Post on 26-Mar-2015
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Solving Systems by Graphing or Substitution.
Objective: To solve a system of linear equations in two variables by
graphing or by substitution.
System of Equations
• A system of equations is a collection of equations in the same variables.
System of Equations
• A system of equations is a collection of equations in the same variables.
• The solution of a system of two linear equations in x and y is any ordered pair, (x, y), that satisfies both equations. The solution (x, y) is also the point of intersection for the graphs of the lines in the system. For example, the ordered pair (2, -1) is the solution of the system below.
xy
xy
35
3
)2(351
321
Example 1
Example 1
Example 1
Try This
• Graph and classify the following system. Then, find the solution from the graph.
42
43
xy
xy
Try This
• Graph and classify the following system. Then, find the solution from the graph.
• Consistent, Independent.
42
43
xy
xy
Try This
• Graph and classify the following system. Then, find the solution from the graph.
• Consistent, Independent.• Solution is the point (0, 4)
42
43
xy
xy
Substitution
• There is also another way to find solutions to a system of equations. This is called substitution.
Example 2
Example 2
Example 2
Example 2
Try This
• Use substitution to solve the system. Check your solution.
428
83
yx
yx
Try This
• Use substitution to solve the system. Check your solution.
• Solve the first equation for y and substitute it into the second equation.
428
83
yx
yx
6
122
46168
4)38(28
38
x
x
xx
xx
xy
Try This
• Use substitution to solve the system. Check your solution.
• Solve the first equation for y and substitute it into the second equation. Find y. Use either equation.
428
83
yx
yx
6
122
46168
4)38(28
38
x
x
xx
xx
xy
26
8)6(3
y
y
26
522
42)6(8
y
y
y
Example 3
• A laboratory technician is mixing a 10% saline solution with a 4% saline solution. How much of each solution is needed to make 500 milliliters of a 6% solution?
Example 3
• A laboratory technician is mixing a 10% saline solution with a 4% saline solution. How much of each solution is needed to make 500 milliliters of a 6% solution?
Example 3
Example 3
Example 3
Example 3
Try This
• If a 7% saline solution and a 4% saline solution are mixed to make 500 milliliters of a 5% solution, how much of each solution, to the nearest milliliter, is needed?
Try This
• If a 7% saline solution and a 4% saline solution are mixed to make 500 milliliters of a 5% solution, how much of each solution, to the nearest milliliter, is needed?
• You need two equations.• x represents the 7% solution.• y represents the 4% solution.
)500(05.04.07.
500
yx
yx
Try This
• If a 7% saline solution and a 4% saline solution are mixed to make 500 milliliters of a 5% solution, how much of each solution, to the nearest milliliter, is needed?
• You need two equations.• x represents the 7% solution.• y represents the 4% solution.• Solve the first equation for y and substitute.
)500(05.04.07.
500
yx
yx
)500(05.)500(04.07.
500
xx
xy
Try This
• If a 7% saline solution and a 4% saline solution are mixed to make 500 milliliters of a 5% solution, how much of each solution, to the nearest milliliter, is needed?
• Solve.
333
167
503.
2504.2007.
)500(05.)500(04.07.
y
x
x
xx
xx
Example 4
Example 4
Example 4
Example 4
Example 4
Homework
• Pages 161-162• 13-41 odd
• You need to do all of these problems to be good at this. Please do that.