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Linear Equations in Two Variables

 Introduction:


 We are already familiar with solving linear equations of one variable. 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 solve it.

In this topic let us discuss with the linear equations with 2 variables, types of solutions and the methods of solving the linear equations in 2 variables.

Systems of linear equations in two variables :

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System of linear equations in two variables  are of the form,

a1 x + b1 y = c1 and
a2 x + b2 y = c2

The above system of linear equations of two variables satisfy any one of the following conditions.
1. One solution.
2. No solution
3. Infinitely many solutions
 
The above system of equations are also called simultaneous linear equations in two variables.

How to solve linear equations in two variables

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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

How to do linear Equations in two variables:


The linear equations in two variables can be solved by any one of the above methods. We should make sure that we frame the equations according to the standard form, ax + by + c = 0

Solving System Linear Equations in Two Variables

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While solving system of linear equations in two variables, we need to remember the following points.
The graph of the linear equations is a straight line.
When we graph the lines on a graph, one of the following conditions will be satisfied.
1. If the lines intersect at a point there will be one and only one solution. The solution is written in the form (x,y).
2. If the lines are parallel there will be no solution, as the lines do not meet.
3. If the lines coincide there will be infinite number of solutions.

Consistency: We say that the system of equations is consistent if it has at least one solution.
Intersecting lines are coinciding lines are the consistent system equations.
Inconsistency: We say that the system of equations is inconsistent if it has no solution.
Parallel lines are inconsistent system of equations.

Linear equations and inequalities in two variables

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Linear Equations in Two Variables:


          When we consider the linear equations in two variables, their solutions will satisfy any one of the following conditions.
1. Only one solution (unique solutions)
2. No solution
3. Infinitely many solutions satisfying the given equation.
we shall discuss the more examples of linear equations in two variables in this section.

Linear Inequalities in two variables:


            When we consider the linear inequalities in two variables, the expressions and the constants are related by the sign >, <,  <= or > =.   
Let us consider the following inequalities.
x+ y <= 4 and 2x + 3y > = 9
graph1


The region common to both the shaded region (checked in black and red ) is the solution to the above inequality.


system of linear equations in two variables

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Linear Equations in two variables can be solved by any one of the above methods specified.
Linear Equations in two variables examples:
Solve : 2x + 3y = 11 ; x - 2y = 2 by the Method of Substitution.
Solution: By the method of substitution we find one of the variable from one of the equation and substitute it in the other equation.
We have, 2x - 3 y = 11 -------------------(1)
and            x - 2y = 2   -------------------(2)
                 x - 2y = 2
=>                   x = 2 + 2y
Substituting      x = 2 - 2y in equation (1), we get
 2 ( 2 + 2y ) + 3y = 11
=>    4 + 4y + 3y = 11
=>            4 + 7y = 11
=>                  7y = 11 - 4 = 7
=>                    y = 7/7 = 1
Substituting       y = 1, in x = 2 + 2y, we get
                        x = 2 + 2(1) = 2 + 2= 4
Hence the solution to the above equation is ( x, y ) = ( 4, 1)

Linear Equations in two variables formula:

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We can derive a formula to solve for the linear equations in two variables.
Cross Multiplication Method:
Consider the two equations,
a1 x + b1 y = c1 and
a2 x + b2 y = c2

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, a1, a2, b1, b2, c1 , c2, and then
substitute in the above equation.

Examples of linear equations in two variables

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Solve for x and y x-y=5 and x+y=4
Step 1 :  x-y =5  implies x=5+y.
Step 2 : Put this in equation 2 that is x+y=4
            5+y+y=4
            5+2y=4
             2y=-1
             y=-1/2.
Step 3 :  x-(-1/2)=5
             x=5-1/2
             x=9/2.
Solution =9/2.

Linear equations in two variables word problems

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Solving linear Equations in two variables:
Example
: The monthly incomes of Kevin and Keydon are in the ratio 3:4 and their monthly expenditures are in the ratio 5 : 7. If each saves 2,500 dollars per month. Find their monthly incomes.
Solution: Let us assume that the monthly incomes expenditures as per the table given below.
   Kevin   Keydon
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)
Multiplying (1) by 4, and (2) by 3,  we get,
                            4 ( 3x - 5y ) = 4(2500)
                  =>       12 x - 20 y = 10000 ------------------------(3)
                               3(4x - 7y) = 3 ( 2500)
                 =>        12 x - 21 y = 7500    --------------------------(4)
Subtracting (3) and (4), 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 Kevin = 3x = 3 ( 5000) = 15,000 dollars
                                          Income of Keydon = 4x = 4( 5000) = 20,000 dollars

Practice Questions:

Solve the following system of linear equations in two variables.

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 m2 . If the length is increased by 3 m and breadth decreased
    by 1 m, the area is decreased by 5 m2. Find (i) the dimensions of the room (ii) the area of the room.
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