DSPM 0850

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DSPM 0850

• Sec. 4.2 Solving a system of equations
  algebraically
• Substitution
• Elimination

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Solving by Substitution

• 2x y 5 (Eqn 1)
• 3x - 2y 4 (Eqn 2)
• This system is a good candidate for solving by
  substitution since in one of the eqns (eqn 1) we
  have a coefficient of 1
• We will solve that eqn for y and substitute our
  result into eqn 2.

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Now Substitute

• 2x y 5 Eqn 1
• Y 5 - 2x
• We now substitute this value for y into eqn 2,
  simplify, and solve for x.
• 3x - 2(5 - 2x) 4 Eqn 2
• 3x - 10 4x 4
• 7x - 10 4
• 7x 14
• X 2

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Substitute Again!

• From solving eqn 2 we got x 2.
• We now substitute our value for x back into Eqn
  1.
• Y 5 - 2x
• Y 5 - 2(2)
• Y 5 - 4 1
• Our solution (2, 1)

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Another Good Candidate

• 3x - 6y 10 eqn 1
• x - 2y 3 eqn 2
• Why a good candidate for substitution?
• From eqn 2 x 3 2y
• Substitute back into eqn 1
• 3(3 2y) - 6y 10
• 9 6y - 6y 10
• 9 10 ??????

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An Inconsistent System

• Since 9 ? 10, we have No Solution!
• How could we have recognized that the system was
  a pair of parallel lines?
• Solve Algebraically by Substitution
• 3x - 6y 9
• x - 2y 3
• The result 9 9 tells us we have an infinite
  number of solutions.

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Another Try!

• 4x - y 1
• 2x 2y 5
• (eqn 1) y 4x - 1
• (eqn 2) 2x 2(4x - 1) 5
• 2x 8x - 2 5
• 10x - 2 5 10x 7 x 7/10
• Y 4(7/10) -1 9/5
• (7/10, 9/5) or (.7, 1.8)

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Solving by Elimination

• 3x - y 12 Eqn 1
• 5x y 12 Eqn 2
• Note that if we added eqn 1 eqn 2, the
  y-variable would be eliminated leaving us with
• 8x 24 and
• X 3

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Solving by Elimination, cont.

• We now substitute our value for x ( 3) back into
  either eqn 1 or eqn 2 whichever would be the
  easiest to work with.
• Eqn 2 y 12 - 5x
• Y 12 - 5(3)
• Y -3
• (3, -3)

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Now you!

• -5x 4y -8
• -x - 4y 10
• (-1/3, -29/12)

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But what if.

• 3x - 6y 3 eqn 1
• 5x - 2y 1 eqn 2
• Here, it is necessary to multiply each eqn by
  something in order to eliminate a variable.
• A possibility

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But what if.

• Mult. Eqn 2 by -3 we get
• 3x - 6y 3 eqn 1
• -3(5x - 2y 1) eqn 2
• 3x - 6y 3 eqn 1
• -15x 6y -3 eqn 2

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Put the oars in the boat

• Adding eqn 1 eqn 2
• -12x 0
• X 0
• And substituting into either eqn 1 or 2
• Y -1/2

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But what if .

• -3x 8y 7 Eqn 1
• 2x 5y 9 Eqn 2

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But what if .

• -3x 8y 7 Eqn 1
• 2x 5y 9 Eqn 2
• One possibility
• Mult eqn 1 by 2 Mult eqn 2 by 3

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But what if .

• -3x 8y 7 Eqn 1
• 2x 5y 9 Eqn 2
• One possibility
• Mult eqn 1 by 2 Mult eqn 2 by 3
• 2(-3x 8y 7) ? -6x 16y 14
• 3(2x 5y 9) ? 6x 15y 27

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But what if .

• -3x 8y 7 Eqn 1
• 2x 5y 9 Eqn 2
• One possibility
• Mult eqn 1 by 2 Mult eqn 2 by 3
• 2(-3x 8y 7) ? -6x 16y 14
• 3(2x 5y 9) ? 6x 15y 27
• Now eliminate and solve
• ( 37/31, 41/31)

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Lets Try

• 0.1x 0.4y 1.3
• 0.3x - 0.2y 1.1
• We can clean up the system by multiplying both
  eqns by 10!
• 1x 4y 13
• 3x - 2y 11
• Solve by elimination

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Lets Try
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Clean it up!

• By multiplying each eqn by its least common
  denominator we can clean up the system.
• 8x - 4y 7
• 10x - 8y 11

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Applications

• A student takes out 2 loans to help pay for
  college. One loan is a 8 simple interest, and
  the other is at 9 simple interest. The total
  amount borrowed was 3500, and the interest after
  1 year for both loans was 294. Find the amount
  of each loan.

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A Quantity-Value Table Helpful
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And the system is

• To develop the system, add column 1 add
  column 3.
• X y 3500
• .08x .09y 294

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Try another?

• Determine the milliliters of 10 sulfuric acid
  and the milliliters of 25 sulfuric acid that
  should be mixed to obtain 20 ml of 18 sulfuric
  acid.

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Now solve

• X y 20
• .10x .25y 3.6

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Homework Sec. 4.2

• 7, 8 13 21, 29, 33, 37, 41, 43, 45, 85, 89, 90.