A system of linear equations, two or more equations, can be seen graphically as the point of intersection of the lines. There are some practical uses for solving systems of equations. For example, suppose a company produces watches. One line represents the revenue for selling a certain number of watches and the other line represents the cost for producing the watches. The break even point, (where sales = costs) is the point of intersection. A graph such as this shows when revenues are greater than, less than or equal to costs of production. But how do we solve systems of linear equations when we don’t have a graph? One such method is elimination.
In this method, we will add the equations or a multiple of the equations together in order to eliminate one variable. This enables us to solve for the other variable. When using this method you want to get the absolute value of the number in front of the variable you wish to eliminate the same (example 5 and -5, -5 and -5 or 5 and 5).
Example: Solve the system using the elimination method.
4x + 6y = -4
3x – 6y = 39
Notice the first equation has a 6y and the second equation has a -6y. By adding the equations together, the y term will be eliminated because 6y + (-6y) = 0. That is the idea of the elimination method.
Add the equations together.
4x + 6y = -4
+ 3x – 6y = 39
7x = 35
x = 5
Now substitute 5 in for x in either equation and solve for y.
4∙5 + 6y = -4
20 + 6y = -4
6y = -24
y = -4
Notice how using the addition method eliminated fractions and made the calculations easier. Sometimes one or both equations must be multiplied before eliminating one of the variables.
Example: Solve the system by the elimination method.
5x – 2y = -8
15x – 6y = -24
To eliminate the x term, multiply both sides of the first equation by 3.
3(5x – 2y) = 3(-8)
15x – 6y = -24
Notice the equation is now the same as the second equation in the system. So when subtracting the
equations, you get 0 =0. This is a true statement, therefore the system has infinitely many solutions.
Example: Solve the system by the elimination method.
(2/3)x + 4y = 10
2x – 6y = 19
To eliminate y, get the least common multiple between 4 and 6, since they are the coefficients of the y terms. Multiples of 4 are 4, 6, 12, 16 and so on. Multiples of 6 are 6, 12, 18 and so on. Notice the least common multiple is 12. In order to eliminate the y terms, multiply the first equation by 3 and the second equation by 2 to get 12y and -12y.
3[(2/3)x + 4y = 10]
2(2x – 6y = 19)
After multiplying, the system becomes
2x + 12y = 30
4x – 12y = 38
Now add the equations to eliminate the y terms.
6x = 68
x = 68/6
x = 34/3
Substitute 34/3 in for x in either equation and solve for y.
2(34/3) -6y = 19
( 68/3) – 6y = 19
-6y = 19 – (68/3)
-6y = (57/3) – (68/3) (change 19 to 57/3 by multiplying 19 by 3/3)
-6y = -11/3
y = 11/18
This answer can easily be verified by substituting back into the original equations. Note that it may be easier to use a calculator to verify. We can solve word problems involving 2 variables by writing and solving a system of equations.
Example: Suppose Tom has 30 coins consisting of dimes and quarters. If he has a total of $5.55, how many of each coin does he have?
Solution:
Let x = number of dimes
y = number of quarters
Number of dimes plus number of quarters equals 30 coins.
x + y = 30
Value of dimes plus value of quarters equals $5.55.
0.10x + 0.25y = 5.55
The system is x + y = 30
0.10x + 0.25y = 5.55
We will solve this system using the elimination method. To eliminate the x terms, multiply the first equation by -0.10. This gives us
-0.10x – 0.10y = -3
0.10x + 0.25y = 5.55
Next, add the two equations to get 0.15y = 2.55, therefore y = 17.
Substitute 17 in for y in the first equation to get 13 for x. Therefore, Tom has 13 dimes and 17 quarters.
This guide should ease any confusion on the topic of solving systems of 2 linear equations using the elimination method.
