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3 Variable Simultaneous Equations – Elimination Method3 Variable Simultaneous Equations – Elimination Method
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Question 1 of 5
1. Question
Solve the following simultaneous equations by elimination.3x-y+2z=54x+2y-z=65x-3y+z=1-
x= (1)y= (2)z= (2)
Hint
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Need TextPlayCurrent Time 0:00/Duration Time 0:00Remaining Time -0:00Stream TypeLIVELoaded: 0%Progress: 0%0:00Fullscreen00:00MutePlayback Rate1x- 2x
- 1.5x
- 1.25x
- 1x
- 0.75x
- 0.5x
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Chapters- Chapters
Elimination Method
- 1) make sure 2 equations have a variable with same coefficients but opposite signs
- 2) add or subtract the equations so that one variable is cancelled
- 3) solve for the variable that remains
- 4) substitute known value to one of the equations to solve for the other variable
First, label the equations 1, 2 and 3 respectively.3x-y+2z = 5 Equation 1 4x+2y-z = 6 Equation 2 5x-3y+z = 1 Equation 3 Next, add equation 2 to equation 3 since their variable z both have 1 as coefficient but with opposite signs. Label the result as equation A.4x+2y-z = 6 + 5x-3y+z = 1 -z+z cancels out 9x-y = 7 Equation A Next, multiply the values of equation 2 by 2 and add the product to equation 1. Label the result as equation B.2×(4x+2y-z) = 2×6 Multiply equation 2 by 2 8x+4y-2z = 12 8x+4y-2z = 12 + 3x-y+2z = 5 +2z-2z cancels out 11x+3y = 17 Equation B Next, multiply the values of equation A by 3 and add the product to equation B to find the value of x.3×(9x-y) = 3×7 Multiply equation A by 3 27x-3y = 21 27x-3y = 21 + 11x+3y = 17 38x = 38 -3y+3y cancels out 38x÷38 = 38÷38 Divide both sides by 38 x = 1 Now, substitute the value of x into equation A to solve for y.9x -y = 7 Equation A 9(1)-y = 7 x=1 9-y = 7 9-y -9 = 7 -9 Subtract 9 from both sides -y = -2 -y ×(-1) = -2 ×(-1) Multiply both sides by -1 y = 2 Finally, substitute the value of x and y into equation 1 to solve for z.3x-y +2z = 5 Equation 1 3(1)-2 +2z = 5 x=1 and y=2 3-2+2z = 5 1+2z = 5 1+2z -1 = 5 -1 Subtract 1 from both sides 2z = 4 2z ÷2 = 4 ÷2 Divide both sides by 2 z = 2 x=1y=2z=2 -
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Question 2 of 5
2. Question
Solve the following simultaneous equations by elimination.3x+5y+z=59x+2y+7z=326x+3y+4z=19-
x= (3)y= (-1)z= (1)
Hint
Help VideoCorrect
Keep Going!
Incorrect
Need TextPlayCurrent Time 0:00/Duration Time 0:00Remaining Time -0:00Stream TypeLIVELoaded: 0%Progress: 0%0:00Fullscreen00:00MutePlayback Rate1x- 2x
- 1.5x
- 1.25x
- 1x
- 0.75x
- 0.5x
Subtitles- subtitles off
Chapters- Chapters
Elimination Method
- 1) make sure 2 equations have a variable with same coefficients but opposite signs
- 2) add or subtract the equations so that one variable is cancelled
- 3) solve for the variable that remains
- 4) substitute known value to one of the equations to solve for the other variable
First, label the equations 1, 2 and 3 respectively.3x+5y+z = 5 Equation 1 9x+2y+7z = 32 Equation 2 6x+3y+4z = 19 Equation 3 Next, multiply the values of equation 1 by 3 and subtract equation 1 from the product. Label the result as equation A.3×(3x+5y+z) = 3×5 Multiply equation 1 by 3 9x+15y+3z = 15 9x+15y+3z = 15 - 9x-2y-7z = 32 9x-9x cancels out 13y-4z = -17 Equation A Then, multiply the values of equation 1 by 2 and subtract equation 3 from the product. Label the result as equation B.2×(3x+5y+z) = 2×5 Multiply equation 1 by 2 6x+10y+2z = 10 6x+10y+2z = 10 - 6x-3y-4z = 19 6x-6x cancels out 7y-2z = -9 Equation B Now, multiply the values of equation B by 2 and subtract equation A from the product to find the value of y.2×(7y-2z) = 2×-9 Multiply equation B by 2 14y-4z = -18 14y-4z = -18 - 13y+4z = 17 -4z+4z cancels out y = -1 Now, substitute the value of y into equation B to solve for z.7y -2z = -9 Equation B 7(-1)-2z = -9 y=-1 -7-2z = -9 -7-2z +7 = -9 +7 Add 7 to both sides -2z = -2 -2z ÷(-2) = -2 ÷(-2) Divide both sides by -2 z = 1 Finally, substitute the value of y and z into equation 1 to solve for x.3x+5y+z = 5 Equation 1 3x+5(-1)+1 = 5 y=-1 and z=1 3x-5+1 = 5 3x-4 = 5 3x-4 +4 = 5 +4 Add 4 to both sides 3x = 9 3x ÷3 = 9 ÷3 Divide both sides by 3 x = 3 x=3y=-1z=1 -
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Question 3 of 5
3. Question
Solve the following simultaneous equations by elimination.x+2y+5z=23x+4y-4z=212x-y+z=3-
x= (3)y= (2)z= (-1)
Hint
Help VideoCorrect
Excellent!
Incorrect
Need TextPlayCurrent Time 0:00/Duration Time 0:00Remaining Time -0:00Stream TypeLIVELoaded: 0%Progress: 0%0:00Fullscreen00:00MutePlayback Rate1x- 2x
- 1.5x
- 1.25x
- 1x
- 0.75x
- 0.5x
Subtitles- subtitles off
Chapters- Chapters
Elimination Method
- 1) make sure 2 equations have a variable with same coefficients but opposite signs
- 2) add or subtract the equations so that one variable is cancelled
- 3) solve for the variable that remains
- 4) substitute known value to one of the equations to solve for the other variable
First, label the equations 1, 2 and 3 respectively.x+2y+5z = 2 Equation 1 3x+4y-4z = 21 Equation 2 2x-y+z = 3 Equation 3 Next, multiply the values of equation 1 by 3 and subtract equation 2 from the product. Label the result as equation A.3×(x+2y+5z) = 3×2 Multiply equation 1 by 3 3x+6y+15z = 6 3x+6y+15z = 6 - 3x-4y+4z = 21 3x-3x cancels out 2y+19z = -15 Equation A Then, multiply the values of equation 1 by 2 and subtract equation 3 from the product. Label the result as equation B.2×(x+2y+5z) = 2×2 Multiply equation 1 by 2 2x+4y+10z = 4 2x+4y+10z = 4 - 2x+y-z = -3 2x-2x cancels out 5y+9z = 1 Equation B Now, multiply the values of equation A by 5 and the values of equation B by 2, then get their difference to find the value of z.5×(2y+19z) = 5×-15 Multiply equation A by 5 10y+95z = -75 2×(5y+9z) = 2×1 Multiply equation B by 2 10y+18z = 2 10y+18z = 2 - 10y-95z = 75 10y-10y cancels out -77z = 77 -77z ÷(-77) = 77 ÷(-77) Divide both sides by -77 z = -1 Now, substitute the value of z into equation B to solve for y.5y+9z = 1 Equation B 5y+9(-1) = 1 z=-1 5y-9 = 1 5y-9 +9 = 1 +9 Add 9 to both sides 5y = 10 5y ÷5 = 10 ÷5 Divide both sides by 5 y = 2 Finally, substitute the value of y and z into equation 1 to solve for x.x+2y+5z = 2 Equation 1 x+2(2)+5(-1) = 2 y=2 and z=-1 x+4-5 = 2 x-1 = 2 x-1 +1 = 2 +1 Add 1 to both sides x = 3 x=3y=2z=-1 -
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Question 4 of 5
4. Question
Solve the following simultaneous equations by elimination.-2x-y+5z=53x+2y-z=7-4x+4y+2z=12-
x= (1)y= (3)z= (2)
Hint
Help VideoCorrect
Well Done!
Incorrect
Need TextPlayCurrent Time 0:00/Duration Time 0:00Remaining Time -0:00Stream TypeLIVELoaded: 0%Progress: 0%0:00Fullscreen00:00MutePlayback Rate1x- 2x
- 1.5x
- 1.25x
- 1x
- 0.75x
- 0.5x
Subtitles- subtitles off
Chapters- Chapters
Elimination Method
- 1) make sure 2 equations have a variable with same coefficients but opposite signs
- 2) add or subtract the equations so that one variable is cancelled
- 3) solve for the variable that remains
- 4) substitute known value to one of the equations to solve for the other variable
First, label the equations 1, 2 and 3 respectively.-2x-y+5z = 5 Equation 1 3x+2y-z = 7 Equation 2 -4x+4y+2z = 12 Equation 3 Next, multiply the values of equation 1 by 2 and add equation 2 to the product. Label the result as equation A.2×(-2x-y+5z) = 2×5 Multiply equation 1 by 2 -4x-2y+10z = 10 -4x-2y+10z = 10 + 3x+2y-z = 7 -2y+2y cancels out -x+9z = 17 Equation A Then, multiply the values of equation 2 by 2 and subtract equation 3 from the product. Label the result as equation B.2×(3x+2y-z) = 2×7 Multiply equation 2 by 2 6x+4y-2z = 14 6x+4y-2z = 14 - 4x-4y-2z = -12 4y-4y cancels out 10x-4z = 2 Equation B Now, multiply the values of equation A by 10 and add the product to equation A to find the value of z.10×(-x+9z) = 10×17 Multiply equation A by 10 -10x+90z = 170 10x-4z = 2 + -10x+90z = 170 10x-10x cancels out 86z = 172 86z ÷86 = 172 ÷86 Divide both sides by 86 z = 2 Now, substitute the value of z into equation B to solve for x.10x-4z = 2 Equation B 10x-4(2) = 2 y=-1 10x-8 = 2 10x-8 +8 = 2 +8 Add 8 to both sides 10x = 10 10x ÷10 = 10 ÷10 Divide both sides by 10 x = 1 Finally, substitute the value of x and z into equation 2 to solve for y.3x+2y-z = 7 Equation 2 3(1)+2y-2 = 7 x=1 and z=2 3+2y-2 = 7 1+2y = 7 1+2y -1 = 7 -1 Subtract 1 from both sides 2y = 6 2y ÷2 = 6 ÷2 Divide both sides by 2 y = 3 x=1y=3z=2 -
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Question 5 of 5
5. Question
Solve the following simultaneous equations by elimination.2a-b-4c=1a+6b-c=83a+4b-2c=11-
a= (3)b= (1)c= (1)
Hint
Help VideoCorrect
Great Work!
Incorrect
Need TextPlayCurrent Time 0:00/Duration Time 0:00Remaining Time -0:00Stream TypeLIVELoaded: 0%Progress: 0%0:00Fullscreen00:00MutePlayback Rate1x- 2x
- 1.5x
- 1.25x
- 1x
- 0.75x
- 0.5x
Subtitles- subtitles off
Chapters- Chapters
Elimination Method
- 1) make sure 2 equations have a variable with same coefficients but opposite signs
- 2) add or subtract the equations so that one variable is cancelled
- 3) solve for the variable that remains
- 4) substitute known value to one of the equations to solve for the other variable
First, label the equations 1, 2 and 3 respectively.2a-b-4c = 1 Equation 1 a+6b-c = 8 Equation 2 3a+4b-2c = 11 Equation 3 Next, multiply the values of equation 2 by 2 and subtract equation 1 from the product. Label the result as equation A.2×(a+6b-c) = 2×8 Multiply equation 2 by 2 2a+12b-2c = 16 2a+12b-2c = 16 - 2a+b+4c = 1 2a-2a cancels out 13b+2c = 15 Equation A Then, multiply the values of equation 2 by 3 and subtract equation 3 from the product. Label the result as equation B.3×(a+6b-c) = 3×8 Multiply equation 2 by 3 3a+18b-3c = 24 3a+18b-3c = 24 - 3a-4b+2c = -11 6x-6x cancels out 14b-c = 13 Equation B Now, multiply the values of equation B by 2 and add equation A to the product to find the value of y.2×(14b-c) = 2×13 Multiply equation B by 2 28b-2c = 26 28b-2c = 26 + 13b+2c = 15 -4z+4z cancels out 41b = 41 41b ÷41 = 41 ÷41 Divide both sides by 41 b = 1 Now, substitute the value of b into equation A to solve for c.13b +2c = 15 Equation A 13(1)+2c = 15 y=1 13+2c = 15 13+2c -13 = 15 -13 Subtract 13 from both sides 2c = 2 2c ÷2 = 2 ÷2 Divide both sides by 2 c = 1 Finally, substitute the value of b and c into equation 2 to solve for a.a+6b-c = 8 Equation 1 a+6(1)-1 = 8 b=1 and c=1 a+6-1 = 8 a+5 = 8 a+5 -5 = 8 -5 Subtract 5 from both sides a = 3 a=3b=1c=1 -
Quizzes
- Solve Simultaneous Equations by Graphing
- Substitution Method 1
- Substitution Method 2
- Substitution Method 3
- Substitution Method 4
- Elimination Method 1
- Elimination Method 2
- Elimination Method 3
- Elimination Method 4
- Nonlinear Simultaneous Equations
- Simultaneous Equations Word Problems 1
- Simultaneous Equations Word Problems 2
- 3 Variable Simultaneous Equations – Substitution Method
- 3 Variable Simultaneous Equations – Elimination Method