Pair of Linear Equations in Two Variables Test - 16

We know that equation of the form y = a is a line parallel to x–axis at a distance ‘a’ from it. y = 0 is the equation of the x–axis and y = –7 is is the equation of the line parallel to the x–axis. So, these two equations represent two parallel lines. Therefore, there is no solution.

x = a is the equation of a straight line parallel to the y-axis at a distance ‘a’ from it. Again, y = b is the equation of a straight line parallel to the x-axis at a distance ‘b’ from it. So, the pair of equations x = a and y = b graphically represents lines which are intersecting at (a, b).

Given equation is 
−5x+7y=2   .....(1)    
We have to find the dependent equation of above equation    
We will check through options
Option A, 10x+14y+4=0
⇒5x+7y=−2
It cannot be written exactly as eq(1). So, it is not a dependent equation of (1).

Option B, −10x−14y+4=0
⇒5x+7y=2
It cannot be written exactly as eq(1). So, it is not a dependent equation of (1).

Option C, −10x+14y+4=0
⇒−5x+7y=−2
It cannot be written exactly as eq(1). So, it is not a dependent equation of (1).

Option D, 10x−14y=−4
⇒−5x+7y=2
It is the same as (1). So, it is a dependent equation of (1).

sol (b and d):

As x = 2, y = –3 is unique solution of system of equations so these values of must satisfy both equations.

(a) x + y = –1 and 2x – 3y = –5

Put x =2 and y = 3 in both the equations.

LHS = x + y ⇒ 2 – 3 = –1 (RHS)

LHS = 2x – 3y ⇒ 2(2) –3(–3) ⇒ 4 + 9 = 13 ≠ RHS

(b) 2x + 5y = –11 and 4x + 10y = –22

Put x = 2 and y = –3 in both the equations.

LHS = 2x + 5y ⇒ 2 × 2 + (–3) ⇒ 4 – 15 = –11 = RHS

LHS = 4x + 10y ⇒ 4(2) + 10(–3) ⇒ 8 – 30 = –22 = RHS

(c) 2x – y = 1 and 3x + 2y = 0

Put x = 2 and y = –3 in both the equations.

LHS = 2x – y ⇒ 2(2) + 3 ⇒ 7 ≠ RHS

LHS = 3x + 2y ⇒ 3(2) + 1(–3) ⇒ 6 – 6 = 0 = RHS

(d) x – 4y – 14 = 0 and 5x – y – 13 = 0

x – 4y = 14 and 5x – y = 13

Put x = 2 and y = –3 in both the equations.

LHS = x – 4y ⇒ 2 – 4(–3) ⇒ 2 + 12 = 14 = RHS

LHS = 5x – y ⇒ 5(2) – (–3) ⇒ 10 + 3 = 13 = RHS

Hence, the pair of equations is (b) and (d).

If (a, b) is the solution of the given equations, then it must satisfy the given equations

so,

a – b = 2 …(i)

a + b = 4 …(ii)

⇒ 2a = 6 [Adding (i) and (ii)]

⇒ a = 3

Now, 3 + b = 4 [From (ii)]

⇒b =1

So, (a, b) = (3, 1).

Let the number of Rs.1 coins = x

and the number of Rs.2 coins = y

So, according to the question

x+y=50......(i)

1x+2y=75.....(ii)

substracting equation (i)from (ii)

y = 25

substituting value of y in (i)

x=25

So, y = 25 and x = 25.

Let the present age of father be x years

and the present age of son be y = years.

∴ According to the question, x = 6y …(i)

Age of the father after four years = (x + 4) years

and the age of son after four years = (y + 4) years

Now, according to the question, x + 4 = 4(y + 4) …(ii)

⇒ x + 4 = 4y + 16

⇒ 6y – 4y = 16 – 4 [∴ x = 6y]

⇒ 2y = 12

⇒ y = 6

∴ x = 6 × 6 = 36 years [From (i)]

and y = 6 years

So, the present ages of the son and the father are 6 years and 36 years respectively.