COMPARING COEFFICIENTS OF LINEAR EQUATIONS IN TWO VARIABLES AND SOLVING

To compare the coefficients of linear equations in two variables, the equations must be in the form.

a1x + b1y + c1  =  0

a2x + b2y + c2  =  0

The following three cases are possible for any given system of linear equations.

(i)  a1/a2    b1/b2, we get a unique solution

(ii)  a1/a2  =  a1/a = c1/c2, there are infinitely many solutions.

(iii)  a1/a2  =  a1/a ≠  c1/c2, there is no solution

Problems :

Which of the following pairs of linear equations are consistent/inconsistent? if consistent, obtain the solution graphically. 

(i)  x + y  =  5

   2 x + 2 y  =  10

Solution :

    x + y - 5  =  0

   2 x + 2 y - 10  =  0 

From the given equations, let us find the values of a1, a2, b1, b2, c1 and c2

a1  =  1, b =  1, c1  =  -5

a2  =  2, b =  2, c2  =  -10

a1/a2  =  1/2  -------(1)

b1/b2  = 1/2  -------(2)

c1/c2  =  -5/(-10)  =  1/2  -------(3)

This exactly matches the condition,

a1/a2  =  b1/b2  =  c1/c2

So, the system of equations will have infinitely many solution.

To draw the graph, let us find x and y intercepts.

x + y - 5  =  0

To find x - intercept :

Put y  =  0

x - 5  =  0

 x  =  5

(5, 0)

To find y - intercept :

Put x  =  0

y - 5  =  0

 y  =  5

(0, 5)

Both equations are representing the same line.

(ii) x - y  =  8

   3 x - 3 y  =  16

Solution :

      x -  y – 8  =  0

     3 x - 3 y -16  =  0

From the given equations, let us find the values of a1, a2, b1, b2, c1 and c2

a1  =  1, b =  -1, c1  =  -8

a2  =  3, b =  -3, c2  =  -16

a1/a2  =  1/3  -------(1)

b1/b2  = (-1)/(-3)  =  1/3  -------(2)

c1/c2  =  -8/(-16)  =  1/2  -------(3)

This exactly matches the condition

 a1/a2  =  b1/b ≠  c1/c2

So, there is no solution. 

(iii)  2 x + y - 6  =  0

       4 x - 2 y - 4  =  0

Solution :

From the given equations, let us find the values of a1, a2, b1, b2, c1 and c2

a1  =  2, b =  1, c1  =  -6

a2  =  4, b =  -2, c2  =  -4

a1/a2  =  2/4  =  1/2  -------(1)

b1/b2  = 1/(-2)  =  -1/2  -------(2)

c1/c2  =  -6/(-4)  =  3/2  -------(3)

This exactly matches the condition a1/a2  ≠  b1/b2

So, it has unique solution.

Graphing 1st equation,

2 x + y - 6  =  0

y  =  -2x + 6

x-intercept :

Put y  =  0

-2x + 6  =  0

-2x  =  -6

x  =  3

(3, 0)

y-intercept :

Put x  =  0

y  =  -2(0) + 6

y  =  6

(0, 6)

Graphing 2nd equation,

4 x - 2 y - 4 = 0

2y  =  4x - 4

y  =  2x - 2

x-intercept :

Put y  =  0

2x - 2  =  0

2x  =  2

x  =  1

(1, 0)

y-intercept :

Put x  =  0

y  =  2(0) - 2

y  =  -2

(0, -2)

The above lines are intersecting at the point (2, 2). So, the solution is x  =  2 and y  =  2.

(iv)  2 x - 2 y - 2  = 0

       4 x - 4 y  - 5 = 0

Solution :

From the given equations, let us find the values of a1, a2, b1, b2, c1 and c2

a1  =  2, b =  -2, c1  =  -2

a2  =  4, b =  -4, c2  =  -5

a1/a2  =  2/4  =  1/2  -------(1)

b1/b2  = -2/(-4)  =  1/2  -------(2)

c1/c2  =  -2/(-5)  =  2/5  -------(3)

This exactly matches the condition a1/a =  b1/b2 ≠ c1/c2

This exactly matches the condition

 a1/a2  =  b1/b ≠  c1/c2

So, there is no solution. 

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