Section 10.2 Solving Linear Systems By Substitution

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Section 10.2 Solving Linear Systems By
Substitution
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Solving Linear Systems

• There are two methods of solving a system of
  equations algebraically
• 1. Substitution
• 2. Elimination

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Substitution

• To solve a system of equations by substitution
• 1. Solve one equation for one of the variables.
• 2. Substitute the value of the variable into the
  other equation.
• Simplify and solve the equation.
• 4. Substitute back into either equation to find
  the value of the other variable.

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Substitution

• Solve the system x - 2y -5
• y x 2
• Substitute (x 2) for y in the first
  equation.
• x - 2y -5
• x - 2(x 2) -5
• x - 2x 4 -5
• -x - 4 -5
• -x -1
• x 1
• Substitute 1 for x in either equation to find
  y.
• y x 2 y 1 2 so y 3
• The solution is (1, 3).

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Substitution

• Lets check the solution. The answer (1, 3)
• must check in both equations.
• x - 2y -5 y x 2
• 1 - 2(3) -5 3 1 2
• -5 -5P 3 3 P

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Substitution

• Solve the system 2p 3q 2
• p - 3q -17
• Notice that neither equation is solved for a
  variable. Since p in the second equation does
  not have a coefficient, it will be easier to
  solve.
• p - 3q -17
• p 3q 17
• Substitute the value of p into the first
  equation, and solve.
• 2p 3q 2 9q 34 2
• 2(3q 17) 3q 2 9q 36
• 6q 34 3q 2 q 4

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Substitution

• Substitute the value of q into the second
  equation to find p.
• p 3q 17
• p 3(4) 17
• p -5
• The solution is (-5, 4).
• Lets check the solution
• 2p 3q 2 p 3q -17
• 2(-5) 3(4) 2 -5 - 3(4)
  -17
• -10 12 2 -5 - 12 -17
• 2 2P
  -17 -17

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Substitution

• Solve the systems by substitution
• 1. x 4
• 2x - 3y -19
• 2. 3x y 7
• 4x 2y 16
• 3. 2x y 5
• 3x 3y 3
• 4. 2x 2y 4
• x 2y 0

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Elimination

• The key to solving a system by elimination is
  getting rid of one variable.
• Lets review the Additive Inverse Property.
• What is the Additive Inverse of 3x? -5y?
  8p? q?
• 
  -3x 5y -8p -q
• What happens if we add two additive inverses?
• We get zero. The terms cancel.
• We will try to eliminate one variable by adding,
  subtracting, or
• multiplying the variable(s) until the two terms
  are additive inverses.
• We will then add the two equations, giving us one
  equation with one variable.
• Solve for that variable.
• Then insert the value into one of the original
  equations to find the other variable.

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Elimination

• Solve the system m n 6
  m - n 5
• Notice that the n terms in both equations are
  additive inverses. So if we add the equations the
  n terms will cancel.
• So lets add solve m n 6
  m - n 5
• 2m 0 11
• 2m 11
• m 11/2 or 5.5
• Insert the value of m to find n 5.5 n 6
• n .5
• The solution is (5.5, .5).

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Elimination

• Solve the system 3s - 2t 10
  4s t 6
• We could multiply the second equation by 2 and
  the
• t terms would be inverses. OR
• We could multiply the first equation by 4 and the
  second equation
• by -3 to make the s terms inverses.
• Lets multiply the second equation by 2 to
  eliminate t. (Its easier.)
• 3s - 2t 10 3s 2t 10
• 2(4s t 6)
  8s 2t 12
• Add and solve 11s 0t 22
• 11s 22
• s 2
• Insert the value of s to find the value of t
  3(2) - 2t 10 t -2
• The solution is (2, -2).

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Elimination

• Solve the system by elimination
• 1. -4x y -12
• 4x 2y 6
• 2. 5x 2y 12
• -6x -2y -14
• 3. 5x 4y 12
• 7x - 6y 40
• 4. 5m 2n -8
• 4m 3n 2