
Linear equation with one variable: has one solution. Example: x + 4 = 12 x + 2x + 6 = 9 simplifies to 3x + 6 = 9 Linear equation with two variables: A linear equation in two variables has infinitely many solutions. Example: x + y = 12 For x = 1, y = 11, x = 2, y = 10, x = 3, y = 9, x = 4, y = 8..... and so on
Linear equation with three variables Method 1: Elimination Method: Solving Linear Equations with two variables. Solve for the values of a and b for the following equations: Example 1: a + b = 12 equation 1 a  b = 4  equation 2. Adding both equations we get a + b = 12 a  b = 4 The terms +b and b cancels, and we get 2a = 16 a = 16/8 = 2 a = 2 substituting the value of a in equation 1 we have a + b = 12 2 + b = 12 b = 12  2 b = 10 The solution of the two equations is a = 2 and b = 10 Example 2: c + 5b = 16 equation 1 c + 2b = 7  equation 2 . Multiplying the equation 2 with (1) we get c + 5b = 16 equation 1 1(c + 2b = 7) => c  2b = 7equation 2 c + 5b = 16 c  2b = 7 The terms c and c cancel each other and we get 3b = 9 b = 9/3 = 3 substituting b in the equation c + 5b = 16 we get c + 5(3) = 16 c + 15 = 16  3 c = 16  15 c = 1 Example 3: 2a + 4b = 12 equation 1 a  b = 3  equation 2 First pick a variable either a or b. Lets pick b. Second find the common multiple of b so we can eliminate that variable and find the value of the other variable. so multiply both sides of the equation 2 by 4 we have 2a + 4b = 12 equation 1 4(a  b = 3) => 4a  4b = 12 equation 2 Now adding equation 1 and 2 we have 2a + 4b = 12 4a  4b = 12 The terms +4b in equation 1 and 4b in equation 2 cancels, and we get 6a = 24 a = 24/6 =4 substituting value of a in equation a  b = 3 we get 4  b = 3 4 3 = b b = 1 Therefore a = 4 and b = 1 Verification: To verify substitute the values of a and b in the equations above 2a + 4b = 12 2a + 4b = 2(4) + 4(1) = 8 + 4 = 12 NOTE: In the above question if you pick b 2a + 4b = 12 equation 1 a  b = 3  equation 2 multiplying equation 2 by 2 we have 2(a  b = 3)  equation 2 2a  2b = 6 multiplying equation by 1 we gave 1(2a  2b = 6) we get 2a + 2b = 6 Solving both equations we have 2a + 4b = 12 equation 1 2a + 2b = 6equation 2  6b = 6 b = 1 Example 4: 2a + b = 4 equation 1 3a + 2b = 3  equation 2. Multiplying the equation 1 with 3 and equation 2 with (2) we get 3(2a + b = 4) equation 1 2(3a + 2b = 3)  equation 2 6a + 3b = 12 equation 1 6a  4b = 6)  equation 2 The terms 6a and 6a cancel each other and we get b = 6 b = 6 substituting b in the equation 2a + b = 4 we get 2a + (6) = 4 2a  6 = 4 2a = 4 + 6 2a = 10 a = 10/2 = 5 Method 2: Substitution Method: Solve the equations: 3x + 5y = 26 y = 2x substitute the second equation in the first 3x + 5(2x) = 26 3x + 10x = 26 13x = 26 x = 2 y = 2x = 2x2 = 4 Solve the equations: x + y = 26  equation 1 x  y = 4 equation 2 x  y = 4 equation 2 can be written as x = y + 4. Substituting x in the equation 1 we have: x + y = 26  equation 1 y + 4 + y = 26 2y + 4 = 26 2y + 4  4 = 26  4 2y = 22 y = 11 x = y + 4 = 11 + 4 = 15 Directions: Solve for the variables using either Elimination or Substitution Method. 