Solve By Addition

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We wii solve Example 11 from the Chapter 2 Roadmap page by addition . Start with the initial system of three linear equations in three variables.

3x + 4y + 2z	=	12	(1)
6x – 4y – 3z	=	5	(2)
5x + 7y – 12z	=	5	(3)

Now, it conveniently turns out that if we add the two equations, left side of (1) to left side of (2) and right side of (1) to right side of (2), the answer that we get will have eliminated the y variable.

(3x + 4y + 2z) + (6x – 4y – 3z)	=	12 + 5	(4)
9x – 1z	=	17	(5)

Now, returning to the original equations we need to use a differrent pair of equations to again eliminate the variable y. We will use equations (2) and (3). However, as they are given they do not have additive inverses for the coefficients of the y variables. So, we will multiply equation (2) by 7 and equation (3) by 4 to give the following:

7(6x – 4y – 3z)	=	(7)(5)	(6)
4(5x + 7y – 12z)	=	(4)(5)	(7)

Which simplifies to the system:

42x – 28y – 21z	=	35	(8)
20x + 28y – 48z	=	20	(9)

We are now ready to add these to eliminate the y variable.

(42x – 28y – 21z) + (20x + 28y – 48z)	=	35 + 20	(10)
62x – 69z	=	55	(11)

Thus we have used two different pairs of the original equation to eliminate the y variable. This has given us equations (5) and (11).

9x – 1z	=	17	(5)
62x – 69z	=	55	(11)

But this is just a case of solving a system of two equations in two variables. We choose one of the variables to eliminate. We will eliminate z. To do this we need only multiply equation (5) by ^–69 to get

^–69(9x – 1z)	=	(^–69)(17)	(12)
^–621x + 69z	=	^–1173	(13)

Then we add equations (13) and (11) to get

(^–621x + 69z) + (62x – 69z)	=	^–1173 + 55	(14)
^–559x	=	^–1118	(15)
x	=	2	(16)

Now that we know the value of x, namely, 2, we substitute that value into either one of the equations the involve only x and z, namely, (5) or (11). In this case we will use equation (5).

9(2) – 1z	=	17	(17)
18 – 1z	=	17	(18)
z	=	17	(19)

Which means that we now know that x=2 and z=1. We can substitute these values into any of the original equations. We will use equation (1).

3(2) + 4y + 2(1)	=	12	(20)
6 + 4y + 2	=	12	(21)
4y + 8	=	12	(22)
4y	=	4	(23)
y	=	1	(24)

Therefore, our solution has x=2, y=1, and z=1 which is the point (2,1,1).

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©Roger M. Palay Saline, MI 48176 January, 2017