Consistent and Inconsistent System of Equations

Ans. Let the cost of the notebook be Rs x

And the cost of the pen be Rs y

As per the condition given in the question,

Cost of the notebook = 2 × Cost of the pen

x = 2y

x – 2y = 0 is the linear equation in two variables to represent the given statement. 

Ans. y = 3x + 5 can also be written as – 3x + y – 5 = 0. This equation is a linear equation in two variables as it is in the form of ax + by + c = 0. Substituting different values for x, we can see that, 

As seen above we can get many numbers of solutions for whichever values we substitute, thus y = 3x + 5 has an infinite number of solutions.

Ans. Here we need to find linear equations which have solutions as (2, 14). Let’s take (2, 14) as (x, y). Simply add them to get the first equation, x + y = 16 (because 2 + 14 = 16). Now subtract them x – y = -12 (because 2 – 14 = - 12). We know that 2 × 7 = 14; thus, we can make another equation 7x = y (because 7 × 2 = 14), that equation can also be written as 7x – y = 0. In this way there can be a number of lines which can have their solution as (2, 14). Thus, we can conclude that the equations of two lines passing through (2, 14) are infinite.

Ans. Let’s consider,

2x − y = 4 … (1)

6x − 3y = 12 … (2)

For equating eqn (1) to eqn (2) multiplying eqn (1) with 3 we get,

6x – 3y = 12 … (3)

Eqn (2) and eqn (3) are same equations, subtracting them will give the answer as ‘0’ 

The equations (2), (3) follow the inconsistent system where,
\(\frac{a1} {a2}\) = \(\frac{b1} {b2}\)≠ \(\frac{c1} {c2}\)

Thus, these equations will not have any solutions, they plot parallel lines when plotted in a graph. Thus, they will have no solutions. 

Ans. 

• 2x+3y=9

2x+3y-9=0

Comparing the above equation with ax + by + c = 0

a = 2

b = 3

c = -9 

• x- y/5-10=0

Comparing the above equation with ax + by + c = 0

a = 1

b = -1/5

c = - 10

• x=3y

x-3y=0

Comparing the above equation with ax + by + c = 0

a = 1

b = -3

c = 0

• 2x= -5y

2x+5y=0

Comparing the above equation with ax + by + c = 0

a = 2

b = 5

c = 0

Ans.

• 2x + 5y = 0 and 4x + 10y = 0

Comparing the first equation with ax + by + c = 0

a₁ = 2

b₁ = 5

c₁ = 0

Comparing the second equation with ax + by + c = 0

a₂ = 4

b₂ = 10

c₂ = 0

Comparing the values from both equations, we can conclude that they are algebraically dependent system

\(\frac{a1}{a2} = \frac{b1}{b2} = \frac{c1}{c2}\)

\(\frac{2}{4} = \frac{5}{10} = \frac{1}{2}\)

Note that here the c1 and c2 are zeros hence they are not considered. The given pair of equations belongs to the dependent system as it satisfies the dependent algebraic representation. 

• 3y + 4x – 2 = 0 and 6y + 3x + 7 = 0

Comparing the first equation with ax + by + c = 0

a₁ = 3

b₁ = 4

c₁ = - 2

Comparing the second equation with ax + by + c = 0

a₂ = 6

b₂ = 3

c₂ = 7

Comparing the values from both equations, we can conclude that they are algebraically independent system

\(\frac{a1}{a2} ? \frac{b1}{b2}\)

\(\frac{3}{6} ? \frac{4}{3}\)

The given pair of equations belongs to the independent system as it satisfies the independent algebraic representation.

Ans.

• x + y = 4

• x – y = 2

• y = 3x

• 3 = 2x + y

Ques. In a Taxi, for the first kilometre, the fare is 8 and for the subsequent distance it is 5 per km. Taking the distance covered as x km and total fare as y, write a linear equation for this information, and draw its graph.  (5 Marks)

Ans. As given in the question, the fare for the first km is Rs 8.

Let the total distance covered by the taxi be ‘x’ km

After the first km, the fare changes to Rs 5 per km for the rest of the distance.

Now, the taxi covers (x-1) distance at Rs 5 per km. while Rs 8 for the first km. Thus, the total fare will be ‘y’

y = 8 + (x - 1) 5

y = 8 + 5x – 5

y = 5x + 3

Ans. Let’s consider,

3x + 2y = 19… (1)

x + y = 8 … (2)

Solving algebraically, we need to first eliminate any one term of the two equations

Thus for equating the equation (2) with eqn (1)

Multiplying eqn (2) with 2 we get the modified equation –

2x + 2y = 16 … (3)

Now subtract (3) from (1) we get,

x = 3 … (4)

substitute eqn (4) in eqn (2)

3 + y = 8

y = 5 

Thus, the correct value of x and y is 3, 4.

Ans. Let’s consider,

x + z = 6 … (1)

z − 3y = 7 … (2)

2x + y + 3z = 15 … (3)

Eqn (1) can also be written as x = 6 – z

We can replace x in eqn (3) by 6 – z

2 (6 - z) + y + 3z = 15

12 – 2z + y + 3z = 15

y + z = 3 … (4)

eqn (4) can also be written as z = 3 – y

we can substitute z = 3 – y in eqn (2)

(3 - y) – 3y = 7

3 – 4y = 7

-4y = 4

y = -1 … (5)

substitute eqn (5) in eqn (4)

-1 + z = 3

z = 4 … (6)

substitute eqn (6) in eqn (1)

x + 4 = 6

x = 2

Thus, the values of x, y, and z are 2, -1, and 4.