Solving Systems of Linear Equations by Elimination

Systems of Linear Equations Lesson #4: Solving Systems of Linear Equations by Elimination So far we have used three methods to solve systems of equations: graphing, inspection, and substitution. In this lesson we will learn another algebraic technique: the method of elimination. This method is particularly useful when the equations involve fractions. Method of Elimination In using the method of elimination, there are four steps which are shown below. Step 1 If necessary, multiply each equation by a constant to obtain coefficients for (or y) that are identical (except perhaps for the sign). Step 2 Add or subtract the two equations to eliminate one of the variables. Step 3 Solve the resulting equation to determine the value of one of the variables. Step 4 Substitute the solution into either of the original equations to determine the value of the other variable. Class Ex. #1 Consider the system of equations: a) Add the two equations. This will eliminate the variable y. 2x+7=13 3x-7y=2 5x = 15 5 5 x=3 b) Use the equation in a) to determine the value of x and hence solve the system. 2x+4y=13 2 (5)+74-13 6+4y=15 y=1 x=3, y=1 e) Verify the solution satisfies both equations. 2x+4y=13 LS 45 2(3)+(1) 13 = 13 3x-74-2 15 PS 3(3)-7(1) 2 2 Chess Ex. #2 Consider the system of equations: 2x+6y=6 2x+3y=4.5 ay-1.5 y=0.5 a) Subtract the two equations. This will eliminate the variable x. b) Use the equation in a) to determine the value of y and hence solve the system. 2x+by=6 2x=3 X = 1.5 2x+3=6 2x+6(0.5)=6 X = 3/2 x=1.5, y=0.5 c) Verify the solution satisfies both equations. 2x+6y=62115610 2(1.5)-6(0.5) 6 =6 2x+3y = 4.5 LS RS 2/15)3(0.5) 4.5 = 4.5 Complete Assignment Questions #1-#3 474 Systems of Linear Equations Lesson #4: Solving Systems... by Elimination Claus Ex. 13 Consider the system of equations: 2x+3y=4 4x-y=22 a) Does adding or subtracting the equations eliminate either of the variables? No. b) Multiply the second equation by 3 and then add the two equations. 3(4x-y=22) 12x-3y=b6 + 2x+34=4 12x-3y=66 14x = 40 c) Solve and verify the system. 14x = 70 x=5 2x+3y=4 2(5)+3y=4 10+3y=4 3y=-6 y=-2 x=5, y=-2 2x+3y=4 LS: 2(5)+3(-2)=4 RS: 4 4x-4-22 LS: 4 (5)-(-2)=22 PS:22 LS-RS LS=PS d) Consider the original system. Multiply the first equation by an appropriate number which will eliminate x by addition or subtraction. Solve the system. 2(2x+3y=4) 4x+by=8 4x+6y= 8 22 7y=-14 y=-2 2x+3y=4 2x+3(-2)=4 2x-6=4 2x = 10 x=5 x=5, y=-2 Clast Ex.M Consider the system of equations: 5a+3b 3 3a-7b=81 a) Choose appropriate whole numbers to multiply each equation so that the system can be solved by eliminating b. equation 1 (x7) equation 2 (x3) b) Solve and verify the system by eliminating b. work 75036 +3) 3(৮) 35a + 2lb = 21 +90-216 = 243 449 = 264 a=6 Sa+3b3 5(6)+36=3 30-36-3 36-27 b=-9 9=6, b=-9 59136-3 LS: 5(6)+3(-9)=3 39-76-81 LS:3(6)-7(-9 75:8) LS = RS LS-PS verified ✓ c) Choose appropriate whole numbers to multiply each equation so that the system can be solved by eliminating a. equation 1 (3), equation 2 (x5) d) Solve the system by eliminating a. Ba3b3 side work 3/5a+3b #3) 5(39-7681) 154-356405 445-396 b=-9 54+3(-1)=3 59-27-3 59 =30 a=6 a=6, b=-9 Systems of Linear Equations Lesson #4: Solving Systems by Elimination 475 Solve the following system using elimination. 4x+2y-13=0, 3x=5y+26 4x+2y=13 (x5) 3x-5y = 26 (x2) 4x+2y = 15 4(4)+2y=13 18+2y=13 2y-5 y=- Class Ex. 36 20x+10y = 65 (add)+ 6x-104 = 52 26x = 117 2 Solve the following system using elimination. *32-2+2=2 (15) *multiplying by 15 and 10 removes the fraction so it is easier to solve 15(2)-15(+2)=15(2) S(x-2)-3(y+2)= 80 5x-10-34-6-30 5x-3y=46 13 (x+1)-(-3)=(<10) 10 [(x+1)]-10 [(4-3)] = 10 (4) 16(x+1)-8(y-3) = 105 6x+6-8y+24 = 105 6x-84 = 75 5x-3y=46 (18) 6x-8y=75 (-) 6x-89=75 6(2)-8y=75 39-84-75 -84-36 y=-1 x,y= (add) + 40x244 = 368 -18x+244-225 22x 143 13 Complete Assignment Questions #4-#12 Assignment 1. In each of the following systems: solve the system using the method of elimination by adding the equations. verify the solution satisfies both equations. a) 8x-y=10/add) 4x+y=14 12x=24 X=2 b) x+2y=3 (add) -x+3y=2 54 #5 c), 4a-3b-2 (add) -4a-b6 -4b=8 b=-2 4x+y=14 4(2)+y=14 y=6 x=2, y=6 x+2y=3 40-3(-2)=2 49+6=2 49--4 a-1, b=-2 ax-y=10 4x+4=14 x+24=3 -X+34°2 LS: (1)+2(1)3 LS-(1)+3(1) 49-36-2 -44-b=6 L514 (-1)-3(-2) 15:-4(-1)-(-2) -6 LS: 8(2)-(6)=10154(2)+(6)=14 RS-10 LS-RS RS: 14 LS-RS verified RS:3 LS-RS 2 RS:2 LS = ES verified =2 RS:2 LS-ES 75:6 verified LS=RS