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Solving systems of linear equations by substitution
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Solving systems of linear equations by substitution

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  • 1. Solving Systems of Linear Equations by Substitution
  • 2. Substitution Method
    Methods of solving systems of linear equations:
  • Substitution Method
    Example:
    Solve the following system using the substitution method.
    2x + y = 4 and 3x + 2y = 7
    Solving the first equation for y:
    2x + y = 4
    y = -2x + 4 Subtracting 2x from both sides.
    Substitute this expression for y into the second equation:
    3x + 2y = 7 The second equation
    3x + 2 ( -2x + 4) = 7 Replace y with its equivalent expression found above.
  • 5. Substitution Method
    3x + 2 ( -2x + 4) = 7 Replace y with its equivalent expression found above.
    3x – 4x + 8 = 7 Distributive Property
    -x + 8 = 7 Simplify
    -x = -1 Subtract 8 from both sides
    x = 1 Divide both sides by negative 1
    Substitute x = 1 into the first equation to solve for y:
    y = -2x + 4 Equation 1 solved for y
    y = -2 (1) + 4 = -2 + 4 = 2
    Our solution is the point (1, 2)
  • 6. Substitution Method
    We can verify that we found the correct answer by verifying that our point, (1, 2) is a solution to both of the equations in the system.
    First equation:
    2x + y = 4
    2(1) + 2 = 4
    4 = 4 True
    Second equation:
    3x + 2y = 7
    3 (1) + 2 (2) = 7
    3 + 4 = 7
    7 = 7 True. Since the point satisfies both equations, it is a solution to the system.
  • 7. Substitution Method
    Solving a System of Linear Equations using the Substitution Method
    Solve one equation for one of the variables
    Substitute the expression obtained in step 1 into the second equation.
    Solve for the remaining variable
    Substitute the value obtained in step 3 back into the equation from step 1 to obtain the second value of the ordered pair.
    Verify the answer by substituting the ordered pair into both equations.
  • 8. Substitution Method
    Example 2:
    Solve the following system of equations using the substitution method:
    y = x + 1 and 2y – 2x = 2
    Since we already have an equation solved for a variable we skip to step 2:
    2 (x + 1) – 2x = 2Substitute expression from first equation.
    2x + 2 – 2x = 2 Distributive property
    2 = 2 Simplify left side.
  • 9. Substitution Method
    Example 2 cont’d:
    This example resulted in an expression that is always true. Every solution to one of the equations is a solution to the other. If we graph these two equations, we find that they are the same line.
    We are interested in the solutions to the system. In this case there would be an infinite number of solutions and they are the same as the set of solutions for either of our equations.
    This is a diagram a system of equations that have infinitely many solutions. This image is for a linear algebra book written by Jim Hefferon released under GFDL and CC-BY-SA-2.5 licenses.
  • 10. Substitution Method
    Example 3:
    Solve the following system of equations using the substitution method:
    y = 2x + 3 and 2y – 4x = 5
    Since we already have and equation solved for a variable we skip to step 2:
    2 (2x + 3) – 4x = 5 Substitute expression from first equation.
    4x + 6 – 4x = 5 Distributive property
    6 = 5 Simplify left side.
  • 11. Substitution Method
    Example 3 cont’d:
    This example resulted in an expression that is never true. The two equations do not share any ordered pairs or points. If we graph these two equations, we find that they are parallel lines like those below.
    Each equation has an infinite number of solutions, however, we are interested in the solutions to the system. In this case there is not a solution to the system.
    This is a diagram a system of equations that have no solutions. This image is for a linear algebra book written by Jim Hefferon released under GFDL and CC-BY-SA-2.5 licenses.