# Linear Equations in Two Variables

We are already familiar with the linear equations of one variable. In a similar way, linear equation in two variables contains two variables where every variable with one degree.
Let us discuss the situation as follows. 4 men and 4 boys can do a piece of work in 3 days while 2 men and 5 boys can finish it in 4 days. How long would it take 1 boy to do it? How long would it take 1 man to do it? .
The above situation seems to be hard to solve. But using the method of solving simultaneous equations of two variables we can easily solve it.
In this topic, let us discuss what are linear equations with 2 variables, general form, types of solutions and the methods of solving the linear system.

## General Form

What is the general form of a linear equation in two variables? It can be written as ax + by = c ; where x, y are variables and a, b, c are non zero constants.

For example:
15x - 17y = 5
-3x + 5 = 2y
3 + 2x = y
s + t = 19
All listed above are linear equations.

## Formula

We can derive a formula to solve for the linear equations in two variables.
Consider two general linear equations,
$a_1$ x + $b_1$y = $c_1$ and
$a_2$ x + $b_2$ y = $c_2$

The solution to the above two equations is given by,

$x = \frac{b_{1}c_{2}- b_{2}c_{1}}{a_{1}b_{2}- b_{2}a_{1}}$

$y = \frac{c_{1}a_{2}- c_{2}a_{1}}{a_{1}b_{2}- b_{2}a_{1}}$

By comparing the given equations to these standard equations we can find the variables, $a_1, a_2, b_1, b_2, c_1 , c_2$, and then
substitute in the above equation.
Note: This result derived using cross multiplication method.

## Methods to Solve Linear Equations

If we look at the methods of solving linear equations in two variables, any one of the following methods can be used.
1.  Substitution Method.
2.  Elimination Method.
3.  Cross multiplication Method
4.. Graphical Method
5.  Matrix Method
6. Method of Determinants (Cramer's Rule)
and many more methods

We should make sure that we frame the equations according to the standard form, ax + by + c = 0.

## How to Solve Linear Equations in Two Variables

There will be infinite solutions for the single equation with 2 variables OR one variable will be in terms of another variable. To find the value for both the variables we always look forward for two equations, called as linear system with two variables.

System of linear equations in two variables are of the form

$a_1$ x + $b_1$y = $c_1$ and
$a_2$ x + $b_2$ y = $c_2$

Where $a_1$$a_2$$b_1$$b_2$$c_1$ and $c_2$ are constants and x, y are variables.
The above system of equations are also called simultaneous linear equations in two variables.

The above system of equations satisfy any one of the following conditions.
1. One solution
2. No solution
3. Infinitely many solutions

Steps to solve:
1. Write equations in the standard form: ax + by + c = 0
2. Choose any of the methods. (listed above)
3. Simplify the equations.

## Examples

Let us illustrate some of the examples to have a clear idea:

Example 1: Solve for x and y. If x - y = 5 and x + y = 4
Solution:
Step 1 :  x - y = 5  implies x = 5 + y
Step 2 : Substitute the value in 2nd equation, we have
(5 + y) + y = 4
5+2y=4
2y=-1
y= -1/2.
Step 3 :  Put the value of y in one of the given equations to get the value of x
Now, x - (-1/2) = 5
x = 5 - 1/2
x = 9/2.

Therefore, (x,y) = (9/2,  -1/2) Answer!

Example 2: Solve 3x - 5y = -11 and 4x + 3y  = 24  using elimination method. Verify your answer.

Solution: 3x - 5y = -11      ---------(1)

4x + 3y  = 24                       ---------(2)

To eliminate the variable y,  let us multiply equation (1) by 3 and multiply equation (2) by 5

3 ( 3x - 5y ) = 3(-11)

5 ( 4x + 3y ) = 5 ( 24 )
This implies
9x - 15y = -33                  -----------(3)

20 x + 15 y = 120           -------------(4)
_____________________
29 x  + 0  = 87
______________________

After Adding (3) and (4) , we have
x = 87/29 = 3

Substituting x= 3 , in Equation (1) we get,

3 (3) - 5 y = -11

9 - 5 y = -11

=> - 5y = -11 - 9 = - 20

=> y = - 20 / - 5 = + 4

Hence the final solution is ( x, y) = ( 3, 4)

Verification:
Substituting the solution in equation (2), we get, {Note: you can choose any of the equations)

4x + 3y = 24

=> 4 ( 3) + 3 ( 4) = 24

=> 12 + 12 = 24

=> 24 = 24

As we have same values on both sides of the equation, our solution is correct.

## Word Problems

Example : The monthly incomes of Kavita and Shyam are in the ratio 3:4 and their monthly expenditures are in the ratio 5 : 7. If each saves Rs. 2,500 per month. Find their monthly incomes.

Solution: Let us assume that the monthly incomes expenditures as per the table given below.
 Kavita Shyam Income 3x 4x Expenditure 5y 7y Savings 3x - 5y 4x - 7y Equations 3x - 5y = 2500 4x - 7y = 2500

Hence we have the Equations,
3x - 5y = 2500      ---------(1)
4x - 7y = 2500      ---------(2)

Use elimination method to solve these equations:
Multiplying (1) by 4, and (2) by 3,  we get,
4 ( 3x - 5y ) = 4(2500)
12 x - 20 y = 10000   ---------(3)
and 3(4x - 7y) = 3 ( 2500)
12 x - 21 y = 7500    -----------(4)

Subtracting (4) from (3), we get,
12 x - 20 y - [ 12 x - 21 y ] = 10000 - 7500
=>  12 x - 20 y - 12 x + 21 y = 2500
=>  y = 2500
Substituting y = 2500 in (1), we get,
3x - 5y = 2500
=> 3x - 5(2500) = 2500
=> 3x - 12500 = 2500
=> 3x = 12500+ 2500 = 15,000
=> x = 15000/3 = 5000
From the above table, we have Income of Kavita = 3x = 3 ( 5000) = Rs15,000
Income of Shyam = 4x = 4( 5000) = Rs. 20,000

## Practice Questions

Solve the following system of linear equations:
1. 3x - 4y = 7 ; x + y = 9 by substitution method.
2. $\frac{2}{x}+\frac{3}{y}= 17$ ; $\frac{1}{x}+ \frac{1}{y}= 7$.
3. If the length and breadth of a room are increased by 1m each, its area is increased by 21 m^2 . If the length is increased by 3 m and breadth decreased by 1 m, the area is decreased by 5 m^2. Find
(i) the dimensions of the room
(ii) the area of the room.