Download Linear equations in two variables Introduction: An

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.