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ACTIVITY 20: Systems of Linear Equations (Section 6.2, pp. 469-474) in Two Variables.

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Presentation on theme: "ACTIVITY 20: Systems of Linear Equations (Section 6.2, pp. 469-474) in Two Variables."— Presentation transcript:

1 ACTIVITY 20: Systems of Linear Equations (Section 6.2, pp. 469-474) in Two Variables

2 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.

3 Example 1: Solve the system Multiply by -3 Multiply by 4 However, 0 is not equal to 3 so there are no solutions to this system!

4 Example 2:  Solve the system 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.

5 Example 3: Show that the system has infinitely many solutions and express them in the ordered pair (parametric) form If we multiply the second equation by 3 then we obtain first equation. Consequently, there are infinitely many solutions.

6 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 Multiply by 2 Multiply by 5

7 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 Multiplying the second equation by -3 we obtain

8 = 25% Now substituting the value for y back into equation one we obtain = 10%

9 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. Let’s right our number as ‘ab’ Notice that ab = 10a + b Reverseing we have ba = 10b + a Consequently, our number is


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