In this section, you will learn how to solve systems of linear equations by the following methods: addition, multiplication/addition, substitution, and determinants. You will be invited to try our quizmasters at the end of each lesson.

Sometimes we are presented with equations that can be solved with little difficulty. If we look at the coefficients on a single letter between the two equations and they are opposites of each other, then we can add the two equations to cancel that letter. We will gain an equation having a single variable that can be solved quite easily. After we obtain the value of one variable, we can then plug that value into an equation and solve for the other letter. Let's look at two examples.

1. -3x + 5y = 12
   3x + 2y = 9

   If we add the two equations, we will get 0x + 7y = 21 or 7y = 21. If we divide both sides by 7, we get y = 3. Next, we substitute 3 in for y into any one of the equations. If we choose to substitute in the first equation, we get -3x + 5(3) = 12 ==> -3x + 15 = 12 ==> -3x = -3 ==> x = 1. So the two lines intersect at (1, 3).

2. 4x + 8y = 4
   7x - 8y = -15

   If we add the two equations, we will get 11x + 0y = -11 or 11x = -11. If we divide both sides by -11, we get x = -1. Next, we substitute -1 in for x into one of the equations of our choice. Using the second equation, we get 7(-1) - 8y = -15 ==> -7 - 8y = -15 ==> -8y = -8 ==> y = 1. So the two lines intersect at (-1, 1).

• Quizmaster: Systems of Linear Equations

Most of the equations that we are presented with do not have coefficients that are opposite each other for a single variable like we saw for the examples above. It is more probable that our equations will need to be manipulated so we may use the addition method. The technique for preparing the equations is sometimes simple and other times more complicated. View the following examples to get a feel for this strategy.

1. 2x + 3y = 5
   x - 5y = -17

   We can see that adding the two equations as they are right now will not cancel any variable, nor leave us with a single variable. However, if the coefficients on the x-terms were opposite in value, we could add the equations for a beneficial end. If we multiply the whole second equation by -2, we will have coefficients on the x-terms that will be opposite in value.

   2x + 3y = 5 x - 5y = -17

   ==>

   2x + 3y = 5 -2(x - 5y = -17)

   ==>

   2x + 3y = 5 -2x + 10y = 34

   Now we can see that the system of two equations and two unknowns can be solved by using the addition method, which is explained above.

2. 6x - 2y = -4
   -4x + 3y = -4

   Like example #1 above, adding the two equations as they are right now will not cancel any variable, nor leave us with a single variable. However, if the coefficients on the x-terms or the y-terms were opposite in value, we could add the equations to end up with an equation with only one variable. This set of equations requires multiplying both equations by values so that the coefficients on either the x-values or the y-values are opposite in value. To achieve these opposite coefficients we could multiply the top equation by 3 and the bottom equation by 2...

   6x - 2y = -4 -4x + 3y = -4

   ==>

   3(6x - 2y = -4) 2(-4x + 3y = -4)

   ==>

   18x - 6y = -12 -8x + 6y = -8

   ...or we could multiply the top equation by 4 and the bottom equation by 6...

   6x - 2y = -4 -4x + 3y = -4

   ==>

   4(6x - 2y = -4) 6(-4x + 3y = -4)

   ==>

   24x - 8y = -16 -24x + 18y = -24

   In either case, we can use the addition method to cancel a single letter from each system, in order to be left with a single variable and one equation.

• Quizmaster: Systems of Linear Equations

The substitution method is one that is used when a coefficient on a variable is equal to one or when all the coefficients are divisible by the coefficient of a variable term. We isolate a variable. Use the expression that results and substitute it for the isolated letter in the other equation. Then solve for the letter that results. Use the following systems of equations as examples:

1. x + 2y = 9
   3x - 8y = -1

   For the first equation, it's easy to solve for x: x + 2y = 9 ==> x = -2y + 9. Now take that expression and substitute it for x in the second equation: 3x - 8y = -1 ==> 3(-2y + 9) - 8y = -1 ==> -6y + 27 - 8y = -1 ==> -14y = -28 ==> y = 2. Then, plug this into an equation to solve for x: x = -2y + 9 = -2(2) + 9 = -4 + 9 = 5. So, the two lines intersect at (5, 2).

2. -5x + 3y = 10
   -3x + y = 6

   Let's solve the second equation for y, since it has a coefficient of 1: -3x + y = 6 ==> y = 3x + 6. We need to substitute this expression into the first equation for y: -5x + 3y = 10 ==> -5x + 3(3x + 6) = 10 ==> -5x + 9x + 18 = 10 ==> 4x + 18 = 10 ==> 4x = -8 ==> x = -2. Use this value and plug it into an equation to solve for y: y = 3x + 6 = 3(-2) + 6 = -6 + 6 = 0. Therefore, the two lines intersect at (-2, 0).

If it becomes difficult to solve for a letter for all equations in a system, it is recommended that another method be chosen to solve the system. The substitution method is a valid method for all systems but the technique is cumbersome for many systems that have coefficients that are indivisible -- relatively prime, to use number theory terminology.

• Quizmaster: Systems of Linear Equations

• Quizmaster: Systems of Linear Equations
• Quizmaster: Evaluating 2x2 Determinants
• Quizmaster: Evaluating 3x3 Determinants

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• Systems of Linear Equations