Download Activity 20:

ACTIVITY 20:
Systems of Linear Equations (Section 6.2, pp. 469-474)
in Two Variables
Number of Solutions of a Linear
System in Two Variables:
For a system of linear equations in two variables,
exactly one of the following is true:
1. The system has exactly one solution.
2. The system has no solution.
3. The system has infinitely many solutions.
Example 1:
Solve the system
3x  6 y  12

4 x  8 y  15
12 x  24 y  48
Multiply by 4
Multiply by -3  12 x  24 y  45
03
However, 0 is not equal to 3 so there are no
solutions to this system!
Example 2:

Solve the system
2 x  y  11
x  2y  4
Let us multiply the first
equation by 2 to cancel the
y’s.
Now we can take this value for x and
substitute it back into either equation.
Using equation 2 we obtain.
26
 2y  4
5
26
 2y  4 
5
4 * 5 26
 2y 

5
5
20 26

5
5
6
 2y 
5
4 x  2 y  22
x  2y  4
5x  26
26
x
5
 2y 
y
6 3

 10 5
 26 3 
 , 
 5 5
Example 3:
Show that the system
12 x  15 y  - 18

4 x  5 y  - 6
has infinitely many solutions and
express them in the ordered pair
(parametric) form
5 y  6  4 x
y
 6  4x
5
If we multiply the
second equation by 3
then we obtain first
equation.
Consequently, there
are infinitely many
solutions.
  6  4t

) | t is any real number 
(t ,
5


Example 4 (Smiley Face Speed):
A smiley face on a river travels downstream between two
points, 20 miles apart, in one hour. The return trip against the
current takes 2.5 hours. What is the smiley face’s speed, and
how fast does the current in the river flow?
Let ‘x’ = the speed of the smiley face
Let ‘y’ = the speed of the water
x  y  20
2.5 x  2.5 y  20
14  y  20
y6
Multiply by 5
Multiply by 2
5 x  5 y  100
5 x  5 y  40
10x  140
140
x
 14
10
Example 5 (Mixture Problem):

A chemist has two large containers of sulfuric acid solution, with different
concentrations of acid in each container. Blending 300 mL of the first
solution and 600 mL of the second solution gives a mixture that is 15% acid,
whereas 100 mL of the first mixed with 500 mL of the second gives a
12.5% acid mixture. What are the concentrations of sulfuric acid in the
original containers?
Let ‘x’ = the concentration of the first acidic solution
Let ‘y’ = the concentration of the second acidic solution
300x  600 y  0.15900  135
100x  500 y  0.125600  75
Multiplying the second equation by -3 we obtain
300 x  600 y  135
 300 x  1500 y  225
 900 y  90
 90
y
 .10 = 10%
 900
Now substituting the value for y back into
equation one we obtain
300x  600.10  135
300x  60  135
300x  75
75
x
 .25 = 25%
300
Example 6 (Number Problem):
The sum of the digits of a two-digit number is 7. When the digits are
reversed, the number is increased by 27. Find the number.
ab 7
Let’s right our number as ‘ab’ Reverseing we have
Notice that ab = 10a + b
ba = 10b + a
10b  a  10a  b  27
ab 7
9a  9b  63

9
a

9
b

27
 9a  9b  27
a5 7
a2
Consequently, our
number is 25
18b  90
90
b
5
18