# DAV Class 8 Maths Chapter 9 Worksheet 1 Solutions

The DAV Class 8 Maths Solutions and DAV Class 8 Maths Chapter 9 Worksheet 1 Solutions of Linear Equations in One Variable offer comprehensive answers to textbook questions.

## DAV Class 8 Maths Ch 9 WS 1 Solutions

Question 1.
Solve the following equations and verify your answer:

(i) $$\frac{p+7}{p-6}=\frac{1}{3}$$
Solution:
$$\frac{p+7}{p-6}=\frac{1}{3}$$
⇒ 3 (p + 7) = 1 (p – 6)
⇒ 3p + 21 = p – 6
⇒ 3p – p = – 6 – 21
⇒ 2p = – 27
⇒ p = $$\frac{27}{2}$$

Verification:
$$\frac{p+7}{p-6}=\frac{1}{3}$$
Put p = $$-\frac{27}{2}$$
$$\frac{\left(\frac{-27}{2}\right)+7}{\left(\frac{-27}{2}\right)-6}=\frac{1}{3}$$
⇒ $$\frac{\frac{-27+14}{2}}{\frac{-27-12}{2}}=\frac{1}{3}$$
⇒ $$\frac{\frac{-13}{2}}{\frac{-39}{2}}=\frac{1}{3}$$
⇒ $$\frac{1}{3}=\frac{1}{3}$$
L.H.S = R.H.S
Hence verified.

(ii) $$\frac{x-2}{6 x+1}=1$$
Solution:
$$\frac{x-2}{6 x+1}=1$$
⇒ x – 2 = 6x + 1
⇒ x – 6x = 1 + 2
⇒ – 5x = 3
∴ x = $$\frac{3}{5}$$

Verification:
Put x = $$-\frac{3}{5}$$ in the given equation $$\frac{x-2}{6 x+1}=1$$,
$$\frac{\frac{-3}{5}-2}{6\left(\frac{-3}{5}\right)+1}$$ = 1
⇒ $$\frac{\frac{-3-10}{5}}{\frac{-18+5}{5}}$$ = 1
⇒ $$\frac{-13}{-13}$$ = 1
∴ 1 = 1
L.H.S = R.H.S
Hence verified.

(iii) $$\frac{3 x}{5 x-5}=-1$$
Solution:
 = – 1
⇒ 3x = – 1 (5x – 5)
⇒ 3x + 5x =5
⇒ 8x = 5
∴ x = $$\frac{5}{8}$$

Verification:
Put x = $$\frac{5}{8}$$ in the given equation $$\frac{3x}{5 x-5}=-1$$,
⇒ $$\frac{3\left(\frac{5}{8}\right)}{5\left(\frac{5}{8}\right)-5}$$ = – 1
⇒ $$\frac{\frac{15}{8}}{\frac{25}{8}-5}$$ = – 1
⇒ $$\frac{\frac{15}{8}}{\frac{-15}{8}}$$ = – 1
⇒ – 1 = – 1
L.H.S = R.H.S
Hence verified.

(iv) $$\frac{2 x-1}{5 x}=-\frac{1}{6}$$
Solution:
$$\frac{2 x-1}{5 x}=-\frac{1}{6}$$
⇒ 6(2x – 1) = – 5x
⇒ 12x – 6 = – 5x
⇒ 17x = 6
⇒ x = $$\frac{6}{17}$$

Verification:
Put x = $$\frac{6}{17}$$ in the given equation $$\frac{2 x-1}{5 x}=\frac{-1}{6}$$,
$$\frac{2\left(\frac{6}{17}\right)-1}{5\left(\frac{6}{17}\right)}=\frac{-1}{6}$$
⇒ $$\frac{\frac{12}{17}-1}{\frac{30}{17}}=\frac{-1}{6}$$
⇒ $$\frac{\frac{12-17}{17}}{\frac{30}{17}}=\frac{-1}{6}$$
⇒ $$\frac{\frac{-5}{17}}{\frac{30}{17}}=\frac{-1}{6}$$
⇒ $$\frac{-1}{6}=\frac{-1}{6}$$
L.H.S = R.H.S
Hence verified.

(v) $$\frac{4 x+1}{3 x-1}=-2$$
Solution:
$$\frac{4 x+1}{3 x-1}=-2$$
⇒ 4x + 1 = – 2 (3x – 1)
⇒ 4x + 1 =- 6x + 2
⇒ 4x + 6x = 2 – 1
⇒ 10x = 1
∴ x = $$\frac{1}{10}$$

Verification:
Put x = $$\frac{1}{10}$$ in the given equation $$\frac{4 x+1}{3 x-1}=-2$$
⇒ $$\frac{4\left(\frac{1}{10}\right)+1}{3\left(\frac{1}{10}\right)-1}$$ = – 2
⇒ $$\frac{\frac{4}{10}+1}{\frac{3}{10}-1}$$ = – 2
⇒ $$\frac{\frac{4+10}{10}}{\frac{3-10}{10}}$$ = – 2
⇒ $$\frac{\frac{14}{10}}{\frac{-7}{10}}$$ = – 2
⇒ – 2 = – 2
L.H.S = R.H.S
Hence verified.

(vi) $$\frac{4 z-3}{2 z+1}=\frac{5}{7}$$
Solution:
$$\frac{4 z-3}{2 z+1}=\frac{5}{7}$$
⇒ 7 (4z – 3) = 5 (2z + 1)
⇒ 28z – 21= 10z + 5
⇒ 28z – 10z = 21 + 5
⇒ 18z = 26
∴ z = $$\frac{26}{18}=\frac{13}{9}$$

Verification:
Put z = $$\frac{13}{9}$$ in the given equation $$\frac{4 z-3}{2 z+1}=\frac{5}{7}$$
⇒ $$\frac{4\left(\frac{13}{9}\right)-3}{2\left(\frac{13}{9}\right)+1}=\frac{5}{7}$$
⇒ $$\frac{\frac{52}{9}-3}{\frac{26}{9}+1}=\frac{5}{7}$$
⇒ $$\frac{\frac{52-27}{9}}{\frac{26+9}{9}}=\frac{5}{7}$$
⇒ $$\frac{\frac{25}{9}}{\frac{35}{9}}=\frac{5}{7}$$
⇒ $$\frac{25}{35}=\frac{5}{7}$$
⇒ $$\frac{5}{7}=\frac{5}{7}$$
L.H.S = R.H.S
Hence verified.

(vii) $$\frac{\frac{2}{5} x+3}{\frac{1}{3} x-1}=\frac{3}{5}$$
Solution:
$$\frac{\frac{2}{5} x+3}{\frac{1}{3} x-1}=\frac{3}{5}$$
⇒ 5 ($$\frac{2 x}{5}$$ + 3) = 3 ($$\frac{1}{3}$$ x – 1)
⇒ 2x + 15 = x – 3
⇒ 2x – x = – 3 – 15
⇒ x = – 18

Verification:
Put x = – 18 in the given equation $$\frac{\frac{2}{5} x+3}{\frac{1}{3} x-1}=\frac{3}{5}$$,
⇒ $$\frac{\frac{2}{5}(-18)+3}{\frac{1}{3}(-18)-1}=\frac{3}{5}$$
⇒ $$\frac{\frac{-36}{5}+3}{-6-1}=\frac{3}{5}$$
⇒ $$\frac{\frac{-36+15}{5}}{-7}=\frac{3}{5}$$
⇒ $$\frac{-21}{-7 \times 5}=\frac{3}{5}$$
⇒ $$\frac{3}{5}=\frac{3}{5}$$
L.H.S = R.H.S
Hence verified.

(viii) $$\frac{2 x-\frac{3}{4}}{3 x+\frac{4}{5}}=\frac{1}{5}$$
Solution:
$$\frac{2 x-\frac{3}{4}}{3 x+\frac{4}{5}}=\frac{1}{5}$$
⇒ 5 (2x – $$\frac{3}{4}$$) = 1 (3x + $$\frac{4}{5}$$)
⇒ 10x – 5 × $$\frac{3}{4}$$ = 3x + $$\frac{4}{5}$$
⇒ 10x – $$\frac{15}{4}$$ = 3x + $$\frac{4}{5}$$
⇒ 10x – 3x = $$\frac{15}{4}$$ + $$\frac{4}{5}$$
⇒ 7x = $$\frac{75+16}{20}$$
⇒ 7x = $$\frac{91}{20}$$
∴ x = $$\frac{91}{20 \times 7}=\frac{13}{20}$$

Verification:
Putting x = $$\frac{13}{20}$$ in the given equation $$\frac{2 x-\frac{3}{4}}{3 x+\frac{4}{5}}=\frac{1}{5}$$
⇒ $$\frac{2\left(\frac{13}{20}\right)-\frac{3}{4}}{3\left(\frac{13}{20}\right)+\frac{4}{5}}=\frac{1}{5}$$
⇒ $$\frac{\frac{13}{10}-\frac{3}{4}}{\frac{13}{20}+\frac{4}{5}}=\frac{1}{5}$$
⇒ $$\frac{\frac{26-15}{20}}{\frac{39+16}{20}}=\frac{1}{5}$$
⇒ $$\frac{\frac{11}{20}}{\frac{55}{20}}=\frac{1}{5}$$
⇒ $$\frac{1}{5}=\frac{1}{5}$$
L.H.S = R.H.S
Hence verified.

(ix) $$\frac{\frac{3}{4} x+1}{x+\frac{1}{5}}=\frac{7}{4}$$
Solution:
$$\frac{\frac{3}{4} x+1}{x+\frac{1}{5}}=\frac{7}{4}$$
⇒ 4 ( + 1) = 7 (x + $$\frac{1}{5}$$)
3x + 4 = 7x + $$\frac{7}{5}$$
3x – 7x = $$\frac{7}{5}$$ – 4
– 4x = $$\frac{7-20}{5}$$
– 4x = $$-\frac{13}{5}$$
∴ x = $$\frac{13}{5 \times 4}=\frac{13}{20}$$
= $$-\frac{13}{20}$$
∴ x = $$-\frac{13}{20}$$

Verification:
Putting the value of x in the given equation $$\frac{\frac{3}{4} x+1}{x+\frac{1}{5}}=\frac{7}{4}$$
⇒ $$\frac{\frac{3}{4}\left(\frac{13}{20}\right)+1}{\frac{13}{20}+\frac{1}{5}}=\frac{7}{4}$$
⇒ $$\frac{\frac{39}{80}+1}{\frac{13}{20}+4}=\frac{7}{4}$$
⇒ $$\frac{\frac{39+80}{80}}{\frac{17}{20}}=\frac{7}{4}$$
⇒ $$\frac{119}{80} \times \frac{20}{17}=\frac{7}{4}$$
⇒ $$\frac{7}{4}=\frac{7}{4}$$
L.H.S = R.H.S
Hence verified.

(x) $$\frac{3 k+5}{4 k-3}=\frac{4}{9}$$
Solution:
$$\frac{3 k+5}{4 k-3}=\frac{4}{9}$$
⇒ 3 (0.5x – 4) = – 5 (2.4x + 6)
⇒ 1.5x – 12 = – 12x – 30
⇒ 15x + 12x = – 30 + 12
⇒ 13.5x = – 18
∴ x = $$\frac{-18}{13.5}$$
= $$\frac{-18 \times 10}{135}=\frac{-4}{3}$$

Verification:
Putting the value of x = $$\frac{-4}{3}$$ in the given equation $$\frac{3 k+5}{4 k-3}=\frac{4}{9}$$

L.H.S = R.H.S
Hence verified.

(xi) $$\frac{0.5 x-4}{2.4 x+6}=\frac{-5}{3}$$
Solution:
$$\frac{0.5 x-4}{2.4 x+6}=\frac{-5}{3}$$
⇒ 3 (0.5x – 4) = – 5 (2.4x + 6)
⇒ 1.5x – 12 = – 12x – 30
⇒ 1.5x + 12x = – 30 + 12
⇒ 13.5x = – 18
∴ x = $$\frac{-18}{13.5}$$
= $$\frac{-18 \times 10}{135}$$
= $$\frac{-4}{3}$$

Verification:
Putting the value of x = $$\frac{-4}{3}$$ in the given equation $$\frac{0.5 x-4}{2.4 x+6}=\frac{-5}{3}$$,
⇒ $$\frac{0.5\left(\frac{-4}{3}\right)-4}{2.4\left(\frac{-4}{3}\right)+6}=\frac{-5}{3}$$
⇒ $$\frac{\frac{-2}{3}-4}{-3.2+6}=\frac{-5}{3}$$
⇒ $$\frac{\frac{-14}{3}}{2.8}=\frac{-5}{3}$$
⇒ $$\frac{-14}{3 \times 2.8}=\frac{-5}{3}$$
⇒ $$\frac{-5}{3}=\frac{-5}{3}$$
L.H.S = R.H.S
Hence verified.

(xii) $$\frac{\frac{x}{3}-\frac{2}{5}}{\frac{3}{4}-2 x}=\frac{16}{15}$$
Solution:
$$\frac{\frac{x}{3}-\frac{2}{5}}{\frac{3}{4}-2 x}=\frac{16}{15}$$
$$15\left(\frac{x}{3}-\frac{2}{5}\right)=16\left(\frac{3}{4}-2 x\right)$$
⇒ 5x – 6 = 12 – 32x
⇒ 5x + 32x = 12 + 6
⇒ 37x = 18.
∴ x = $$\frac{18}{37}$$

Verification:
Putting the value of x in the given equation $$\frac{\frac{x}{3}-\frac{2}{5}}{\frac{3}{4}-2 x}=\frac{16}{15}$$

L.H.S = R.H.S
Hence verified.

(xiii) $$\frac{(1-2 x)+(1+2 x)}{(4 x+1)+(x-3)}=\frac{1}{2}$$
Solution:
$$\frac{(1-2 x)+(1+2 x)}{(4 x+1)+(x-3)}=\frac{1}{2}$$
⇒ $$\frac{1-2 x+1+2 x}{4 x+1+x-3}=\frac{1}{2}$$
⇒ $$\frac{2}{5 x-2}=\frac{1}{2}$$
⇒ 5x – 2 = 4
⇒ 5x = 2 + 4
⇒ 5x = 6
∴ x = $$\frac{6}{5}$$

Verification:
Putting the value of x = $$\frac{6}{5}$$ in the given equation $$\frac{(1-2 x)+(1+2 x)}{(4 x+1)+(x-3)}=\frac{1}{2}$$,

L.H.S = R.H.S
Hence verified.

(xiv) $$\frac{x^2-(x+2)(x+3)}{7 x+1}=\frac{2}{3}$$
Solution:
$$\frac{x^2-(x+2)(x+3)}{7 x+1}=\frac{2}{3}$$
⇒ $$\frac{x^2-\left(x^2+2 x+3 x+6\right)}{7 x+1}=\frac{2}{3}$$
⇒ $$\frac{x^2-\left(x^2+5 x+6\right)}{7 x+1}=\frac{2}{3}$$
⇒ $$\frac{x^2-x^2-5 x-6}{7 x+1}=\frac{2}{3}$$
⇒ $$\frac{-5 x-6}{7 x+1}=\frac{2}{3}$$
⇒ – 15x – 18 = 14x + 2
⇒ – 15x – 14x = 2 + 18
⇒ – 29x = 20
⇒ x = $$-\frac{20}{29}$$

Verification:
Put x = $$-\frac{20}{29}$$ in the given equation $$\frac{x^2-(x+2)(x+3)}{7 x+1}=\frac{2}{3}$$,

L.H.S = R.H.S
Hence verified.

Question 2.
Find the positive value of x for which the given equation is satisfied.

(i) $$\frac{3-x^2}{8+x^2}=\frac{-3}{4}$$
Solution:
$$\frac{3-x^2}{8+x^2}=\frac{-3}{4}$$
⇒ 4 (3 – x2) = – 3 (8 + x2)
⇒ 12 – 4x2 = – 24 – 3x2
⇒ – 4x2 + 3x2 = – 24 – 12
⇒ – x2 = – 36
⇒ x2 = 36
⇒ x = ± 6
∴ Positive value of x is 6.

(ii) $$\frac{y^2+6}{8 y^2+3}=\frac{1}{5}$$
Solution:
$$\frac{y^2+6}{8 y^2+3}=\frac{1}{5}$$
5 (y2 + 6) = 1 (8y2 + 3)
5y2 + 30 = 8y2 + 3
5y2 – 8y2 = 3 – 30
– 3y2 = – 27
y2 = $$\frac{-27}{-3}$$ = 9
y2 = 9
∴ y = ± 3.
∴ Positive value of x is 3.

(iii) $$\frac{x^2-9}{5+x^2}=\frac{5}{9}$$
Solution:
$$\frac{x^2-9}{5+x^2}=\frac{5}{9}$$
9 (x2 – 9) = 5 (5 + x2)
9x2 – 81 = 25 + 5x2
9x2 – 5x2 = 25 + 81
4x2 = 106
x2 = $$\frac{106}{4}$$
x2 = $$\frac{53}{2}$$
∴ x = ± $$\sqrt{\frac{53}{2}}$$
∴ Positive value of x is $$\sqrt{\frac{53}{2}}$$.

(iv) $$\frac{y^2+4}{3 y^2+7}=\frac{1}{2}$$
Solution:
$$\frac{y^2+4}{3 y^2+7}=\frac{1}{2}$$
2 (y2 + 4) = 3y2 + 7
2y2 + 8 = 3y2 + 7
2y2 – 3y2 = – 8 + 7
– y2 = – 1
y2 = 1
∴ y = ± √1
= ± 1
∴ Positive value of y = 1.

### DAV Class 8 Maths Chapter 9 Worksheet 1 Notes

• An algebraic equation is an equality involving variables. Here we will deal with the equations having only one variable.
e.g. 3x + 5 = x + 3.
• The algebraic equations with one variable are called as linear equations.
• In an equation, the values of the expressions on the L.H.S and R.H.S are equal which happens to be true only for certain values of the variable. These values are the solutions of the expression.
• To get a balance in the linear equation, any kind of operation such as addition, subtraction, multiplication or division is made on the both sides of the sign ‘=‘.

Example 1.
Solve: $$\frac{3}{2}$$ x – 7 = 5.
Solution:
$$\frac{3}{2}$$ x – 7 = 5
⇒ $$\frac{3}{2}$$ x = 5 + 7
[Transposing 7 to RHS]
⇒ $$\frac{3}{2}$$ x = 12
⇒ x = 12 ÷ $$\frac{3}{2}$$
= 12 × $$\frac{2}{3}$$ = 8
Hence x = 8.

Check :
$$\frac{3}{2}$$ x – 7 = 5
Put x = 8,
$$\frac{3}{2}$$ (8) – 7 = 5
⇒ 12 – 7 = 5
⇒ 5 = 5
∴ LHS = RHS
Hence verified.

Example 2.
Solve: $$\frac{3 x-5}{x+4}$$ = 5, x ≠ – 4.
Solution:
$$\frac{3 x-5}{x+4}$$ = 5
Multiplying both sides by (x + 4), we get
$$\frac{3 x-5}{x+4}$$ × (x+4) = 5 (x + 4)
3x – 5 = 5x + 20
3x – 5x = 20 + 5
– 2x = 25
x = $$\frac{25}{-2}$$
= – 12 $$\frac{1}{2}$$.
Hence x = – 12 $$\frac{1}{2}$$

Example 3.
Solve the equation $$\frac{7 x-5}{4 x+2}=\frac{8}{7}$$ and check your answer.
Solution:
$$\frac{7 x-5}{4 x+2}=\frac{8}{7}$$
Cross-multiplying, we get
7 (7x – 5) = 8 (4x + 2)
⇒ 49x – 35 = 32x + 16
⇒ 49x – 32x = 35 + 16
⇒ 17x = 51
⇒ x = $$\frac{51}{17}$$
⇒ x = 3
Hence x = 3.

Check: –
$$\frac{7 x-5}{4 x+2}=\frac{8}{7}$$
Put x = 3, we have
⇒ $$\frac{7(3)-5}{4(3)+2}=\frac{8}{7}$$
⇒ $$\frac{21-5}{12+2}=\frac{8}{7}$$
⇒ $$\frac{16}{14}=\frac{8}{7}$$
⇒ $$\frac{8}{7}=\frac{8}{7}$$
⇒ LHS = RHS.
Hence verified.

Example 4.
The perimeter of a rectangle is 13 cm and its width is 2$$\frac{3}{4}$$ cm. Find its length.
Solution:
Let the length of the rectangle be x cm.
Perimeter of the rectangle = 2 [Length + Width]
= 2 [x + 2 $$\frac{3}{4}$$]
= 2 [x + $$\frac{11}{4}$$]
But the given perimeter = 13 cm
∴ 2 [x + $$\frac{11}{4}$$] = 13
2x + 2 × $$\frac{11}{4}$$ = 13
2x + $$\frac{11}{2}$$ = 13
= 2x = 13 – $$\frac{11}{2}$$
= 2x = $$\frac{15}{2}$$
x = $$\frac{15}{2 \times 2}$$
= $$\frac{15}{4}=3 \frac{3}{4}$$ cm
Hence the required length = 3 $$\frac{3}{4}$$ cm.

Example 5.
The sum of three consecutive multiples of 11 is 363. Find these multiples.
Solution:
Let the three consecutive multiples of 11 be x, x + 11 and x + 22.
∴ x + (x + 11) + (x + 22) = 363
⇒ x + x + 11 + x + 22 = 363
⇒3x + 33 = 363
⇒ 3x = 363 – 33
⇒ 3x = 330
⇒ x = $$\frac{330}{3}$$
∴ x = 110
Hence the required multiples are 110, 110 + 11 = 121 and 110 + 22 = 132.

Example 6.
The number of boys and girls in a class are in the ratio 7 : 5. The number of boys is 8 more than the number of girls. What is the total strength of the class?
Solution:
Let the number of boys and girls in the class be 7x and 5x respectively.
∴ 7x – 5x = 8
⇒ 2x = 8
∴ x = 4
Number of boys = 7 × 4 = 28
Number of girls = 5 × 4 = 20
Total strength of the class = 28 + 20 = 48

Example 7.
Ram is twice as old as Reena. Five years ago, his age was three times Reena’s age. Find their present ages.
Solution:
Let Reena’s present age be x years
∴ Ram’s present age = 2x years.
5 years ago Reena’s age = (x – 5) years
5 years ago Ram’s age = (2x – 5) years
As per the condition,
2x – 5 = 3 (x – 5)
⇒ 2x – 5 = 3x – 15
⇒ 2x – 3x = – 15 + 5
⇒ – x = – 10
∴ x = 10
Hence Reena’s present age = 10 years
and Ram’s present age = 10 × 2 = 20 years.

Example 8.
The sum of digits of a two digit number is 9. If 27 is added to the number, the places of the digits are reversed, find the number.
Solution:
Let unit place of the two digit number is x
∴ Its ten’s place = 9 – x
∴ Number = x + 10 (9 – x)
= x + 90 – 10x
= 90 – 9x
Number obtained by reversing the digits = 10x + (9 – x)
= 10x + 9 – x
= 9x + 9
As per the condition,
90 – 9x + 27 = 9x + 9
– 9x – 9x = 9 – 27 90
– 18x = – 108
18x = 108
∴ x = $$\frac{108}{18}$$ = 6
∴ Unit place = 6
and Ten’s place = 9 – 6 = 3
Hence the required number = 6 + 3 × 10
= 6 + 30 = 36.