Substitution Method in Linear Equation

Cross Multiplication Method

Elimination Method

Substitution Method

Under this method, we substitute the value of one given variable in terms of the other given variable and find the solution. Lets understand this method with the help following examples: Example 1: Solve the following pair of examples: 2x + y = 10 5x + 2y = 23 Solution: Label the given equations as shown below: 2x + y = 10 ... (equation 1) 5x + 2y = 23 ...(equation 2) Pick any of the given equation and write one variable in terms of the other variable. Here let's take equation 1 and we get: 2x + y = 10 ... (equation 1) or y = (10 - 2x) ... (equation 3) Now, substitute the value of y from (equation 3) to (equation 2) and we get: 5x + 2(10 - 2x) = 23 Solve brackets and we get: 5x + 20 - 4x = 23 Combine like terms and we get: x + 20 = 23 subtract 20 from both sides and we get: x = 3 Now, put the value of x in (equation 3) and we get: y = (10 - 2 X 3) Solve bracket and we get y = 4 Hence, the solution is x = 3 and y = 4 Example 2: Solve the following pair of examples: 4x - 8y = 24 x + 10y = 18 Solution: Label the given equations as shown below: 4x - 8y = 24 ... (equation 1) x + 10y = 18 ... (equation 2) Pick any of the given equation and write one variable in terms of the other variable. Here let's take equation 2 and we get: x + 10y = 18 ... (equation 2) or x = (18 - 10y) ... (equation 3) Now, substitute the value of x from (equation 3) to (equation 1 )and we get: 4 (18 - 10y) - 8y = 24 Solve brackets and we get: 72 - 40y - 8y = 24 Combine like terms and we get: 72 - 48y = 24 Subtract 72 from both sides and we get: -48y = -48 Divide both sides by -48 and we get: y = 1 Now, put the value of y in (equation 3) and we get: x = (18 - 10 X 1) Solve bracket and we get x = 8 Hence, the solution is x = 8 and y = 1