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

1584

1263

yx

yxMultiply by -3Multiply by 4 482412 yx

452412 yx

30

However, 0 is not equal to 3 so there are no solutions to this system!

Example 2:

Solve the system

42

112

yx

yx Let us multiply the first equation by 2 to cancel the y’s. 4 2

2224

yx

yx

265 x

5

26x

425

26 y

Now we can take this value for x and substitute it back into either equation. Using equation 2 we obtain.

5

2642 y

5

26

5

5*42 y

5

26

5

202 y

5

62

y

10

6

y5

3

5

3,

5

26

Example 3:Show that the system

has infinitely many solutions and express them in the ordered pair (parametric) form

6- y 5 x 4

18- y 15 x 12If we multiply the second equation by 3 then we obtain first equation. Consequently, there are infinitely many solutions.

5

46 xy

number realany is |)5

46,( t

ttxy 465

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 faceLet ‘y’ = the speed of the water

20 yx205.25.2 yx Multiply by 2

Multiply by 5 10055 yx4055 yx

14010 x

10

140x 14

2014 y6y

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

90015.0600300 yx 600125.0500100 yx

13575

Multiplying the second equation by -3 we obtain

135600300 yx2251500300 yx

90900 y

900

90

y 10.

= 25%

Now substituting the value for y back into equation one we obtain

13510.600300 x13560300 x

75300 x300

75x 25.

= 10%

Example 6 (Number Problem):

The sum of the digits of a two-digit number is 7. When the digits arereversed, the number is increased by 27. Find the number.

7 ba

271010 baab

2799 ba7 ba 6399 ba

2799 ba

9018 b

18

90b 5

75 a

2a 25

Let’s right our number as ‘ab’Notice that ab = 10a + b

Reverseing we haveba = 10b + a

Consequently, our number is