Download Solving Linear Systems Using Algebra Methods

Advanced Algebra
Section 3.2: Solving Linear Systems Using Algebra Methods
In section 3.1 we solved linear systems by the graphing method which is not the most practical
method to find the point of intersection when it’s a fraction or decimal without using a
graphing calculator.
So in this section we will learn 3 algebraic methods to solve linear systems. They are
Substitution , _______________________________
Elimination by Addition/Subtraction , __________________________.
Elimination by Multiplication
____________
The substitution method is a great method to use when one of the coefficients of a
variable is 1 or -1. If you can’t find any variable with this type of coefficient, you will have
to solve the problem working with fractions.
Steps:
Get the variable with the coefficient of 1 by itself.
1. __________________________________________________________________
Take the equation from step 1, & substitute it into the other equation for that variable.
2. __________________________________________________________________
Solve the equation from step 2 for the variable.
3. __________________________________________________________________
Take the answer from step 3 and put it into the equation from step 1 & solve it.
4. ___________________________________________________________________
Put your answers for each variable into an ordered pair (x, y) & this is the pt. of
5. ___________________________________________________________________
intersection.
__________________
Examples: Solve the system .
1.
y = 2x
2x + 5y = -12
2x + 5(2x) = -12
2x + 10x = -12
y = 2x
y = 2(-1)
12x = -12
y = -2
12
12
x = -1
(-1, -2) pt. of intersection
2.
x+y=6
3x + y = 15
x+y=6
-y -y
x = -y + 6
3(-y + 6) + y = 15
xy6
-3y + 18 + y = 15
3
x 6
2
-2y + 18 = 15
-18
-18
3 12
x 
2 2
-2y = -3
-2
y
-2
3
2
x
9
2
9 3
 ,  pt. of int er sec tion
2 2
Elimination by Addition/Subtraction method when the coefficients of either
We will use the _______________________________
the x or y variables are opposites or the same number.
EX:
x – 4y = 6
3x + 4y = 10
** The coefficients of the of the y variables are opposites of each other.
Eliminate by Addition!
6x + 5y = 4
6x – 7y = -20
** The coefficients of the x variables are the same numbers.
Eliminate by Subtraction!
Examples:
3.
Solve the linear system.
-3x + 4y = 12
3x - 4y = -18
Coefficients of the x & y variables are opposites,
Eliminate by Addition.
 3 x  4 y 12
 3 x  4 y   18
0 6

Lines are parallel.
4.
3x + y = 6
4x + y = 7
The coefficients of the y variable are the same, Eliminate by Subtraction.
3x  y  6
 4x  y  7
3x  y  6
3x  y  6
  4 x  y  7
 x  1
1 1
x 1
3(1)  y  6
3 y  6
3
3
y 3
1,3
elimination by
We will use our last algebraic method to solve a linear system which is ___________
multiplication
______________.
Steps:
1. Pick one of the variables that you want to elimination.
2. Multiply one or both of the equations by a number so that the coefficients of one
of the variables will be opposites or the same.
3. Solve the rest of the problem like we did with addition/subtraction method.
Example:
5.
Solve the linear system.
3x + 5y = 11
2x + 3y = 7
2x  3y  7
2 x  3(1)  7
2x  3 7
3 3
2x  4
2
2
2  3x  5 y  11(2)
6 x  10 y  22
3(2 x  3 y)  7(3)
6 x  9 y   21
y 1
Solution:
(2,1)
x2
6.
3x – y = 2
x + 2y = 3
2(3x  y )  2(2)
6x  2 y  4
x  2y 3
x  2y 3
7x  7
7
7
x 1
(1,1)
x  2y 3
1 2 y  3
1
1
2y  2
2 2
y 1
Lesson 3.2 Practice C in the
packet: 1-20 all