Pair of Linear Equations In Two Variables Important Questions

Sample Questions

Ques: The price of 1kg of grapes and 2 kg of apples on a day were found to be Rs.160. After a month, the price of 2 kg of grapes and 4 kg of apples are Rs.300. Represent the situation algebraically.

Ans. Let the price of 1 kg of apples be ‘Rs. x’.

And, let the price of 1 kg of grapes be ‘Rs. y’.

According to the question, the algebraic representation is

2x + y = 160

And 4x + 2y = 300

For, y = 160 − 2x, the solution table will be;

For y = (300 – 4x)/ 2, the solution table will be;

Ques: Find out the number of solutions for the given pair of linear equations y=o, y=-5. 

Ans. As y = 0 and y = -5 are two Parallel lines, hence there is no solution.

Ques: If the pair of equations has no solution 4x – 3y = 9, ky + 2x = 11 then what is the suitable value of k? 

Ans. We have, 4x – 3y = 9 and 2x + ky = 11

Ques: Comment on the consistency of the given pair of linear equations:

3x + 2y = 8, 5x – 4y = 9 

Ans. 3/6 ≠ 2/ -4

a1 /a2 ≠ b1/ b2

Thus, the pair of linear equations is consistent. 

Read More: Substitution method

Ques: Draw the graph of 2y = 4x – 6; 2x = y + 3 and then determine if this system of linear equations has a unique solution or not.

Ans. Since both, the lines coincide.

Therefore there are infinitely many solutions.

Ques: Solve the given pair of equations for the values of x and y: 

\(\frac{a^2}{x}\) – \(\frac{b^2}{x}\) = 0

\(\frac{a^2b}{x}\) + \(\frac{b^2a}{x}\) = a+b,

x ≠ 0; y ≠ 0

Ans. Putting 1/x = p, 1/y = q in the above equations.

Ques: Solve for x and y:
10/(x+y) + 2/(x−y) =4; 15/(x+y) −5/(x−y) =−2
x + y ≠ 0
x – y ≠ 0

Ans.

By using cross-multiplication method, we get

Adding eq. 3 and 4, we get

2x = 6

x = 3

Substituting the value of x in eq. 3, we get

y = 2

x = 3, y = 2.

Ques: Find out the value of x and y from the pair of linear equations given below:
141x + 93y = 189;
93x + 141y = 45 

Ans. 141x + 93y = 189 … (i)
93x + 141y = 45 … (ii)

Multiplying eq. (i) by 93 and (ii) by 141.

93 (141x + 93y = 189)
141 (93x + 141y = 45 )

13113x + 8649y = 17577
13113x + 19881y = 6345

By subtracting these equations,

-11232y = 11232

y = -1

Putting the value of y in eq. (i)

141x + 93y = 189
141x + 93 (-1) = 189
141x - 93 = 189
141x = 189 + 93
141x = 282
x = 2

Read More: Cross-multiplication of solving linear equations

Ques: Solve by elimination:
3x = y + 5
5x – y = 11

Ans. We have, 3x = y + 5, and 5x – y = 11

in equation (i) if we put the value of x, we get

3x – y = 5 ⇒ 3(3) – y = 5

9 – 5 = y ⇒ y = 4

∴ x = 3, y = 4

Ques: Solve: (x/a ) + (y/b) = a + b; (x/a2) + (y/b2) =2, a, b ≠ 0 

Ans. The given equation is,

(x/a ) + (y/b) - a + b = 0;

(x/a²) + (y/b²) - 2 = 0

For cross multiplication, we use

 \(\frac{x}{b_1c_2 - b_2c_1}\)= \(\frac{-y}{c_2a_1 - c_1a_2}\)= \(\frac{1}{a_1b_2 - a_2b_1}\)

Now, comparing the above two equations with the general form, we get

Ques: A woman earns ₹600 per month less than her husband. One-tenth of the husband’s salary and l/6th of the woman’s salary amount to ₹1,500, which is saved every month. Find their incomes. 

Ans. Let woman’s monthly income = ₹x

Then husband’s monthly income = ₹(x + 600)

we can write,

1/10 (x + 600) + 1/6 (x) = ₹1,500

{3(x+600)+5x}/30 = ₹1,500

3x + 1,800 + 5x = ₹45,000

8x = ₹45,000 – ₹1,800

x = ₹343,2008 = ₹5,400

Woman’s income = ₹x = ₹5,400

husband’s income = ₹(x + 600) = ₹6,000

Ques: A man’s age is currently twice the summation of the ages of his 2 children. After 20 years, his age will be equal to the summation of the ages of his children. Find the age of the man. 

Ans. Let the current ages of his children be x years and y years.

Then the current age of the man = 2(x + y) …(i)

After 20 years, his children’s ages will be

(x + 20) and (y + 20) years

After 20 years, man’s age will be 2(x + y) + 20

We can write,

⇒ 2x + 2y + 20 = x + y + 40

⇒ x + y = 20 …[From (i)

∴ Present age of the man = 2(20) = 40 years

Read More: Difference between linear and non-linear equations

Ques: A two-digit number is seven times the sum of its digits and the difference between the number and that formed by reversing the digits is 18. Find the given number. 

Ans. Let the tenth place digit be y and the unit's place digit be x.

Therefor original number = x + 10y

and the reversed number = 10x + y

we can write,

x + 10y = 7(x + y)

x + 10y = 7x + 7y

⇒ 10y – 7y = 7x – x

⇒ 6x = 3y ⇒ 2x = y …(i)

(x + 10y) – (10x + y) = 18

⇒ 9y – 9x = 180

⇒ – x + y = 2 …[Dividing both side by 9]

⇒ 2x – x = 2 …[From (i)

∴ x = 2

Putting the value of ‘x’ in equation (i), we get y = 2(2) = 4

∴ Required number = x + 10y

= 2 + 10(4) = 42

Ques: In still water, a boat can travel at a speed of 15km/h. If it goes 30 km upstream and returns at the same point in 4 hours 30 minutes. Find the speed of the stream. 

Ans. Suppose, the speed of the stream = x km/hr

and the Speed of the boat in still water = 15 km/hr

then, the speed of the boat upstream = (15 – x) km/hr

and the speed of the boat downstream = (15 + x) km/hr

∴ Speed of the stream = 5 km/hr

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