TOPICS

Unit-2(Examples)

Linear Equations in One Variable

**Example-1 :-** Find the solution of 2x – 3 = 7.

2x - 3 = 7 2x = 7 + 3 2x = 10 x = 10/2 x = 5 Verification : L.H.S = 2x - 3 = 2 x 5 - 3 = 10 - 3 = 7 = R.H.S

**Example-2 :-** Solve 2y + 9 = 4.

2y + 9 = 4 2y = 4 - 9 2y = -5 y = -5/2 Verification : L.H.S = 2y + 9 = 2 x (-5/2) + 9 = -5 + 9 = 4 = R.H.S

**Example-3 :-** Solve x/3 + 5/2 = -3/2.

x/3 + 5/2 = -3/2 x/3 = -3/2 - 5/2 x/3 = (-3 - 5)/2 x/3 = -8/2 x/3 = -4 x = -4 x 3 x = -12 Verification : L.H.S = x/3 + 5/2 = -12/3 + 5/2 = -4 + 5/2 = (-8 + 5)/2 = -3/2 = R.H.S

**Example-4 :-** Solve 15/4 - 7x = 9.

15/4 - 7x = 9 15/4 - 9 = 7x 7x = (15 - 36)/4 7x = -21/4 x = -21/(4 x 7) x = -3/4 Verification : L.H.S = 15/4 - 7x = 15/4 - 7 x (-3/4) = 15/4 + 21/4 = (15 + 21)/4 = 36/4 = 9 = R.H.S

**Example-5 :-** What should be added to twice the rational number -7/3 to get 3/7.

Let the number added is X in given rational number. According to question : X + 2 x (-7/3) = 3/7 X + (-14/3) = 3/7 X = 3/7 + 14/3 X = (9 + 98)/21 X = 107/21 So, 107/21 be added to twice the rational number -7/3 to get 3/7.

**Example-6 :-** The perimeter of a rectangle is 13 cm and its width is 2 ^{3}⁄_{4} cm. Find its length.

Let the length of the rectangle to be X cm. The perimeter of the rectangle = 2 × (length + width) = 13 cm width = 2^{3}⁄_{4}cm = 11/4 cm Then, 2 x (X + 11/4) = 13 2x + 22/4 = 13 2x = 13 - 11/2 2x = (26 - 11)/2 x = 15/4 Length of rectangle is 15/4 cm.

**Example-7 :-** The present age of Sahil’s mother is three times the present age of Sahil. After 5 years their ages will add to 66 years. Find their present ages.

Let Sahil’s present age be x years. So, mother's age is 3x years. According to question : (x + 5) + (3x + 5) = 66 4x + 10 = 66 4x = 66 - 10 4x = 56 x = 56/4 x = 14 Then, Sahil's present age be x = 14 years. And mother's age is 3x = 3 x 14 = 42 years.

**Example-8 :-** Bansi has 3 times as many two-rupee coins as he has five-rupee coins. If he has in all a sum of ₹ 77, how many coins of each denomination does he have?

Let the number of five-rupee coins that Bansi has be X. Then the number of two-rupee coins he has is 3 times X or 3X. The amount Bansi has: (i) from 5 rupee coins, ₹ 5 × X = ₹ 5X (ii) from 2 rupee coins, ₹ 2 × 3X = ₹ 6X Hence the total money he has = ₹ 11X But this is given to be ₹ 77; therefore, 11X = 77 X = 77/11 X = 7 Thus, number of five-rupee coins = X = 7 and number of two-rupee coins = 3X = 21

**Example-9 :-** The sum of three consecutive multiples of 11 is 363. Find these multiples.

Let three Consecutive numbers x, x + 11, x + 22 which are multiple of 11. Sum of multiples of 11 = 363 x + x + 11 + x + 22 = 363 3x + 33 = 363 3x = 363 - 33 3x = 330 x = 330/3 x = 110 Now, first number is x = 110, second number is x + 11 = 110 + 11 = 121, third number is x + 22 = 110 + 22 = 132

**Example-10 :-** The difference between two whole numbers is 66. The ratio of the two numbers is 2 : 5. What are the two numbers?

Since the ratio of the two numbers is 2 : 5, we may take one number to be 2x and the other to be 5x. The difference between the two numbers is (5x – 2x). It is given that the difference is 66. Therefore, 5x – 2x = 66 3x = 66 x = 22 Since the numbers are 2x and 5x, they are 2 × 22 = 44 and 5 × 22 = 110, respectively. The difference between the two numbers is 110 – 44 = 66 as desired.

**Example-11 :-** Deveshi has a total of ₹ 590 as currency notes in the denominations of ₹ 50, ₹ 20 and ₹ 10. The ratio of the number of ₹ 50 notes and ₹ 20 notes is 3:5. If she has a total of 25 notes, how many notes of each denomination she has?

Let the number of ₹ 50 notes and ₹ 20 notes be 3x and 5x, respectively. But she has 25 notes in total. Therefore, the number of ₹ 10 notes = 25 – (3x + 5x) = 25 – 8x. The amount she has from ₹ 50 notes : 3x × 50 = ₹ 150x from ₹ 20 notes : 5x × 20 = ₹ 100x from ₹ 10 notes : (25 – 8x) × 10 = ₹ (250 – 80x) Hence the total money she has = 150x + 100x + (250 – 80x) = ₹ (170x + 250) But she has ₹ 590. Therefore, 170x + 250 = 590 170x = 590 – 250 170x = 340 x = 340/170 x = 2 The number of ₹ 50 notes she has = 3x = 3 × 2 = 6 The number of ₹ 20 notes she has = 5x = 5 × 2 = 10 The number of ₹ 10 notes she has = 25 – 8x = 25 – (8 × 2) = 25 – 16 = 9

**Example-12 :-** Solve 2x – 3 = x + 2.

2x – 3 = x + 2 2x - x = 2 + 3 x = 5 Verification : L.H.S = 2x – 3 = 2 x 5 - 3 = 10 - 3 = 7 R.H.S = x + 2 = 5 + 2 = 7 L.H.S = R.H.S

**Example-13 :-** Solve 5x + 7/2 = 3x/2 - 14.

5x + 7/2 = 3x/2 - 14 5x - 3x/2 = -14 - 7/2 (10x - 3x)/2 = (-28 - 7)/2 7x/2 = -35/2 7x = -35 x = -35/7 x = -5 Verification : L.H.S = 5x + 7/2 = 5 x (-5) + 7/2 = -25 + 7/2 = (-50 + 7)/2 = -43/2 R.H.S = 3x/2 - 14 = [3 x (-5)]/2 - 14 = -15/2 -14 = (-15 - 28)/2 = -43/2 L.H.S = R.H.S

**Example-14 :-** The digits of a two-digit number differ by 3. If the digits are interchanged, and the resulting number is added to the original number, we get 143. What can be the original number?

Let us take the two digit number such that the digit in the units place is b. The digit in the tens place differs from b by 3. Let us take it as b + 3. So the two-digit number is 10 (b + 3) + b = 10b + 30 + b = 11b + 30. With interchange of digits, the resulting two-digit number will be 10b + (b + 3) = 11b + 3. If we add these two two-digit numbers, their sum is (11b + 30) + (11b + 3) = 11b + 11b + 30 + 3 = 22b + 33. It is given that the sum is 143. Therefore, 22b + 33 = 143 22b = 143 – 33 22b = 110 b = 110/22 b = 5 The units digit is 5 and therefore the tens digit is 5 + 3 which is 8. The number is 85.

**Example-15 :-** Arjun is twice as old as Shriya. Five years ago his age was three times Shriya’s age. Find their present ages.

Let us take Shriya’s present age to be x years. Then Arjun’s present age would be 2x years. Shriya’s age five years ago was (x – 5) years. Arjun’s age five years ago was (2x – 5) years. It is given that Arjun’s age five years ago was three times Shriya’s age. Thus, 2x – 5 = 3(x – 5) 2x – 5 = 3x – 15 15 – 5 = 3x – 2x 10 = x x = 10 So, Shriya’s present age = x = 10 years. Therefore, Arjun’s present age = 2x = 2 × 10 = 20 years.

**Example-16 :-** Solve (6x + 1)/3 + 1 = (x - 3)/6.

(6x + 1)/3 + 1 = (x - 3)/6 (6x + 1 + 3)/3 = (x - 3)/6 (6x + 4)/3 = (x - 3)/6 6 x (6x + 4) = 3 x (x - 3) 36x + 24 = 3x - 9 36x - 3x = -9 - 24 33x = -33 x = -1 Verification : L.H.S = (6x + 1)/3 + 1 = [6 x (-1) + 1]/3 + 1 = (-6 + 1)/3 + 1 = -5/3 + 1 = (-5 + 3)/3 = -2/3 R.H.S = (x - 3)/6 = (-1 - 3)/6 = -4/6 = -2/3 L.H.S = R.H.S

**Example-17 :-** Solve 5x – 2(2x – 7) = 2(3x – 1) + 7/2.

5x – 2(2x – 7) = 2(3x – 1) + 7/2 5x - 4x + 14 = 6x - 2 + 7/2 x - 6x = -14 - 2 + 7/2 -5x = -16 + 7/2 -5x = (-32 + 7)/2 -5x = -25/2 x = 5/2 Verification : L.H.S = 5x – 2(2x – 7) = 5 x 5/2 - 2(2 x 5/2 - 7) = 25/2 - 2(5 - 7) = 25/2 + 4 = (25 + 8)/4 = 33/4 R.H.S = 2(3x – 1) + 7/2 = 2(3 x 5/2 - 1) + 7/2 = 2(15/2 - 1) + 7/2 = 2(15 - 2)/2 + 7/2 = 13 + 7/2 = (26 + 7)/2

**Example-18 :-** Solve (x + 1)/(2x + 3) = 3/8.

(x + 1)/(2x + 3) = 3/8 8 (x + 1) = 3 (2x + 3) 8x + 8 = 6x + 9 8x - 6x = 9 - 8 2x = 1 x = 1/2 Verification : L.H.S = (x + 1)/(2x + 3) = (1/2 + 1)/(2 x 1/2 + 3) = (1 + 2)/2 ÷ (1 + 3) = 3/2 ÷ 4 = 3/8 = R.H.S

**Example-19 :-** Present ages of Anu and Raj are in the ratio 4:5. Eight years from now the ratio of their ages will be 5:6. Find their present ages.

Let the present ages of Anu and Raj be 4x years and 5x years respectively. After eight years, Anu’s age = (4x + 8) years; After eight years, Raj’s age = (5x + 8) years. Therefore, the ratio of their ages after eight years = (4x + 8)/(5x + 8) This is given to be 5 : 6 Therefore, (4x + 8)/(5x + 8) = 5/6 6 (4x + 8) = 5 (5x + 8) 24x + 48 = 25x + 40 24x - 25x = 40 - 48 -x = -8 x = 8 Therefore, Anu’s present age = 4x = 4 × 8 = 32 years Raj’s present age = 5x = 5 × 8 = 40 years

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