Lesson Video: Systems of Linear Equations | Nagwa Lesson Video: Systems of Linear Equations | Nagwa

Lesson Video: Systems of Linear Equations Mathematics

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

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Video Transcript

In this video, we’ll learn how to solve systems of linear equations using elimination or substitution.

A system of equations is simply a collection of equations, usually in the same variables. When we’re asked to solve a system of equations, we’re looking for the collection of solutions that make all of the equations in the system true at the same time. For this reason, systems of equations are often referred to as simultaneous equations. Although a system of simultaneous equations can contain any number of equations with any number of variables of various degrees, in this video, we’ll be restricting our attention to systems of linear equations in two variables.

We recall that a linear equation contains no terms of degree higher than one. In this lesson, we’ll be looking at pairs of linear equations. The following is an example of a pair of simultaneous linear equations: 𝑥 plus 𝑦 equals three and 𝑥 minus 𝑦 equals one. To solve this system, we are looking for the pair of unknowns, 𝑥 and 𝑦. We are able to find exact values for these unknowns because we have two distinct linear equations, as shown in our sketch. We observe that this system has the solution 𝑥 equals two, 𝑦 equals one, because these are the values that solve the equations simultaneously. In the diagram, we see this solution matches the coordinates of their point of intersection. We check our solution algebraically by substituting two for 𝑥 and one for 𝑦 to see if those values make both equations true simultaneously.

In this video, we’ll learn two methods for solving linear systems. The first method is called elimination. In our first example, we observe the coefficient of 𝑦 in the second equation is of equal magnitude but opposite sign to the coefficient of 𝑦 in the first equation. This means that if we add the left-hand side of the second equation to the left-hand side of the first equation, we get two 𝑥, because the terms in 𝑦 will cancel each other out. In other words, 𝑦 is eliminated. The resulting expression contains only one variable, 𝑥. If we are to add a quantity 𝑥 minus 𝑦 to the left side of the first equation, we must add an equal quantity to the right side in order to maintain equality.

Fortunately, the second equation tells us that 𝑥 minus 𝑦 is equal to one. So, we can maintain equality by adding one to the right-hand side of the first equation. We can also think of this as adding the left-hand sides of both equations and adding the right-hand sides of both equations. When we add one to the right side of the first equation, we get a sum of four. Now that we are sure the left side is equal to the right side again, we can solve the resulting equation for 𝑥. After dividing each side of the equation by two, we get 𝑥 equals two. Now, we can find the value of 𝑦 by substituting 𝑥 equals two back into either of our starting equations.

Let’s use the first equation. So, that’s two plus 𝑦 equals three. So, 𝑦 equals one. An important last step is to check our solution by substituting 𝑥 equals two and 𝑦 equals one into the equation we did not use: 𝑥 minus 𝑦 equals one. Since two minus one equals one, we have established that 𝑥 equals two and 𝑦 equals one solves both of our equations simultaneously.

Suppose we have a pair of simultaneous equations with the same coefficient and the same sign on one of the variables. For example, five 𝑥 plus three 𝑦 equals two and two 𝑥 plus three 𝑦 equals 11. We can do essentially the same thing as above. But this time, we subtract the corresponding sides of one equation from the other instead of adding them. If we’d like to keep the like terms stacked, it can be helpful to distribute the negative sign through the second equation, then add the like terms vertically. In this case, 𝑦 is eliminated. And we solve to get 𝑥 equals negative three. Again, we substitute this 𝑥-value into either starting equation. This time, we choose the first equation. So, we have 𝑦 equals seventeen-thirds.

Finally, we substitute 𝑥 and 𝑦 into the second equation to check that the equations are consistent and that 𝑥 equals three and 𝑦 equals seventeen-thirds simultaneously solves both of them. This check confirms we’ve found the unique solution to this system of equations.

We can also use the method of elimination to solve systems of linear equations when none of the coefficients are equal. Consider, for example, the system negative seven 𝑥 plus six 𝑦 equals negative eight and six 𝑥 plus 12𝑦 equals 24. In this case, we notice that the 𝑦-terms could be eliminated if we multiply the first equation through by negative two and then add it to the second equation.

Alternatively, we might multiply by positive two then subtract the equations. Carefully multiplying through the entire equation by negative two gives 14𝑥 minus 12𝑦 equals 16. We note that by multiplying by a negative, we now have coefficients with opposite signs of the same magnitude for the 𝑦-terms. So, we can add the corresponding sides of the simultaneous equations together in order to eliminate 𝑦. Then, we solve to get 𝑥 equals two. As before, we substitute the 𝑥-value back into one of our starting equations to find 𝑦. In this case, we find that 𝑦 equals one, as shown. Finally, we substitute into the first equation to verify our solution.

As we’ll see in our next example, sometimes, neither of the variables in either equation is a direct multiple of the other. So more calculations will need to be done to use the elimination method.

Use the elimination method to solve the simultaneous equations: four 𝑥 minus two 𝑦 equals four; five 𝑥 plus three 𝑦 equals 16.

In this example, the coefficients of the corresponding variables in the two linear equations are neither equal nor integral multiples of one another. This means we will have to do a little work before we can eliminate one of the variables. Since adding is generally easier than subtracting, we’d like to eliminate the variable that already has opposite signs. In our case, that is the 𝑦-term. However, we need the coefficients of 𝑦 to have the same magnitude. We can do this by multiplying the first equation through by three and the second equation through by two. As shown, the coefficients of 𝑦 are now negative six and positive six. So, by adding the corresponding sides of the simultaneous equations, we can eliminate 𝑦.

And we are left with an equation involving only one variable, 𝑥. By dividing through by 22, we get 𝑥 equals two. Since these are two variable equations, we must find both 𝑥 and 𝑦 to solve the system. So, we can substitute 𝑥 equals two into either of the starting equations. In this case, we’ll substitute two for 𝑥 in the second equation. Solving the second equation for 𝑦 gives the value of two. Finally, we check our solution, 𝑥 equals two and 𝑦 equals two, by substituting it into the other equation we started with, four 𝑥 minus two 𝑦 equals four. This indicates that our calculations were sound and that the pair of values 𝑥 equals two and 𝑦 equals two simultaneously solve the two equations.

We can now summarize this discussion into a “How To” for solving pairs of simultaneous linear equations by elimination.

Let’s suppose we’re asked to solve a pair of simultaneous linear equations 𝑎𝑥 plus 𝑏𝑦 equals 𝑐 and 𝑑𝑥 plus 𝑒𝑦 equals 𝑓. First, we check if one variable can easily be eliminated by adding or subtracting the equations. If not, choose one variable to be eliminated and multiply one or both equations so that the coefficients of this variable are the same or one is the negative of the other. Add or subtract the equations together to eliminate one variable and rearrange to get the other variable. Substitute this value into one of the two equations to get the eliminated variable. Finally, we check our answer by substituting both variables into the other equation and verifying that both sides are the same.

Now, we’ll turn to the second method of solving systems of linear equations: the substitution method. The method of substitution is very handy when one of the variables is the subject of one of the equations in the system, as in the next example, or when one of the equations can be easily rearranged to make a variable the subject.

Solve the simultaneous equations 𝑦 plus four 𝑥 equals negative eight and 𝑦 equals five 𝑥 plus 10.

Here, we have a pair of simultaneous equations in which 𝑦 is the subject of the second equation. This means we can directly substitute the expression five 𝑥 plus 10 for 𝑦 in the first equation. So, we have five 𝑥 plus 10 plus four 𝑥 equals negative eight. Combining the 𝑥-terms on the left, we have nine 𝑥 plus 10 equals negative eight. Solving for 𝑥 gives a value of negative two. We can now substitute 𝑥 equals negative two into either of the starting equations to solve for 𝑦. Using the second equation may be easiest since the subject is already 𝑦. This leads us to the value of 𝑦, which is zero.

Finally, we substitute our solution 𝑥 equals negative two and 𝑦 equals zero into the other equation to check for mistakes. We see that zero plus four times negative two equals negative eight, so our answer is correct. We conclude that the pair of values 𝑥 equals negative two and 𝑦 equals zero solve the simultaneous equations 𝑦 plus four 𝑥 equals negative eight and 𝑦 equals five 𝑥 plus 10.

The method of substitution can also be used when neither variable is the subject of either equation, as given. In this case, we will need to do a little rearrangement first. For example, consider the equations three 𝑥 plus 12𝑦 equals 21 and two 𝑥 minus four 𝑦 equals 11. Here, rearranging the first equation to make 𝑥 the subject can be achieved by first dividing through by three then subtracting four 𝑦 from both sides of the equation. So, we have 𝑥 equal to seven minus four 𝑦. We can now substitute 𝑥 equals seven minus four 𝑦 into the second equation and solve for 𝑦. That is, two times seven minus four 𝑦 minus four 𝑦 equals 11. Distributing two into the parentheses gives 14 minus eight 𝑦 minus four 𝑦 equals 11. As shown, combining like terms and solving for 𝑦 gives a value of one-fourth.

To find the value of 𝑥, we substitute 𝑦 equals one-fourth into either of the starting equations. If we substitute one-fourth into the first equation, we find that 𝑥 equals six. As always, the final thing to do is to check whether our solution is correct when substituted into the other equation. As shown, 𝑥 equals six and 𝑦 equals one-fourth is a solution to the second equation as well. Therefore, it is the solution to the simultaneous pair of equations.

Let’s now summarize how to use the method of substitution step by step. Suppose we’re asked to solve a pair of simultaneous linear equations: 𝑎𝑥 plus 𝑏𝑦 equals 𝑐 and 𝑑𝑥 plus 𝑒𝑦 equals 𝑓. First, we rearrange one of the two equations to make one of the two variables the subject. Then, we substitute the resulting expression for that variable everywhere in the other equation and rearrange to solve for the remaining variable, then substitute this value back into either of the starting equations and solve for the variable we made the subject. Finally, we check our answer by substituting both variables into the other equation and verifying that both sides are the same.

Now that we have seen how to solve systems of linear equations using methods of elimination and substitution, we’ll end with an example of the reverse process, calculating the coefficients in a system of equations when the solution is given.

Consider the simultaneous equations 𝑎𝑥 plus 𝑏𝑦 equals 12, 𝑥 minus two 𝑏𝑦 equals negative 10, where 𝑎 and 𝑏 are constants. Given that 𝑥 equals two, 𝑦 equals three is a solution to the pair of simultaneous equations, find the values of 𝑎 and 𝑏.

We are told that 𝑥 equals two and 𝑦 equals three solve the simultaneous equations 𝑎𝑥 plus 𝑏𝑦 equals 12 and 𝑥 minus two 𝑏𝑦 equals negative 10, which simply means that these equations are both true for these values. We can therefore substitute in two and three for 𝑥 and 𝑦 in the first equation, like so. We have two 𝑎 plus three 𝑏 equals 12. And then, we do the same for the second equation, which simplifies to two minus six 𝑏 equals negative 10. This single-variable equation could be solved for 𝑏 as follows. We subtract two from each side of the equation then divide by negative six, finding that 𝑏 equals two.

We can now substitute 𝑏 equals two into the equation two 𝑎 plus three 𝑏 equals 12 that we got from the first equation. So, that’s two 𝑎 plus three times two equals 12. Solving this equation reveals that 𝑎 equals three. Putting our answers together, we’ve solved for the missing coefficients: 𝑎 equals three and 𝑏 equals two.

Let’s finish by recapping a few important concepts from this video.

We learned that we can solve simultaneous equations by the method of elimination or, alternatively, by the method of substitution. For elimination, first, we rearrange both equations into the form 𝑎𝑥 plus 𝑏𝑦 equals 𝑐. Then, we check to see if one variable can be easily eliminated by adding or subtracting the equations. If not, we choose a variable to be eliminated and multiply one or both equations so that the coefficients of this variable are the same or one is the negative of the other. Once the coefficients of one of the variables are suitably equalized, we add or subtract the equations together to eliminate that variable. And then, we rearrange to solve for the other. Having found the value of one variable, we substitute it back into either of the starting equations to get the eliminated variable.

For the substitution method, we rearrange one of the equations to make one of the variables the subject and then substitute the resulting expression for that variable into the other equation. Rearranging the resulting equation allows us to solve for the remaining variable. We can then substitute this value back into either of the starting equations to solve for the variable that was made the subject. In both methods, the last thing to do is substitute the solution into the other starting equation to check that we have indeed found the pair of values that solves both equations simultaneously.

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