COMPARING COEFFICIENT OF LINEAR EQUATIONS IN TWO VARIABLES AND SOLVING

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

Problems 1 :

x + y = 5 and 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, b₁  =  1, c₁  =  -5

a₂  =  2, b₂  =  2, c₂  =  -10

a_1/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

Both equations are representing the same line.

Problems 2 :

x - y = 8 and 3x - 3y = 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, b₁  =  -1, c₁  =  -8

a₂  =  3, b₂  =  -3, c₂  =  -16

a_1/a₂  =  1/3  -------(1)

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

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

This exactly matches the condition

 a_1/a₂  =  b₁/b₂  ≠  c₁/c₂

So, there is no solution. 

Problems 3 :

2x + y - 6 = 0 and 4x - 2y - 4 = 0

Solution :

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

a₁  =  2, b₁  =  1, c₁  =  -6

a₂  =  4, b₂  =  -2, c₂  =  -4

a_1/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 1^st equation,

2 x + y - 6  =  0

y  =  -2x + 6

Graphing 2^nd equation,

4 x - 2 y - 4 = 0

2y  =  4x - 4

y  =  2x - 2

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

Problems 4 :

2x - 2y - 2 = 0

4x - 4y - 5 = 0

Solution :

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

a₁  =  2, b₁  =  -2, c₁  =  -2

a₂  =  4, b₂  =  -4, c₂  =  -5

a_1/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/a₂  =  b₁/b₂ ≠ c₁/c₂

This exactly matches the condition

 a_1/a₂  =  b₁/b₂  ≠  c₁/c₂

So, there is no solution. 

Problems 5 :

The pair of equations ax + 2y = 7 and 3x + by = 16 represent parallel lines if

(a) a = b     (b) 3a = 2b     (c) 2a = 3b    (d) ab = 6

Solution :

ax + 2y = 7------(1)

3x + by = 16 ------(2)

Since the given lines are parallel, they must have same slope and different y-intercepts.

2y = -ax + 7

y = (-a/2)x + (7/2)

Slope = -a/2

by = -3x + 16

y = (-3/b) x + (16/b)

Slope = -3/b

-a/2 = -3/b

ab = 6

So, option d is correct.

Problems 6 :

Using the following equations:

(4/x) + 6y = 10 and (1/x) – 6y = 5

find the value of p if p = 3x.

(a) 1     (b) 2     (c) 3    (d) 4

Solution :

(4/x) + 6y = 10 ------(1)

(1/x) – 6y = 5 ------(2)

Let a = 1/x, then 

4a + 6y = 10 ------(1)

a – 6y = 5 ------(2)

Finding the value of a in terms of y,

a = 6y + 5

Applying the value of a, we get

4(6y + 5) + 6y = 10

24y + 20 + 6y = 10

30y = 10 - 20

30y = - 10

y = -10/30

y = -1/3

a = 6(-1/3) + 5

a = -2 + 5

a = 3

1/x = 3

x = 1/3

p = 3x ==> 3(1/3)

p = 1

Option a is correct.

Problems 7 :

The value of k for which the system of equations

x + y – 4 = 0 and 2x + ky = 3

has no solution, is

(a) – 2     (b) ≠ 2    (c) 3    (d) 2

Solution :

x + y – 4 = 0 and 2x + ky = 3

y = -x + 4 ----(1)

ky = -2x + 3

y = (-2/k) x + (3/k) ----(2)

Since the system has no solution, they must be parallel and they will never intersect.

Equating the slopes, we get

-1 = -2/k

k = 2

So, option d is correct.

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