times 3y, which is 15y, and 5 times negative 15, which is negative 75. I multiply this equation by 5, so that way I could get a 10x here. So what number should we multiply this 5x by, so that way we get a number here that would eliminate with the 10x? Take some time to think about it.

## Multiply Both Equations

We should multiply this equation by negative 2. If I distribute this negative 2 to each term, then I can see that I were to get negative 2 times 5x, which is negative 10x. Negative 2 times positive 2y, which is negative 4y. And negative 2 times positive 1, which is negative 2. And look at that, we're ready to do elimination, if positive 10x and negative 10x. These terms have the same absolute value of 10 but one is positive and the other's negative, so we know when we add them they sum to 0.

## Eliminate

Alright, so finish out this systems problem. I want you to find the solution, or the point of intersection for the equations. Put your answers here.

## Eliminate

If we add the two equations together, the x terms eliminate and add to zero. 15y and negative 4y equals 11y and negative 75 minus 2 makes negative 77. We can divide both sides by 11, so one 1y is equal to negative 7. We found y so now let's go and find x. Remember we could use either of the original equations to find the other variable. Usually it's best to use the original equations, because sometimes we might have made a mistake when multiplying, so we don't want to use these multiplied equations. I'm going to use the second equation, 5x plus 2y equals 1. This is a great opportunity to pause the video. Maybe you can try to find the value of x by plugging in to this first equation instead. See if we get the same result. So I can take the value of y, which is negative 7, and plug it in for y in this second equation. So I get 5x plus 2 times negative 7 equals 1. 2 times negative 7 is negative 14. Then we add 14 to both sides of the equation. So 5x is equal to 15. We divide both sides by 5, so x is equal to 3. This is great. We found our solution.

## Check it Out

Here's our original systems again. And this time, I want you to make sure that you can check to know if you're right. Check the solution by plugging in the appropriate values into each equation. Then, carry out the math to fill in the rest of the boxes. You should wind up with true statements at the end. They should check.

## Check it Out

We know x should be replaced with positive 3. So I replace x with positive 3 and this x with positive 3, y is negative 7. So we replace both of these y's with negative 7. 2 times 3 is 6, and 3 times negative 7 is negative 21. If I add these two together, I get negative 15. And look at that, this side does check. 5 times 3 is 15. And 2 times negative 7 is negative 14. If we add these together we get 1 which is equal to 1. Great. We know we have the right solution.

## Elimination or Substitution 1

Let's move away from solving systems of equations and think about which method would work best. So here's a system of equations, and I want you to think about, how should we solve this? So what do you think? Should we graph the equations as lines and find the point of intersection? Should we use substitution to find the point of intersection? Or should we use elimination? Take some time to think about it and actually try solving it using these different methods. You'll find out which one's best.

## Elimination or Substitution 1

These equations are not in slope-intercept form so it would be pretty tough to use graphing. We'd have to do a lot of rearranging and then plot points. If we were graphing, we could have also made tables, but that could take a lot of work, too. So, I'm going to take out graphing. If we want to use substitution, we need one of the equations solved for one variable. This one is almost solved for x. All I would need to do would be to subtract 3 from both sides. Then, I could perform substitution. So, x could equal 2y minus 3 and now, I'd be ready to substitute this in to my first equation for x. That's a pretty great choice. If you were leaning towards elimination, you might have thought to multiply the second equation by negative 5. If we distribute the negative 5, we would get negative 15 minus 5x minus 10y. Now, we could add these equations together. I would want to switch the places of these two terms so that way I could add my x terms together more easily. Now, I'm actually ready to do elimination. And I would get negative 4y minus 15 equals negative 10y plus 9. Notice I can't combine these, because their not like terms. This has a y and this doesn't. This also has a y and this doesn't. This isn't so great with elimination. I'm having to do a lot of steps, and I'm having to reorder a lot of terms. So, that's why I think substitution is the best here.

## Elimination or Substitution 2

Here's another system of equations. I want you to think about which method would be best to solve it, graphing, substitution, or elimination. Again, make sure you're trying to actually solve these using the different methods. Find out which one's easiest.

## Elimination or Substitution 2

For this one, I think we should use elimination. If I multiply the first equation by positive 2, I could wind up with a positive 12 x, which would be the opposite in sign of this negative 12 x. This would let me eliminate the x variables. 2 times 6 x is 12 x. 2 times negative 5 y is negative 10 y. And 2 times positive 3 is positive 6. So, I have the new first equation, and then I just rewrite my second equation right under it. So, elimination would probably be the best here. We choose elimination over the other two because, with substitution, we don't have any single variable solved for. And I know if I solve for these variables, I'd have to divide by coefficients, like 12 or 10. That could create fractions in my substitution, which would make my math more complicated. I know graphing would be more difficult, too, because I would either have to create tables, or solve these equations for y and put them in slope intercept form.

## Elimination Nonsense

Let's continue solving the system and see what happens. If I add the x terms together I get 0, if I add the y terms together I also get 0, and then 6 and 5 make 11. Hm, this is weird, 0 equals 11. This can't be a true statement, we know this isn't right. So something else is going on with these 2 equations. If we remember that the solution to a system of equations is a point of intersection, then we know that this doesn't make sense. There's actually no point that works as the intersection for both of these 2 lines. If I solve each equation for y, I would wind up with these 2 equations. These parallel lines will never intersect. So whenever we connect this math to the graph, we can tell that there's no solution here. There's no ordered pair, x and y, that will make both of these equations true. And that makes sense, because there's no point of intersection.

## Elimination Practice

Alright, here's one last problem for you to solve. Try solving the system using any method you prefer. Write your answers here, good luck.

## Elimination Practice

Did you get 18 5th and 3 5th? If you did, excellent work. This is a really tough problem. Even if you didn't get this answer. If you got part of the solution right, or some of the variables right, great work. I'm going to first solve this problem using substitution. So, if you chose this method, stay with me. If you tried elimination, fast forward until you see the word, elimination, here. Also at any point if you find a correction to your work, pause the video and then try and keep solving. Finding the types of mistakes you make is really important so that we can improve on our math skills. Try and think about the mistake you made and where you went wrong. Thinking about that mistake can prevent you from making it in the future. I think it's easier to solve for x using this equation. I just need to add y to both sides. So I have x equals 3 plus y, and remember I could also write this as y plus 3. Addition is commutative. Now that I've solve for x I can replace this x with the equivalent expression.3 plus y. So I have 3 times 3 plus y plus 2y equals 12, I distribute the positive 3 to get 9 plus 3y plus 2y equals 12. 3y and 2y make positive 5y. And then I just carry down the terms. So 9 plus 5y equals 12. I can subtract 9 from both sides, so 5y is equal to 3. And finally I divide both sides by 5, so y equals 3 fifths. Now I can substitute the value of y, which is 3 fifths, into the second equation. I chose the second equation, because these coefficients were 1. It looked a little bit more simple than the first equation. So I plus in the value of 3 5th for y. Then I need to add 3 5th to both side to isolate x. So x is equal to 3 plus 3 5th. We know we can write this as a mixed number. The mixed number is 3 and 3 fifths. We have the whole number and then the fraction. Converting this to an improper fraction we would have 18 fifths. This is some amazing math work for substitution. If we use elimination I can multiply the second equation by a positive 2. This will allow me to eliminate the y variable. You might have chosen to multiply this equation by negative 3. That would've let you eliminate the x variable. That's okay, that could work too. 2 times x is 2 x. 2 times negative y is negative 2 y. And 2 times 3 is 6. So I get 5 x is equal to 18, and the y's eliminate. To solve for x, we divide both sides by 5. So 1 x is equal to our x value. I'm going to use the second equation to find the value of y. I take equal to, this improper fraction. 5 goes into 18 3 times and I have a remainder of 3, or 3 5th. So I have to take away some number from 3 and 3 5th to get 3. Well, this number has to be 3 5th. If I just think about it I know why it has to be 3 5th. There's our solution.

## Solving Systems of Equations

We've learned a lot of methods to solve systems of equations. We looked at graphing, substitution and elimination. I hope you've gained a greater appreciation of how to solve these problems. With each of these methods there are some advantages and some disadvantages. When solving try and think about what would work best for each problem.