the student will be able to: solve systems of equations using elimination with multiplication....
TRANSCRIPT
The student will be able to:
solve systems of equations using elimination with
multiplication.
Objective
So far, we have solved systems using graphing, substitution, and elimination.
This goes one step further and show how to use ELIMINATION with multiplication.
What happens when the coefficients are not the same?
We multiply the equations to make them the same! You’ll see…
Solving Systems of Equations
1) Solve the system using elimination.
2(2) + 2y = 6
4 + 2y = 6
2y = 2
y = 1
2x + 2y = 6
3x – y = 5Multiply the bottom
equation by 2
2x + 2y = 6
(2)(3x – y = 5)
2x + 2y = 6(+) 6x – 2y = 10 8x = 16 x = 2
(2, 1)
x + 4y = 7
4x – 3y = 9
2) Solve the system using elimination.
Multiply the top equation by -4
(-4)(x + 4y = 7)
4x – 3y = 9
-4x – 16y = -28 (+) 4x – 3y = 9
-19y = -19 y = 14x – 3y = 9
4x – 3(1) = 9
4x – 3 = 9
4x = 12
x = 3
(3, 1)
Which variable is easier to eliminate?
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21 22 23 24 25 26 27 28 29 30 31 32
3x + y = 44x + 4y = 6
y is easier to eliminate. Multiply the 1st equation by (-4)
3x + 4y = -14x – 3y = 7
3) Solve the system using elimination.
3(1) + 4y = -1
3 + 4y = -1
4y = -4
y = -1
Multiply both equations
(3)(3x + 4y = -1)
(4)(4x – 3y = 7)
9x + 12y = -3 (+) 16x – 12y = 28
25x = 25 x = 1
(1, -1)
4) What is the best number to multiply the top equation by to eliminate the x’s?
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21 22 23 24 25 26 27 28 29 30 31 32
3x + y = 46x + 4y = 6
Multiply the 1st equation by (-2) (-2)(3x + y = 4) -6x -2y = -8
6x + 4y = 6
2y = -2
y = 1
6x + 4y = 66x + 4(1) = 6 6x = 2
x = 2/6x = 1/3
(1/3, 1)
5) Solve using elimination.
2x – 3y = 1x + 2y = -3
Multiply 2nd equation by (-2) to eliminate the x’s. (-2)(x + 2y = -3) -2x – 4y = 6
1st equation 2x – 3y = 1
-7y = 7
y = -1
2x – 3y = 1
2x – 3(-1) = 1
2x + 3 = 1
2x = -2
x = -1(-1, -1)
2(4x + y = 6 ) -8x – 2y = 13
8x + 2y = 12 -8x – 2y = 13
0 = 25
Add down to eliminate x.
But look what happens, y is eliminated too.
We now have a false statement, thus the system has no solution, it is inconsistent.
5) Solve using elimination.
6) Elimination
Solve the system using elimination.
5x – 2y = –15 3x + 8y = 37
20x – 8y = –60 3x + 8y = 37
23x = –23 x = –1
Since neither variable will drop out if the equations are added together. Multiply one or both ofthe equations by a constant to make one of the variables have the same number with opposite signs.
The best choice is to multiply the top equation by 4 since only one equation would have to be multiplied. Also, the signs on the y-terms arealready opposites.
(4)
6) Elimination
3x + 8y = 37 (second equation)
3(–1) + 8y = 37–3 + 8y = 37 8y = 40 y = 5
The solution is (–1, 5)
To find the second variable, substitute in any equation that contains two variables.
x = –1
7) Elimination
Solve the system using elimination.
4x + 3y = 8 3x – 5y = –23
20x + 15y = 40 9x – 15y= –69
29x = –29 x = –1
For this system, we must multiply both equations by a different constant in order to make one of the variables “drop out.”
It would work to multiply the top equation by –3 and the bottom equation by 4 OR to multiply the top equation by 5 and the bottom equation by 3.
(5)(3)
7) Elimination
4x + 3y = 84(–1) + 3y = 8
–4 + 3y = 8 3y = 12 y = 4
The solution is (–1, 4)
x = –1
8) Elimination Method
Solve the system.
To eliminate x, multiply equation (1) by –2 and
equation (2) by 3 and add the resulting equations.
3696
286
yx
yx
2
3417
y
y
1232
143
yx
yx (1)
(2)
8) Elimination Method
Substitute 2 for y in (1) or (2).
The solution is (3, 2)3
93
1)2(43
x
x
x
9) Solving an Inconsistent System
Solve the system
Eliminate x by multiplying (1) by 2 and adding the result to (2).
Solution set is .
746
423
yx
yx (1)
(2)
746
846
yx
yx
150 Inconsistent System
10) Solving a System with Dependent Equations
Solve the system.
Eliminate x by multiplying (1) by 2 and adding the result to (2).
Each equation is a solution of the other.
Infinite solutions.
42824
yxyx (1)
(2)
428
428
yx
yx
00