NCERT Solutions for Pair of Linear Equations in Two Variables Chapter 3 Ex 3.3 Class 10 Maths has all the solutions to the questions provided in the NCERT Book of the latest edition.

Students are advised to practice all the questions to get good marks in the board examination.

Textbook | NCERT |

Class | 10 |

Subject | Mathematics |

Chapter | 3 |

Exercise | 3.3 |

Chapter Name | Pair of Linear Equations in Two Variables |

## Class 10 Maths Chapter 3 Exercise 3.3 Pair of Linear Equations in Two Variables NCERT Solution

Solve the following pair of linear equations by the elimination method and the substitution method :

(i) x + y = 5 and 2x – 3y = 4

Answer

$$2x-3y=4$$ —-(i)

$$x+y=5$$ —-(ii)

Multiply eq(ii) by 3

$$3x+3y=15$$ —-(iii)

Adding eq(i) and eq(ii), we get

$$5x=19$$

⇒ $$x=\frac{19}{5}$$

Put in eq(ii), we get

$$\frac{19}{5}+y=5$$

⇒ $$y=5-\frac{19}{5}$$

⇒ $$y=\frac{6}{5}$$

(ii) 3x + 4y = 10 and 2x – 2y = 2

Answer

$$3x+4y=10$$ —-(i)

$$2x-2y =2$$

⇒ $$2(x-y)=2$$

⇒ $$x-y=1$$ —-(ii)

Multiply eq(ii) by 4, we get

$$4x-4y=4$$ —-(iii)

Adding eq(i) and eq(iii)

$$7x=14$$

⇒ $$x=2$$

Putting in eq(ii), we get

$$2-y=1$$

⇒ $$y=1$$

(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7

Answer

$$3x-5y=4$$ —-(i)

$$9x-2y=7$$ —-(ii)

Multiply eq(i) by 3, we get

$$9x-15y=12$$ —-(iii)

Subtracting eq(iii) from eq(ii)

$$9x-2y=7$$ |

$$9x-15y=12$$ |

$$13y=-5$$ |

⇒ $$y=\frac{-5}{13}$$

Putting value of y in eq(i), we get

$$3x+\frac{25}{13}=4$$

⇒ $$3x=4-\frac{25}{13}$$

⇒ $$x=\frac{27}{39}$$

⇒ $$x=\frac{9}{13}$$

(iv) $$\frac{x}{2}+\frac{2y}{3}=-1$$ and $$x-\frac{y}{3}=3$$

Answer

$$\frac{x}{2}+\frac{2y}{3}=-1$$ —-(i)

$$x-\frac{y}{3}=3$$ —-(ii)

Multiply eq(ii) by 2, we get

$$2x-\frac{2y}{3}=6$$ —-(iii)

Adding eq(ii) and eq(iii), we get

$$\frac{5x}{2}=5$$

⇒ $$x=2$$

Putting value of x in eq(i), we get

$$1+\frac{2y}{3}=-1$$

⇒ $$\frac{2y}{3}=-2$$

⇒ $$y=-3$$

Form the pair of linear equations in the following problems, and find their solutions(if they exist) by the elimination method :

(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes $$\frac{1}{2}$$ if we only add 1 to the denominator. What is the fraction?

Answer

Let Numerator of the fraction be x

and Denominator of the fraction = y

Then, the fraction = $$\frac{x}{y}$$

A/Q,

$$\frac{x+1}{y-1}=1$$

⇒ $$x+1=y-1$$

⇒ $$x-y=-2$$ —-(i)

and

$$\frac{x}{y+1}=\frac{1}{2}$$

⇒ $$2x=y+1$$

⇒ $$2x-y=1$$ —-(ii)

Subtract eq(i) from eq(ii), we get

$$x=3$$

Putting value of x in eq(i), we get

$$3-y=-2$$

⇒ $$y=5$$

Then the fraction = $$\frac{3}{5}$$

(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

Answer

Let the age of Nuri be x years

And the age of Sonu = y years

Five years ago,

Age of Nuri = $$(x-5)$$ years

Age of Sonu = $$(y-5)$$ years

A/Q,

$$x-5=3(y-5)$$

⇒ $$x-3y = -10$$ —-(i)

After ten years,

Age of Nuri = $$x+10$$

Age of Sonu = $$y+10$$

A/Q,

$$x+10=2(y+10)$$

⇒ $$x-2y=10$$ —-(ii)

Subtract eq(ii) from eq(i), we get

$$-y=-20$$

⇒ $$y=20$$

Putting value of y in eq(ii), we get

$$x-40=10$$

⇒ $$x=50$$

$$\therefore$$ Age of Nuri = x = 50 years

Age of Sonu = y = 20 years

(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Answer

Let the units place digit be x

Tens place digit = y

Two digit no. = $$10y+x$$, Reversing no. = $$10x+y$$

A/Q,

$$x+y=9$$ —-(i)

and

$$9(10y+x)=2(10x+y)$$

⇒ $$90y+9x=20x+2y$$

⇒ $$11x-88y=0$$

⇒ $$x-8y=0$$ —-(ii)

Subtract eq(i) from eq(ii), we get

$$-9y=-9$$

⇒ $$y=1$$

Putting value of y in eq(ii), we get

$$x=8$$

$$\therefore$$ Required no. = $$10y+x=10+8=18$$

(iv) Meena went to a bank to withdraw ₹ 2000. She asked the cashier to give her ₹ 50 and ₹ 100 notes only. Meena got 25 notes in all. Find how many notes of ₹ 50 and ₹ 100 she received.

Answer

Let no. of ₹50 rupee note be x

No. of ₹100 rupee note be y

A/Q,

$$x+y=25$$ —-(i)

and

$$50x+100y=2000$$

⇒ $$50(x+2y)=2000$$

⇒ $$x+2y=40$$ —-(ii)

Subtract eq(ii) from eq(i)

$$y=15$$

Putting value of y in eq(i), we get

$$x=10$$

(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹ 27 for a book kept for seven days, while Susy paid ₹ 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Answer

Let the fixed charge be ₹ x

Charge for each extra day = ₹ y

A/Q,

$$x+(7-3)y=27$$ —-(i)

and

$$x+(5-3)y=21$$ —-(ii)

Subtract eq(ii) from eq(i), we get

$$2y=6$$

⇒ $$y=3$$

Putting in eq(ii), we get

$$x=15$$

Hope NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3, helps you in solving problems. If you have any doubts, drop a comment below and we will get back to you.