Lesson Video: Solving Systems of Linear Equations by Omitting a Variable Mathematics • 8th Grade

In this video, we will learn how to solve systems of linear equations by omitting a variable.

17:46

Video Transcript

In this video, we will learn how to solve systems of linear equations using elimination.

Linear equations are equations in which the highest power of each variable that appears is one. And we also donโ€™t have any terms in which different variables are multiplied together. For example, the equation two ๐‘ฅ plus three ๐‘ฆ equals seven is a linear equation. It involves two variables or unknowns, ๐‘ฅ and ๐‘ฆ. And as it involves two variables, we canโ€™t solve this equation on its own; we need more information.

If we were given a second linear equation also involving these same two variables such as the equation five ๐‘ฅ minus ๐‘ฆ equals nine, then we now have whatโ€™s called a system of linear equations. In general, we need the same number of equations as there are variables. So in this system, we have two variables ๐‘ฅ and ๐‘ฆ and two equations. So weโ€™d be able to solve it in order to find the values of ๐‘ฅ and ๐‘ฆ that work in both equations. There are a range of methods that we can use to do this. In this video, weโ€™re going to focus on the elimination method. Letโ€™s look then at a series of different examples, beginning with this one.

Using elimination, solve the simultaneous equations three ๐‘ฅ plus two ๐‘ฆ equals 14, six ๐‘ฅ minus two ๐‘ฆ equals 22.

Now, simultaneous equations are just another way of saying a system of equations. We can see that we have two equations. And theyโ€™re each in the same two variables, ๐‘ฅ and ๐‘ฆ. Weโ€™re told that we need to use the method of elimination to answer this question. So letโ€™s see what this looks like. The principle of this method is that we can eliminate or get rid of one of the two variables from our two equations. We can choose to eliminate either ๐‘ฅ or ๐‘ฆ. But to make things easier, we notice that in this question, we have two ๐‘ฆ in each equation. But in equation 1, two ๐‘ฆ is being added to the ๐‘ฅ-term and in equation 2 itโ€™s being subtracted from the ๐‘ฅ-term.

The key thing we need to spot is that if we were to add these two entire equations together, then weโ€™ll eliminate the ๐‘ฆ-term. Letโ€™s see what that looks like. On the left-hand side, three ๐‘ฅ plus six ๐‘ฅ gives nine ๐‘ฅ. We then have positive two ๐‘ฆ plus negative two ๐‘ฆ. Thatโ€™s two ๐‘ฆ minus two ๐‘ฆ, which is equal to zero. On the right-hand side, we have 14 plus 22, which is equal to 36. So weโ€™ve eliminated the ๐‘ฆ-variables and created an equation in ๐‘ฅ only. Nine ๐‘ฅ is equal to 36.

Now that our equation is in terms of ๐‘ฅ only, itโ€™s straightforward to solve to find the value of ๐‘ฅ. We have nine ๐‘ฅ equals 36, so we need to divide both sides of the equation by nine. Doing so gives ๐‘ฅ equals four. So weโ€™ve found the value of one of our two variables. Next, we need to find the value of our ๐‘ฆ-variable. And to do this, we can substitute the value weโ€™ve found for ๐‘ฅ into either of our two equations. It really doesnโ€™t matter which we choose. Iโ€™m going to choose to use equation 1 simply because the coefficient of ๐‘ฆ is positive in this equation, so it will make things a little easier.

So substituting ๐‘ฅ equals four gives three times four plus two ๐‘ฆ is equal to 14. Three times four is, of course, equal to 12. So we have the equation 12 plus two ๐‘ฆ equals 14, which is an equation in ๐‘ฆ only. To solve, we first need to subtract 12 from each side to give two ๐‘ฆ is equal to two and then divide each side of the equation by two to give ๐‘ฆ is equal to one. So weโ€™ve also found the value of ๐‘ฆ. And therefore, weโ€™ve solved the simultaneous equations. Our solution is a pair of values: ๐‘ฅ is equal to four and ๐‘ฆ is equal to one.

Now itโ€™s always a good idea to check our answer where we can. And in order to do this, weโ€™re going to substitute the pair of values we found for ๐‘ฅ and ๐‘ฆ into whichever equation we didnโ€™t use when determining the second value. So weโ€™re going to substitute into equation 2. Substituting ๐‘ฅ equals four and ๐‘ฆ equals one on the left-hand side gives six multiplied by four minus two multiplied by one. Thatโ€™s 24 minus two, which is equal to 22. And that is indeed the value that we should have on the right-hand side of the equation. So this confirms that our solution is correct.

The key principle of the method of elimination in this question then was to notice that we had almost the same coefficient of ๐‘ฆ in each equation, but one was positive and one was negative. We, therefore, found that if we were to add the two equations together, this would eliminate the ๐‘ฆ-variables, leaving an equation in ๐‘ฅ only. Our solution is ๐‘ฅ equals four and ๐‘ฆ equals one.

In our next example, weโ€™ll see how the method of elimination works if we canโ€™t eliminate one variable by adding the two equations together.

Use the elimination method to solve the simultaneous equations three ๐‘Ž plus two ๐‘ equals 14, four ๐‘Ž plus two ๐‘ equals 16.

So we have a pair of simultaneous equations or a system of linear equations in two variables ๐‘Ž and ๐‘. And weโ€™re told that we must use the elimination method in order to solve this system of equations. Weโ€™ll label our two equations as equation 1 and equation 2 for ease of referencing them. And looking at the two equations, we notice, first of all, that they have exactly the same coefficient of ๐‘. They both have positive two ๐‘. Now your first thought may be that we can, therefore, eliminate the ๐‘-variable by adding the two equations together. But letโ€™s see what that looks like.

On the left-hand side, three ๐‘Ž plus four ๐‘Ž gives seven ๐‘Ž. We then have positive two ๐‘ plus another positive two ๐‘, which gives positive four ๐‘. And on the right-hand side, we have 14 plus 16, which is equal to 30. So we have the equation seven ๐‘Ž plus four ๐‘ equals 30. This equation still involves both variables, so we havenโ€™t achieved our aim of eliminating one, which means that adding the two equations together wasnโ€™t the correct step to take.

Instead, letโ€™s try subtracting one equation from the other. And as the coefficient of the other valuable, ๐‘Ž, is greater in equation 2 than it is in equation 1, Iโ€™m going to try subtracting equation 1 from equation 2. On the left-hand side, four ๐‘Ž minus three ๐‘Ž gives ๐‘Ž. We then have two ๐‘ minus two ๐‘. So that cancels out to zero. And on the right-hand side, 16 minus 14 is two. So we have ๐‘Ž equals two. Weโ€™ve eliminated the ๐‘-variable. And in fact, we found the solution for ๐‘Ž at the same time. The correct way to eliminate one variable then was to subtract one equation from the other. And the reason for this is that the coefficients of the variable we were trying to eliminate, that is the ๐‘โ€™s, are identical in both equations and they have the same sign.

There is a helpful acronym that we can use to help us remember this, SSS. It stands for if we have the same signs, then we subtract. We must remember that it is the signs of the variable we are looking to eliminate that is important. So itโ€™s the ๐‘โ€™s that we were interested in here. As the signs of the ๐‘โ€™s were the same, we eliminated them by subtracting one equation from the other.

Now that we found the value of ๐‘Ž, we need to find the value of ๐‘, which we can do by substituting our value of ๐‘Ž into either of the two equations. Letโ€™s choose equation one. We have three multiplied by two plus two ๐‘ is equal to 14. Thatโ€™s six plus two ๐‘ equals 14. And subtracting six from each side gives two ๐‘ is equal to eight. We then solve for ๐‘ by dividing each side of the equation by two, giving ๐‘ equals four.

So we have our solution to the simultaneous equations: ๐‘Ž is equal to two and ๐‘ is equal to four. But we should check our answer, which we can do by substituting the pair of values we found into the other equation. Thatโ€™s equation 2. Substituting ๐‘Ž equals two and ๐‘ equals four into the left-hand side of equation 2 gives four times two plus two times four. Thatโ€™s eight plus eight, which is equal to 16, the value on the right-hand side of equation 2. So this confirms that our solution is correct.

We need to remember then that helpful acronym SSS, which stands for if the signs of the variable we want to eliminate are the same, then we subtract. Of course, the reverse is also true. If the signs of the variable we want to eliminate are different, then we add. Our solution to this set of simultaneous equations which weโ€™ve checked is ๐‘Ž equals two and ๐‘ equals four.

Now, in the two examples weโ€™ve seen so far, weโ€™ve been able to eliminate one of the variables straightaway by adding or subtracting the original two equations. Sometimes, though, there may be an extra step needed before we can do this, which weโ€™ll see in our next example.

Using elimination, solve simultaneous equations five ๐‘ฅ minus four ๐‘ฆ equals 21, four ๐‘ฅ plus 12๐‘ฆ equals 32.

So weโ€™re asked to solve this system of equations using the elimination method, which means weโ€™re looking to eliminate either the ๐‘ฅ- or ๐‘ฆ-variable by adding or subtracting our two equations. However, if we were to try this as the equations currently are, weโ€™d find that in both cases we still have ๐‘ฅ- and ๐‘ฆ-variables in the equation we are left with. If we were to add, weโ€™d have the equation nine ๐‘ฅ plus eight ๐‘ฆ equals 53. And if we were to subtract, weโ€™d have the equation ๐‘ฅ minus 16๐‘ฆ equals negative 11. So this hasnโ€™t actually helped.

So why hasnโ€™t it worked? Well, in order to use the method of elimination, weโ€™re looking for the coefficients of one of the variables to be the same in both equations, or at least to have the same magnitude such as positive and negative three. But in this problem, this isnโ€™t the case. We have a coefficient of five for ๐‘ฅ in the first equation and four in the second. And we have a coefficient of negative four for ๐‘ฆ in the first equation and 12 in the second. So simply adding or subtracting these equations as they currently are doesnโ€™t eliminate either variable.

Weโ€™re told, though, that we need to use the method of elimination. So what should we do? Well, what weโ€™re going to do is manipulate these equations slightly so that we do have the same coefficient, or at least the same magnitude of coefficient for one of the variables. To achieve this, we use the equality property of multiplication, which says that if we multiply both sides of an equation by the same value, then the equality is still true. Weโ€™re looking for a value that we can multiply one equation by which will then give the same coefficient, or same-size coefficient, for one of the variables in both equations.

Looking at the ๐‘ฆ-variables in our two equations, we see that four is a factor of 12. So if we were to multiply equation 1 by three, then we would have negative 12๐‘ฆ. And so the magnitude of the coefficients of ๐‘ฆ would be the same in both equations. Letโ€™s try that then. Letโ€™s multiply equation 1 by three. So we multiply everything in equation 1 by three, giving 15๐‘ฅ minus 12๐‘ฆ is equal to 63. The coefficients of ๐‘ฆ in our two equations are now the same size but with different signs, which means we can eliminate the ๐‘ฆ- variables by adding our two equations together.

When we do, we have 15๐‘ฅ plus four ๐‘ฅ, which gives 19๐‘ฅ; negative 12๐‘ฆ plus 12๐‘ฆ, which gives zero; and on the right-hand side 63 plus 32, which is 95. So weโ€™ve eliminated the ๐‘ฆ-variable from our equations, giving a single equation in ๐‘ฅ, which we can solve by dividing both sides by 19. Doing so gives the solution for ๐‘ฅ. ๐‘ฅ is equal to five.

We can then find the value of ๐‘ฆ by substituting ๐‘ฅ equals five into any of the three equations, either of the original two or the equation we created when we multiplied equation 1 by three. Iโ€™m going to use equation 2 as all of the coefficients are positive in this equation. Doing so gives four times five, which is 20, plus 12๐‘ฆ equals 32. We can then subtract 20 from each side and divide by 12 to give ๐‘ฆ equals one. So we have our solution: ๐‘ฅ equals five and ๐‘ฆ equals one. But of course, we should check it. Iโ€™m going to choose to check by substituting the values into equation 1. Thatโ€™s five ๐‘ฅ minus four ๐‘ฆ equals 21. Substituting ๐‘ฅ equals five and ๐‘ฆ equals one gives five times five minus four times one. Thatโ€™s 25 minus four, which is indeed equal to 21. So this confirms that our solution is correct.

The key stage in this question, then, was to notice that we couldnโ€™t eliminate either variable by adding or subtracting the two equations in their original form. Instead, we had to first multiply one equation by a constant to create an equivalent equation in which the coefficient of ๐‘ฆ was the same magnitude as it was in the second equation. Only then could we use the method of elimination to solve these simultaneous equations.

Now it may not always be possible to just multiply one equation by a constant. Instead, we may need to multiply both equations by different constants. Letโ€™s have a look at an example of this.

Using elimination, solve the simultaneous equations four ๐‘ฅ plus six ๐‘ฆ equals 40 and three ๐‘ฅ plus seven ๐‘ฆ equals 40.

In the two equations weโ€™ve been given, the coefficients of ๐‘ฅ are different and the coefficients of ๐‘ฆ are also different, which means we canโ€™t eliminate one variable by just adding or subtracting the equations together. We also notice that neither of the coefficients of ๐‘ฅ are factors of the other and neither of the coefficients of ๐‘ฆ are factors of each other. We want to create equations in which the coefficients of either variable are the same or at least the same but with different signs. So how are we going to do this?

Well, weโ€™re going to have to multiply both equations by some constant. Iโ€™m going to choose to multiply equation 1 by three and equation 2 by four because this will create 12๐‘ฅ in each equation. We could also have chosen to multiply equation 1 by seven and equation 2 by six as this would create 42๐‘ฆ in each equation. It doesnโ€™t matter which variable we choose to eliminate.

Now that we have our two new equations, we have 12๐‘ฅ in each. And as the signs are the same, we can eliminate the ๐‘ฅ-variable by subtracting. Iโ€™m actually going to subtract the top equation from the bottom one because the coefficient of ๐‘ฆ is greater in the second equation. We have 12๐‘ฅ minus 12๐‘ฅ, which cancels out; 28๐‘ฆ minus 18๐‘ฆ, which gives 10๐‘ฆ; and 160 minus 120, which is 40. So weโ€™ve eliminated the ๐‘ฅ-variable from our equation. We can then solve for ๐‘ฆ by dividing both sides of this equation by 10, giving ๐‘ฆ equals four.

To solve for ๐‘ฅ, we substitute this value of ๐‘ฆ into any of our four equations. Iโ€™m going to choose equation 1. Doing so gives a straightforward linear equation for ๐‘ฅ which we can solve by subtracting 24 and dividing by four to give ๐‘ฅ equals four. So we have our solution. Both ๐‘ฅ and ๐‘ฆ are equal to four. As always, we should check our solution by substituting our values into any of the other equations. Iโ€™ve used equation 2. And it confirms that our solution is correct.

The key step in this question then was to multiply both equations by some number to create the same coefficient of one of the variables. We could then use the method of elimination to eliminate this variable and solve our simultaneous equations.

Letโ€™s review some of the key points from this lesson. Our aim in this method is to eliminate one variable by adding or subtracting the two equations. If the coefficients of the variable we wish to eliminate have different signs, we add the two equations together. And if the signs are the same, then we subtract the equations, which we can remember using the acronym SSS. We also saw that in some cases we may need to multiply one or both equations by a constant before we can add or subtract to eliminate one variable.

Once weโ€™ve eliminated one variable and found the value of the other, we need to substitute this value back into one of our equations in order to find the value of the variable we eliminated. And we should always check our answer by substituting both values into whichever equation we didnโ€™t use when working out the second value.

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