Algebra I : Semester II : Solving Systems

Sections:

Introduction  |   Section 1  |   Section 2  |   Section 3  |  Section 4  |  Section 5

Section Two

Part 1  |  Part 2  |  Part 3  |  Part 4

Algebra 1: Section 2: Solving Systems Elimination and Multiplication   Click here for a presentation on Elimination and Multiplication The method shown in the presentation will always work. However, you may want to use multiplication on just one equation instead.  Elimination with Multiplication First, check to see if a variable can be eliminated by addition or subtraction. Multiply the two sides of the first equation by the leading coefficient of the second. Multiply the two sides of the second equation by the leading coefficient of the first. If necessary, change all terms in the second equation to opposites. Then use elimination as usual. Example 1: Using Multiplication to Help Solve a System Solve the system. 3a + 2b = -12 5a – b = -20 Multiply the second equation by two to eliminate b. So, 13a = -52 a = -52/13 = -4 Going back to the original equation: 3a + 2b = -12 3(-4) + 2b = -12 -12 + 2b = -12 2b = -12 + 12 2b = 0 b = 0/2 = 0 The ordered pair is (-4, 0). Check this by substituting into the two original equations. Feel free to use the method in Example 1 when you feel it is best. However, the method used in the presentation will always work on any two variable system. You won’t have to examine the equations first. Example 2: Using the Rule Solve the system. 4c – 6d = 14 5c + 2d = 8 Steps     First, check to see if a variable can be eliminated by addition or subtraction.   4c – 6d = 14 5c + 2d = 8 Neither addition nor subtraction will eliminate a variable. Multiply the two sides of the first equation by the leading coefficient of the second. Multiply the two sides of the second equation by the leading coefficient of the first.   5(4c – 6d) = 5(14) 4(5c + 2d) = 4(8) which gives 20c – 30d = 70 20c +  8d  = 32   If necessary, change all terms in the second equation to opposites.    20c – 30d =  70 -20c – 8d   = -32   Use elimination as usual.  20c – 30d =  70 -20c – 8d   = -32            -38d = 38                 d = 38/-38 = -1 4c – 6d = 14 4c – 6(-1) = 14 4c + 6 = 14 4c = 14 – 6 4c = 8 c = 8/4 = 2 The solution is (2, -1). Check by substituting into the original equations. Quick Practice Solve the system using elimination. Problem Solution Details   6x + 4y = 20 4x – 2y = 4       5a – 3b = 3 2a + 6b = 30       2x + 3y = 12 4x + 6y = 24       3m – 4n = 1 2m + n = -3       Problem Solution Details   6x + 4y = 20 4x – 2y = 4 (2,2) 4(6x + 4y) = 4(20) 6(4x - 2y) = 6(4) 24x + 16y = 80 24x - 12y = 24 24x + 16y = 80 -24x + 12y = -24             28y = 56      y = 56/28 = 2 6x + 4y = 20 6x + 4(2) = 20 6x + 8 = 20 6x = 12 x = 2 5a – 3b = 3 2a + 6b = 30 (3, 4) 2(5a - 3b) = 2(3) 5(2a + 6b) = 5(30) 10a - 6b = 6 10a + 30b = 150 10a – 6b = 6 -10a – 30b = -150          -36b = -144     b = -144/-36 = 4 5a – 3b = 3 5a – 3(4) = 3 5a – 12 = 3 5a = 15 a = 3 2x + 3y = 12 4x + 6y = 24 Infinitely many 2(2x + 3y) = 2(12) 4x + 6y = 24 4x + 6y = 24 4x + 6y = 24 If you subtract you get 0 = 0.  They are the same equation. 3m – 4n = 1 2m + n = -3 (-1, -1) 2(3m - 4n) = 2(1) 3(2m + n) = 3(-3) 6m - 8m = 2 6m + 3n = -9 6m – 8n = 2 -6m – 3n = 9        -11n = 11    n = 11/ -11 = -1 3m – 4n = 1 3m – 4(-1) = 1 3m + 4 = 1 3m = -3 m = -1 Homework 3 (10 points) Click "Solving Systems" and then find the 6-HW3 link. You may do this more than one time to improve your score.  Now go on to the next part

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