Theory:

Methods of solving a pair of linear equations
In the previous topic, we have learnt how to find the solution of the pair of linear equations using the graphical method. But, for some pair of linear equations, the solution would be a fractional value. Therefore, these values will not be as accurate in the graph. The exact value can be determined using other methods of algebraically solving the pair of linear equations.
 
They are:
 
1. Substitution method
 
2. Elimination method
 
3. Cross-Multiplication method
Substitution method
Let us learn how to solve the system of linear equations using the substitution method.
The steps to solve the system of linear equations using the substitution method is given by:
 
Step 1: From the given two equations, consider an equation. Find the value of one of the variables in terms of the other.
 
Step 2: Substitute the value (obtained in Step 1) in the other equation.
 
Step 3: Now, simplify the equation and find the value of the other unknown variable.
 
Step 4: Substituting the obtained value (variable in Step 3) in the equation (obtained in Step 1), we get the value of the first unknown variable.
To understand the concept clearly, let us consider an example.
Example:
Solve the simultaneous linear equations \(2x + y = 8\) and \(x + 2y = 10\) by substitution method.
 
Solution:
 
Step 1: Consider an equation and find the value of one of the variables in terms of the other.
 
Consider the equation \(2x + y = 8\)
 
Subtract \(2x\) from both the sides, we get:
 
\(2x + y - 2x = 8 - 2x\)
 
\(y = 8 - 2x\)
 
Step 2: Substitute the value of \(y\) in the other equation.
 
\(x + 2(8 - 2x) = 10\) (Substituting \(y = 8 - 2x\))
 
Step 3: Solve the above equation and find the value of the variable \(x\).
 
\(x + 16 - 4x = 10\)
 
\(-3x + 16 = 10\)
 
\(-3x = 10 - 16\)
 
\(-3x = -6\)
 
\(x = 2\)
 
Therefore, the value of one of the variables is \(x = 2\).
 
Step 4: Substitute the value of \(x\) in the equation obtained in step 1.
 
Thus, substituting \(x = 2\) in the equation \(y = 8 - 2x\), we have:
 
\(y = 8 - 2(2)\)
 
\(y = 8 - 4 = 4\)
 
Therefore, the value of the other variable is \(y = 4\).
 
Hence, the solution of the given system of equations is \(x = 2\) and \(y = 4\).
 
Verification:
 
Substitute the solutions \(x = 2\) and \(y = 4\) in equation (\(1\)), we get:
 
\(\text{LHS} = 2(2) + 4 = 4 + 4 = 8 = \text{RHS}\)
 
Similarly, substituting \(x = 2\) and \(y = 4\) in equation (\(2\)), we have:
 
\(\text{LHS} = 2 + 2(4) = 2 + 8 = 10 = \text{RHS}\)
 
Since in both the equations \(\text{LHS} = \text{RHS}\), \(x = 2\) and \(y = 4\) is the solution to the given equations.
Important!
1. In any of the steps, if we get a false equation like \(0 = 1\), then the system has no solution.
 
2. If we get an equation like \(0 = 0\), then the system has infinitely many solutions.