Transcript Solving a system of Linear Equations by Elimination.

Solving a system of Linear
Equations by Elimination.
MPM2D
Elimination
• Recall:
From section 1.4, we discovered that given a linear
system such as
x + 2y = 4 (1)
x – y = 1 (2)
• We can add or subtract the equations to make a new
equation. For example: subtracting (2) from (1) gives….
x + 2y = 4 (1)
subtract
x–y=1
(2)
gives
3y = 3
(3)
Elimination
• Notice that the x-term was eliminated!
• We discovered that the linear system made of the
equations (1) and (2), is equivalent to the linear systems
made up of the equations (1) and (3) and the equations (2)
and (3).
• So we can use equation (3) to solve for y. That is,
3y = 3
y=1
(3)
Elimination
• Now we can substitute y=1 into either equation (1) or (2) to
solve for x:
x–y=1
(2)
x–1=1
x=1+1
x=2
• Now we have to check if the solution (2, 1) satisfies
equation (1) and (2).
Elimination
• As we used (2) to find x, we sub the point (2, 1) into (1):
x + 2y = 4 (1)
2 + 2(1) = 4?
2+2=4
YES!
Therefore (2, 1) is the solution.
Elimination
• Example 1 Solving by Addition
• Solve by elimination. Check the solution.
3x + 2y = 19 (1)
5x – 2y = 5
(2)
• Solution
• Adding the equations eliminates the y terms and creates an
equation with one variable!
Elimination
3x + 2y = 19
added to
(1)
5x – 2y = 5 (2)
8x = 24
Solve for x:
x = 24/8
x=3
Sub this into (1) or (2)
(1)
3(3) + 2y = 19
9 + 2y = 19
2y = 19 – 9
y = 10/2 = 5
Elimination
Check (3, 5) in equation (2)
5x – 2y = 5
5(3) – 2(5) = 5?
15 – 10 = 5?
5 = 5 Yes!
Therefore (3, 5) is the solution for this system.
Elimination
• Example 2 Solving Using Multiplication and Subtraction
• Solve by elimination. Check the solution.
3x + 2y = 2
4x + 5y = 12
• Solution
• Neither addition nor subtraction will eliminate a variable.
Use multiplication to write a system of equivalent
equations, so that a variable can be eliminated by addition
or subtraction.
Elimination
• Example 2
Solving Using Multiplication and Subtraction
Method 1: Eliminating y
Write (1): 3x + 2y = 2 ①
Write (2): 4x + 5y = 12 ②
Multiply (1) by 5:
15x + 10y = 10
Multiply (2) by 2: 8x + 10y = 24
Subtract ③ and ④
③
④
7x = – 14
x=–2
Sub x = – 2 into either equation to find y, and check the point.
Elimination
• Example 2
Solving Using Multiplication and Subtraction
Method 2: Eliminating x
Write (1): 3x + 2y = 2 ①
Write (2): 4x + 5y = 12 ②
Multiply (1) by 4:
Multiply (2) by 3:
Subtract ③ and ④
12x + 8y = 8
12x + 15y = 36
③
④
– 7y = – 28
y=4
Sub y = 4 into either equation to find x, and check the point.
Elimination
• Example 3 Rewriting Equations
• Solve and check:
0.6x – 0.3y = 2.4 ①
– 0.4y + 0.7x – 2.9 = 0 ②
• Solution
• Rewrite the equations with like terms in the same column.
0.6x – 0.3y = 2.4 ①
0.7x – 0.4y = 2.9 ②
Elimination
• Example 3
Rewriting Equations
0.6x – 0.3y = 2.4 ①
0.7x – 0.4y = 2.9 ②
• To clear the decimals, multiply each equation by 10.
6x – 3y = 24 ③
7x – 4y = 29 ④
• Strategy: it’s easier to eliminate the y-terms as they have a
lower common factor.
Elimination
• Example 4 Solving Systems of Equations With Fractions
• Solve and Check:
x
2

y
 4 ①
x
3
8

y
 2
②
2
• Solution
• To clear fractions, multiply each equation by its lowest
common denominator.
① x 8 and② x 6 to create 2 equivalent equations.
(Equivalent system)
Class/Homework
• 2D Math Power 10
• Pg. 30-33 #1bd, 2ac, 3a, 4a, 5bdf, 6cegi, 7a, 8a, 9ab,
10, 14, 16
• 2DE Principles of Math 10
• Pg 40 #1cd, 2d, 4b, 5cd, 7d, 8, 10, 12b, 15