y - DoeringGrade8ACC

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I. Model Problems.
II. Practice
III. Challenge Problems
VI. Answer Key
Web Resources
Systems of Linear Equations
www.mathwarehouse.com/algebra/linear_equation/systems-of-equation/
Interactive System of Linear Equations:
www.mathwarehouse.com/algebra/linear_equation/systems-of-equation/interactivesystem-of-linear-equations.php
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I. Model Problems
Systems of linear equations can be solved by graphing. To solve by
graphing, graph both of the linear equations in the system. The solution
to the system is the point of intersection of the two lines.
Example 1 Solve the system by graphing:
y=x+5
y = 2x
Graph both lines. The graph is shown below:
Notice that the intersection of the two lines is at the point (5, 10).
The solution is x = 5, y = 10, or (5, 10).
Sometimes the lines do not intersect. This occurs when the lines
graphed are parallel. In this case, the system of equations is said to have
no solutions.
Example 2 Solve the system by graphing:
y = 2x + 10
y = 2x – 5
Notice that the slopes of these lines are equal, so they are parallel. This
is confirmed by graphing:
There is no solution to the system of equations.
Sometimes the two equations in the system will yield the graph of the
same line. In this case the system is said to have “infinitely many”
solutions.
Example 3 Solve the system by graphing:
y = 3x + 10
2y – 20 = 6x
The graph is shown:
As you can see both equations yield the same graph.
There are infinitely many solutions to the system.
II. Practice
Solve each system of linear equations by graphing. Use estimation to
calculate solutions that are not integers. If there is no solution or
infinitely many solutions, so state.
 y  3x
1. 
y  x  4
y  x  2
2. 
 y  2x  5
 y  3x  2
3. 
 y  5x  10
 y  12 x  5
4. 
 y  2x  10
 y  3x  16
5. 
 y  5x
 y  3x  1
6. 
2
12
 y   5 x  5
 y  2x  5
7. 
 y  5x  9
 y  13 x  1
8. 
2
 y   3 x
2 y  6x  4
9. 
 y  3x  2
 y  2x  16
10. 
 y  14x
 y  5x  2
11. 
y  x  3
 y  2x  10
12. 
4 y  8x  16
 y  2x  10
13. 
y  x  3
 y  6x  5
14. 
3y  18x  15
 y   23 x  15
15. 
3y  2x  7
 y  4x  5
16.  1
5
 2 y  2x  2
 y   23 x  12
17. 
1
5
 y  4 x  6
3x  4 y  10
18.  9
 2 x  6x  15
III. Challenge Problems
19. Explain how you can tell if a system of linear equations has no
solutions by analyzing the slope of each line.
_________________________________________________________
20. Consider the following system:
y = ax + b
y = cx + d
If a and b are both positive and c and d are both negative, in which
quadrant is the solution to the system? Is there more than one possible
answer?
_________________________________________________________
21. Correct the Error.
2x  3y  18
Question: Solve 
2x  5y  10
Solution:
Since the coefficient of x for both equations is 2, the slope for both
lines is equal to 2. Therefore, the lines are parallel. Systems of parallel
lines do not have any solutions, so there is no solution.
What is the error? Explain how to solve the problem.
_________________________________________________________
_________________________________________________________
IV. Answer Key
1. (2, 6)
2. (-7, -9)
3. (6, 20)
4. (-6, 2)
5. (2, -10)
6. (1, 2)
7. (2, 1)
8. (1, -2/3)
9. infinitely many solutions
10. (1, 14)
11. (5/4, 17/4)
12. no solutions
13. (5/4, 15/2)
14. infinitely many solutions
15. no solutions
16. infinitely many solutions
17. (-1/5, 4/5)
18. infinitely many solutions
19. If the slopes are equal, the lines are parallel (if they are not the same
line). If the lines are parallel, they do not intersect and there are no
solutions.
20. The solution would be in either Quadrant II or Quadrant III. The
exact location of the solution depends on the exact values of a, b, c and
d.
21. The student incorrectly stated that the slope for both lines equals 2.
The student needs to write each equation in slope-intercept form (y =
mx + b) in order to state that the coefficient of x equals the slope. The
equations in slope-intercept form are y = (-2/3)x + 6 and y = (-2/5)x + 2.
The slopes are not equal. The lines intersect at the point (-5, 4).