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Topic: Solving Systems of Equations by Substitution

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Use the solution for Y to find the answer for X using substitution again. ... If one of the equations in a system is already solved for X or Y then it is born ... – PowerPoint PPT presentation

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Title: Topic: Solving Systems of Equations by Substitution


1
Topic Solving Systems of Equations by
Substitution
  • X Y 5
  • X Y -7

2
Step One
  • To solve the system by substitution method
  • X Y 5
  • X Y -7
  • Solve one of the equations for X or Y. (It
    doesnt matter which one you choose. Pick the
    variable you think is easiest!)
  • If X Y -7
  • then X Y 7 ( just add y to both sides)

3
Step Two
  • X Y 5
  • X Y -7 so X Y 7
  • Substitute the value for X in for the X in the
    second equation.
  • X Y 7 so in the equation X Y 5 when you
    substitute Y 7 in for X you get the equation (Y
    7) Y 5.

4
Step Three
  • Solve the equation for Y.
  • ( Y 7 ) Y 5
  • 2Y 7 5 (combine like terms)
  • 2Y 12 (add 7 to both sides)
  • Y 6 (divide both sides by 2)

5
Step Four
  • Use the solution for Y to find the answer for X
    using substitution again. (You may use either
    equation for the last substitution.)
  • Remember Y 6
  • If X Y 7, then using substitution
  • X 6 7 and so X -1
  • The solution to the system is (-1,6)

6
Summary
  • Solving by Substitution Method
  • Sample X Y 5
  • X Y -7
  • X Y 7 ( Solve for X in the
    2nd equation)
  • (Y 7) Y 5 (Substitute into the
    1st equation)
  • 2Y 7 5 (Solve)
  • 2Y 12
  • Y 6
  • so X 6 7
  • and X -1.
  • Solution to the system is (-1,6)

7
If one of the equations in a system is already
solved for X or Y then it is born for the
Substitution Method!!
  • If 4X 3Y 12 and Y 2X 5 then
  • 2X 5 would be substituted for Y in the
    first equation.
  • 4X 3(2x 5) 12
  • 4X 6X 15 12
  • 10X 15 12
  • 10X 27
  • X 2.7
  • So if X 2.7 then Y 2(2.7) 5
  • Y .4 and the solution to the system is (2.7,
    -.4)
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