Download Solving Systems of Linear Equations By Elimination

Solving Systems of Linear
Equations By Elimination
Unit 2
Lesson 3
Text Topic 3-2 pp.77-82
Warm-Up (SAT Question)
𝑥 2 + 𝑦 2 = 153
𝑦 = −4𝑥
If (𝑥, 𝑦) is a solution to the system of equations above,
what is the value of 𝑥 2 ?
Hint: Use substitution to find x first.
Essential Question

How is solving by elimination similar to
solving by substitution?
What is Elimination?

To eliminate means to get rid of or
remove.

You solve equations by eliminating one
of the variables (x or y) using the
addition and or subtraction of equivalent
equations.
Steps for Solving by Elimination

*Line up like terms vertically between the
two equations before starting (STANDARD FORM).

1. Choose a variable to eliminate.
2. Eliminate that variable by adding or subtracting one
equation from the other. (Sometimes you have to
multiply first.)

TIP: Use the coefficients of the equations as your multipliers, if
necessary.

3. Solve the new equation.
 4. Plug in your answer to find the other variable (or do
elimination for the other variable).
 5. Check your answer & write it as an ordered pair.
Example 1
Solve the following system of linear
equations by elimination.
(1)
2x – 3y = 15
(2)
5x + 3y = 27
Add equation (1)
7x + 0y = 42

to equation (2)
7x = 42  By eliminating y, we
x=6
can now solve for x
Example 1
Substitute x= 6 into
equation (1) to solve for y
2x – 3y = 15
2(6) – 3y = 15
12 – 3y = 15
– 3y = 15 – 12
– 3y = 3
y = -1
Check your solution x = 6
and y = -1 in equation (2)
5x + 3y = 27
5(6) + 3(-1) = 27
30 – 3 = 27
27 = 27
Therefore, the solution set = {(6,-1)}
Example 2
5x + 4y = -28 (1)
3x + 10y = -13 (2)
 If we were to add these equations we would obtain
8x + 14y = -41
 Even though we have only one equation now, we still
have 2 variables.
 We need to multiply the equations by values that will
allow us to eliminate either x or y. (Hint: use the coefficients)
Example 2
(1)
(2)
5x + 4y = -28
3x + 10y = -13
 If we multiply equation (1) by 5 and equation (2) by -2,
we be able to eliminate y using a 20 and -20.
(1) x 5
25x + 20y = -140 (3)
(2) x -2
-6x – 20y = 26 (4)
Add (3) & (4)  19x = -114
x = -6
 When you change the
equations you need to
renumber them.
Example 2
Substitute x = -6 into equation (1)
5x + 4y = -28
Check your answer x = -6 and
y = ½ into equation (2)
5(-6) + 4y = -28
-30 + 4y = -28
3(-6) + 10(½) = -13
4y = -28 +30
4y = 2
2
y
4
1
y
2
-18 + 5 = -13
-13 = -13
Therefore, the solution set = {(-6, ½)}
Assignment
20.
21.
22.
23.
24.
25.
26.
27.
19.
TB pg. 81 (19-27)