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### High School Mathematics1.23 Linear Equation - Three Variables

 Example:1 Solve the following linear equations in three variables: 2x + y -z = 4 -------(1) x + y + z = 10 -------(2) 2z = 4 -------(3) Solution: From equation 3 we have 2z = 4 then z = 4/2 = 2. Therefore z = 2. substitute z = 2 in equation 1 and equation 2. i.e 2x + y -z = 4 => 2x + y - 2 = 4 => 2x + y = 6. and x + y + z = 10 => x + y + 2 = 10 =>x + y = 8 Solve these equations 2x + y = 6 and x + y = 8 2x + y = 6 -x - y = -8 -------------- x = -2 i.e x = -2. Substituting values x = -2 and z = 2 in equation 2. i.e x + y + z = 10 => -2 + y + 2 = 10 i.e y = 10 Therefore x = -2, y = 10 and z = 2 Verification: If we substitute x = -2, y = 10 and z = 2 in 2x + y - z = 4, then we have 2x + y - z = 2(-2) + 10 - 2 = -4 + 10 - 2 = -6 + 10 = 4 Hence verified. Example: Solve the following linear equations in three variables. a + b = 12 -------(1) b + c = 6-------(2) c + a = 8-------(3) Solution: From equation 1 we have, a + b = 12 => b = 12 - a substitute b = 12 - a in equation 2 12 - a + c = 6 => - a + c = - 6 => - a + c = - 6. Also equation 3 is c + a = 8. Add both equations. - a + c = - 6 ------(2) a + c = 8-------(3) ---------------- 2c = 2 therefore c = 1 substituting c = 1 in equation a + c = 8 we get a + c = 8 => a + 1 = 8 => a = 7. substituting c = 1 and a = 7 in equation 1(2 or 3) we get a + b = 12 => 7 + b = 12 => b = 5. Therefore a = 7, b = 5 and c = 1. Verification: If we substitute a = 7, b = 5 and c = 1 in a + b = 12 and b + c = 6, then we have a + b = 7 + 5 = 12 and b + c = 5 + 1 = 6. Hence verified. Example: Solve the following linear equations in three variables: a + b + c = 10 --------(1) a + 2b + 3c = 5 ------(2) 2a + b + 5c = 4-------(3) Solution: a + b + c = 10--------equation 1 a = 10 - b - c a + 2b + 3c = 5 ------equation 2 (10 - b - c) + 2b + 3c = 5 => 10 - b - c + 2b + 3c = 5 => 10 + b + 2c = 5 => b + 2c = -5 2a + b + 5c = 4---------equation 3 2(10 - b - c) + b + 5c = 4 =>20 - 2b - 2c + b + 5c = 4 => 20 - b + 3c = 4 => -b + 3c = -16 b + 2c = -5 -b + 3c = -16 solving these equations we will find b and c. b + 2c = -5 -b + 3c = -16 ----------------------- 0 + 5c = 21 Hence c = 21/5 substituting in 1 we get a + b = -1/5 Directions: Solve the following equations with three variables. Also write at least 5 examples of your own.
 Q 1: Solve the linear equations. x + y + z = 14 x - y + 2z = 12 x - y - z = 3x = 17/2, y = 5/2 and z = 3x = 1/2, y = 1/2 and z = 2x = 2, y = 10 and z = 19inconsistent system Q 2: Solve the following linear equations. x - 2y + 3z = 9 -x + 3z = -4 2x - 5y + 5z = 17x = 53/8, y = 1/8 and z = 7/8inconsistent systemx = -1, y = 1 and z = 2x = 51/3, y = -1/8 and z = -5/8 Q 3: Solve the following linear equations. a + b + c = 10 a + 2b + 3c = 5 2a + b + 5c = 4a = 54/5, b = 17/5 and c = -21/5a = 15, b = 2 and c = -8inconsistent systema = 27/2, b = -10/2 and c = -7/2 Q 4: Solve the following linear equations. x - 2y + 3z = 9 y + 3z = 5 2x - 5y + 5z = 17x = -1, y = 1 and z = 2x = 1, y = -1 and z = -2inconsistent systemx = 1, y = -1 and z = 2 Q 5: Solve the following linear equations. x - 2y + 3z = 9 y + 3z = 5 z = 2x = -1, y = -2 and z = 2inconsistent systemx = 1, y = -3 and z = 2x = 1, y = -1 and z = 2 Q 6: Solve the following linear equations. x - 3y + z = 1 2x - y - 2z = 2 x + 2y - 3z = -1x = 1, y = -1 and z = 3inconsistent systemx = 1, y = -1 and z = -2x = -1, y = -1 and z = 2 Question 7: This question is available to subscribers only! Question 8: This question is available to subscribers only!