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Linear equations in two variables Introduction: An equation of the form ax + by +c = 0, where a, b and c are arbitrary constants and a and b are non-zero is called a general linear equation in the two variables x and y. Every linear equation in two variables has an unlimited number of solutions. A solution to a system of two simultaneous linear equations in two variables is an ordered pair of numbers which satisfy both the equations. There are two methods to solve a system of simultaneous linear equations: 1. Substitution method 2. Elimination method Substitution method: Step 1: Solve one of the given equations for one of the variables Step 2: Substitute that value of the variable in the other equation. Step 3: Solve the resulting single variable equation. Substitute this value into either of the two original equations, and solve it to find the value of the second variable. Elimination method: Step 1: Multiply one or both equations (if necessary) by suitable numbers to transform them so that addition or subtraction will eliminate one variable. Step 2: Solve the resulting single variable equation and substitute this value into either of the two original equations, and solve it to find the value of the second variable. Example 1: Solve the following system of linear equation by substitution method 4x – 3y = 8 x – 2y = -3. Solution: The given equations are 4x – 3y = 8 …..(i) x – 2y = -3 …..(ii) We can solve either equation for either variable. But to avoid fractions, we solve the second equation for x, x = 2y – 3 …..(iii) Substituting this value of x in equation (i), we get 4 (2y – 3) – 3y = 8 8y – 12 – 3y = 8 5y = 20 y=4 Substituting this value of y in equation (ii), we get x – 2 × 4 = -3 x – 8 = -3 x=5 Hence the solution is x = 5, y = 4. Example 2: Solve the system of simultaneous linear equations by elimination method 4x – 3y – 11 = 0 6x + 7y – 5 = 0. Solution: The given equations are 4x – 3y – 11 = 0 6x + 7y – 5 = 0 ….(i) ….(ii) Multiplying the first equation by 3 and the second equation by 2, making both coefficients of the x terms 12 (L.C.M. of 4 and 6), we get 12x – 9y – 33 = 0 ….(iii) 12x + 14y – 10 = 0 ….(iv) Subtracting (iv) from (iii), we get -23y – 23 = 0 -23y = 23 y = -1 Substituting y = -1 in equation (i), we get 4x – 3 (-1) – 11 = 0 => 4x + 3 – 11 = 0 4x – 8 = 0 => 4x = 8 x = 2. Hence, the solution is x = 2, y = -1. Practice problems: Problem 1: Solve the following system of linear equation by substitution method: 8x + 5y = 9 3x + 2y = 4. Problem 2: Solve by elimination method: 83x – 67y = 383 67x – 83y = 367. Problem 3: Solve 4 x 6 y 15, 3 x 4 y 7.