Download Solving Systems by Graphing or Substitution.

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.
 y  x 3

 y  5  3x
 1  2  3

  1  5  3(2)
Example 1
Example 1
Example 1
Try This
• Graph and classify the following system. Then, find
the solution from the graph.
 y  3x  4

 y  2 x  4
Try This
• Graph and classify the following system. Then, find
the solution from the graph.
 y  3x  4

• Consistent, Independent.
 y  2 x  4
Try This
• Graph and classify the following system. Then, find
the solution from the graph.
 y  3x  4

• Consistent, Independent.
 y  2 x  4
• Solution is the point (0, 4)
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.
 3x  y  8

8x  2 y  4
Try This
• Use substitution to solve the system. Check your
solution.
 3x  y  8

8x  2 y  4
• Solve the first equation for y and substitute it into the
second equation.
y  8  3x
8 x  2(8  3 x )  4
8 x  16  6 x  4
2 x  12
x  6
Try This
• Use substitution to solve the system. Check your
solution.
 3x  y  8

8x  2 y  4
• Solve the first equation for y and substitute it into the
second equation. Find y. Use either equation.
y  8  3x
8 x  2(8  3 x )  4
3(6)  y  8
8 x  16  6 x  4
y  26
2 x  12
x  6
8(6)  2 y  4
2 y  52
y  26
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?
x  y  500
• You need two equations.
• x represents the 7% solution. .07 x  .04 y  .05(500)
• y represents the 4% solution.
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?
x  y  500
• You need two equations.
• x represents the 7% solution. .07 x  .04 y  .05(500)
• y represents the 4% solution.
• Solve the first equation for y y  500  x
.07 x  .04(500  x)  .05(500)
and substitute.
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.
.07 x  .04(500  x)  .05(500)
.07 x  20  .04 x  25
.03 x  5
x  167
y  333
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.