Basic Algebra/Systems of Linear Equations/Solving Linear Systems by Substitution

VocabularyEdit

LessonEdit

To solve a system of linear equations without graphing, you can use the substitution method. This method works by solving one of the linear equations for one of the variables, then substituting this value for the same variable in the other linear equation and solving for the other variable. It does not matter which equation you choose first, or which variable you solve for first; the values for both variables will be the same.

Example ProblemsEdit

For example, given the system of linear equations:

2x-3y=-2

4x+y=24


The first step would be to choose one of the equations and solve it for either x or y. In the second equation y is not multiplied by a constant so it can be isolated in less steps.

4x+y=24

y=24-4x


Now you have a value, 24-4x, for y. Substitute this value of y into the first equation.

2x-3y=-2

2x-3(24-4x)=-2


Solve this equation for the variable x.

2x-3(24-4x)=-2

2x-72+12x=-2

14x-72=-2

14x=70

x=5


We now have a value for x that can be substituted into either equation to solve for y.

4x+y=24

4(5)+y=24

20+y=24

y=4


The solution to this system of linear equations is x=5, y=4. This can also be written as (x,y)=(5, 4).


NOTE: If we substitute the value of x into the other equation, the value of y will remain the same.

2x-3y=-2

2(5)-3y=-2

10-3y=-2

-3y=-12

y=4

Practice GamesEdit

[1] Practice Problems


[2] Math Drills

Practice ProblemsEdit

Solve the system of linear equations.

1)

7x+2y=16

-21x-6y=24


2)

9x+4y=20

8x-10y=60


3)

t-2r=3

5r-4t=6


4)

3v-w=7

2v+3w=1


5)

p-4q=-21

3p+q=2

Last modified on 14 May 2010, at 19:29