Systems of Equations

1. Systems of Equations

2. Solving for one variable We have a systematic process for solving equations with only one variable: get all the xterms on one side of the equal sign and get all the numbers on the other: Example: 3x + 2 = 2x – 1 -2x -2x x + 2 = -1 -2 -2 x = -3 But, when we have to solve for two variables, shoving all the x terms on one side and all the numbers on the other doesn’t make sense anymore, because we don’t have anywhere for the y terms to go!

3. Dealing with two variables When we have two (or more) variables, we also have two (or more) equations. This is so we can solve for both xand yand actually get numbers for each of them. A system of equations looks like this: ax+by=c dx+ey=f , where x and y are the variables, and athroughfare numbers. To solve systems, we need to get rid of one variable and solve for the other. How do we do this? We just have to combine both equations, making sure we eliminate one variable as we do so, and then solve for the one that’s left. This process is best explained through example.

4. Quick Example Solve the system of equations: 2x + y = 4 4x – 3y = -2 First, we need to eliminate either the y’s or the x’s from this set of equations. It doesn’t really matter which one we eliminate, but since one of the y’s has a minus sign in front of it, we can get rid of the y’sfirst by multiplying the first equation by 3. 3 (2x + y) = 3(4) 6x + 3y = 12 Now our system of equations is 6x + 3y= 12 4x– 3y = -2 We can see that adding these two equations together will get rid of the y terms, so let’s do that: 6x + 3y = 12 + 4x– 3y= -2 10x = 10

5. Now we’re left with one equation and one unknown, our x. Let’s solve for it. 10x = 10 x = = 1 We’ve got one of our variables! All that’s left to do is find out what y equals. To do that, we just plug in our x value into either one of the original equations and solve for y. Just because I want to, let’s use the first one: 2(1) + y = 4 2 + y = 4 -2 -2 y = 2 So, our solution is x = 1, y = 2. Just to be sure, let’s plug in our x = 1 value into the second equation to make sure we did our math right: 4(1)– 3y = -2 4 – 3y = -2 -4 -4 -3y = -6 y= 2 Our answer checks out, so indeed the solution isx= 1,y = 2. 

6. Quick Example 4x + 2y = 12 -2x – y = -4 Solve the system of equations: If we multiply the second equation by 2, we can then add the two equations together to get rid of the x variable, or we can multiply by 2 to get rid of the y variable. In any case, let’s multiply the second equation by 2: 2 (-2x – y) = 2 (-4) -4x – 2y = -8 Now let’s add the two equations together: 4x + 2y = 12 -4x – 2y = -8 + 0 = 4 Obviously, this is not true. What just happened?!??!!?

7. Not all systems are solvable! Sometimes, there are systems of equations that have no solution, or the solution you get makes no sense (like the last example). But don’t worry! You don’t always have to try to solve them to find out no solution exists. There are a couple of ways to tell just by looking at your system. Tipoff #1: More variables than equations. If you have a system with two variables but only one equation, there is no one answer to the solution. Example: Suppose we are given the equation 4x + 3y = 7. Just by eyeballing it, we can see that x= 1, y= 1 is a solution: 4(1) + 3(1) = 4 + 3 = 7. But, we have a different solution if we choose, say, x = 2: 4(2) + 3y = 7 8 + 3y = 7 3y = -1 y = - Basically, there are infinitely many solutions to systems of equations with less equations than variables!

8. Tipoff #2: More equations than variables. When we have more equations than variables, the system is said to be overdetermined, meaning we’re given more information than we need to solve the system of equations. Example: We are given the system 4x+ y = 5 2x – 4y = -2 2x + 3y = 4 Let’s solve for the first two equations, and then we’ll see if that answer is consistent with the third equation: 4x + y = 5 2x – 4y = -2 I’ll multiply the second equation by 2, and then subtract it from the first one, to get rid of the x’s: 2 (2x– 4y) = 2(-2) 4x– 8y = -4 4x+ y= 5 - 4x– 8y = -4 (Now we’re subtracting this from the first equation) -(-9y) = 9 9y = 9 So y =

9. Now we will solve for x. Let’s plug it in the first equation: 4x + y = 5 4x + 1 = 5 -1 -1 4x= 4 x = 1 Now we check our answer with the second equation to make sure this answer is okay: 2x – 4y = -2 2(1) – 4(1) = -2 2 – 4 = -2 So, x = 1, y = 1 is a solution for just these two equations. If it’s a solution to the third one, it’s a solution to the entire system. 2x + 3y = 4 2(1) + 3(1) = 4 2 + 3 = 4 5 = 4 This means the system has no solution! While it may work out sometimes that there is a solution, in general there won’t be one for systems with more equations than unknowns.

10. Tipoff #3: One equation is a constant multiple of another. This one is less obvious than the others, but it will also affect whether the system has one solutions or not. One equation is a constant multiplier of another if you can factor out a constant from it and get the other equation. Example: Consider the system x + 3y = 9 2x + 6y = 18 2x + 6y = 18 2 is a common factor in each term of the second equation, so let’s divide both sides of that equation by 2: 2x + 6y = 18 2 2 x + 3y = 9 This is exactly the same as the first equation! So, equation 2 in the system is a constant multiple of the first equation. Essentially, having two equations that are constant multiples is the same as having just one of those equations, since they are both going to give the same solution to the system.

11. Quick Example -3x+ 5y = 12 3x+ 2y = 5 6x+ 4y = 10 Solve the system of equations: Let’s look at this system: there are three equations, but only two unknowns. We could be in for some trouble! Actually, let’s look at these equations a little closer. It appears that the final equation is a multiple of the second one! Every term in there is 2 times the stuff in the second equation: 2(equation #2) → 2 (3x + 2y) = 2(5) 6x + 4y = 10 ← equation #3 So, it’s as though we have just two equations, like we should have, since every solution of the second equation will be a solution of the third (the two equations are, in a sense, the same). That means we have to solve the system -3x+ 5y = 12 3x+ 2y = 5

12. Let’s just add the two equations together, since they both have 3x terms, and one of those terms has a minus sign: -3x+ 5y = 12 3x+ 2y = 5 + 7y = 17 17 7 y = Gross. It looks like we’re not going to get whole numbers this time. That’s okay, though. We can just solve for x like we usually do, and as long as the answer works for the other equation we’ve solved the system. Let’s plug y = 17/7 into the first equation to see what we get for x: -3x + 5(17/7) = 12 Here, I multiplied 12 by 7/7 so I can subtract 85/7 from it. -3x + 85/7 = 12 -3x = 12 – 85/7 = 84/7 – 85/7 = -1/7 x = (-1/7)/(-3) = 1/21 Now it’s time to cross our fingers and hope this solution works for the other equation! I multiplied 34/7 by 3/3 so I can add these terms together. 3(1/21) + 2(17/7) = 5 = 3/21 + 102/21 = 105/21 = 5 3/21 + 34/7 Whew! Our solution works, and so x = -1/7, y = 17/7, as awful as that is, is our solution to the system. 