Sections 1.1 & 1.2 Intro to Systems of Linear Equations & Gauss Jordan Elimination.
Solving Systems of Equations Using Elimination Students will be able to solve systems of equations...
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Transcript of Solving Systems of Equations Using Elimination Students will be able to solve systems of equations...
Solving Systems of Solving Systems of Equations Using Equations Using
EliminationElimination
Solving Systems of Solving Systems of Equations Using Equations Using
EliminationElimination
Students will be able to solve Students will be able to solve systems of equations using systems of equations using
elimination elimination
FHS Systems of Equations 2
Solving Systems - Elimination
• Another way to solve systems of equations is with the elimination method.
• With elimination, you get rid of (eliminate)one of the variables by adding or subtracting equations.
• The elimination method is sometimes called the addition and subtraction
method.
FHS Systems of Equations 3
Using Elimination - Example 1
Use elimination to solve the system of equations.
4 2 1
2
8
3 4y
x y
x
Step 1: Find the value of one variable.
3x + 2y = 4
4x – 2y = -18
7x = -14
x = -2
The y-terms have opposite coefficients.
Add the equations to eliminate y.
First part of the solution.
FHS Systems of Equations 4
Example 1 (cont.)
Step 2: Substitute the x-value into one of the original equations to solve for y.
3(-2) + 2y = 4
2y = 10
y = 5 Second part of the solution.
The solution is the ordered pair (-2, 5).
Substitute -2 in for x.
FHS Systems of Equations 5
What if Elimination Does Not Work?
• When you cannot eliminate one of the variables by just adding or subtracting the two equations, it is still possible to solve the system.
• Sometimes you can multiply one or both of the equations by some number that would make elimination possible.
FHS Systems of Equations 6
Use elimination to solve the system of equations.
Example 2 - Using Elimination
Step 1: To eliminate x, multiply both sides of the first equation by 2 and both sides of the second equation by –3.
Add the equations.
First part of the solution: y = –5
1
3 9
3 5
2
6
x y
x y
2(3x + 5y) = 2(–16)
–3(2x + 3y) = –3(–9)
6x + 10y = –32
–6x – 9y = 27
→
FHS Systems of Equations 7
Example 2 (Cont.)
Second part of the solution
Step 2: Substitute the y-value into one of the original equations to solve for x.
The solution for the system is (3, –5).
3x + 5(–5) = –16
3x – 25 = –16
3x = 9
x = 3