RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Linear Equations in One Variable Ex 9.2

Other Exercises

Solve each of the following equations and also check your result in each case :
Question 1.
$$\frac { 2x + 5 }{ 3 }$$ = 3x – 10
Solution:
$$\frac { 2x + 5 }{ 3 }$$ = $$\frac { 3x – 10 }{ 1 }$$
By cross multiplication
⇒ 2x + 5 = 3 (3x – 10)
⇒ 2x + 5 = 9x – 30
⇒ 5 + 30 = 9x – 2x (By transposition)
⇒ 35 = 7x
⇒ x = 5

Question 2.
$$\frac { a – 8 }{ 3 }$$ = $$\frac { a – 3 }{ 2 }$$
Solution:

Question 3.
$$\frac { 7y + 2 }{ 5 }$$ = $$\frac { 6y – 5 }{ 11 }$$
Solution:

Question 4.
x – 2x + 2 – $$\frac { 16 }{ 3 }$$ x + 5 = 3 – $$\frac { 7 }{ 2 }$$ x.
Solution:

Question 5.
$$\frac { 1 }{ 2 }$$ x + 7x – 6 = 7x + $$\frac { 1 }{ 4 }$$
Solution:

Question 6.
$$\frac { 3 }{ 4 }$$ x + 4x = $$\frac { 7 }{ 8 }$$ + 6x – 6
Solution:

Question 7.
$$\frac { 7 }{ 2 }$$ x – $$\frac { 5 }{ 2 }$$ x = $$\frac { 20 }{ 3 }$$ x + 10
Solution:
$$\frac { 7 }{ 2 }$$ x – $$\frac { 5 }{ 2 }$$ x = $$\frac { 20 }{ 3 }$$ x + 10

Question 8.
$$\frac { 6x + 1 }{ 2 }$$ + 1 = $$\frac { 7x – 3 }{ 3 }$$
Solution:

Question 9.
$$\frac { 3a – 2 }{ 3 }$$ + $$\frac { 2a + 3 }{ 2 }$$ = a + $$\frac { 7 }{ 6 }$$
Solution:

Question 10.
x – $$\frac { x – 1 }{ 2 }$$ = 1 – $$\frac { x – 2 }{ 3 }$$
Solution:

Question 11.
$$\frac { 3x }{ 4 }$$ – $$\frac { x – 1 }{ 2 }$$ = $$\frac { x – 2 }{ 3 }$$
Solution:

Question 12.
$$\frac { 5x }{ 3 }$$ – $$\frac { x – 1 }{ 4 }$$ = $$\frac { x – 3 }{ 5 }$$
Solution:
$$\frac { 5x }{ 3 }$$ – $$\frac { x – 1 }{ 4 }$$ = $$\frac { x – 3 }{ 5 }$$

Question 13.

Solution:

Question 14.

Solution:

Question 15.

Solution:

Question 16.
0.18 (5x – 4) = 0.5x + 0.8
Solution:

Question 17.

Solution:

Question 18.

Solution:

Question 19.

Solution:

Question 20.

Solution:

Question 21.

Solution:

Question 22.

Solution:

Question 23.

Solution:

Question 24.
(3x – 8) (3x + 2) – (4x – 11) (2x + 1) = (x – 3) (x + 7)
Solution:
(3x – 8) (3x + 2) – (4x – 11) (2x + 1) = (x – 3) (x + 7)
⇒ (9x² + 6x – 24x – 16) – (8x² + 4x – 22x – 11) = x² + 7x – 3x – 21
⇒ 9x² + 6x – 24x – 16 – 8x² – 4x + 22x + 11 = x² + 4x – 21
⇒ 9x² – 8x² – x² + 6x – 24x + 22x – 4x – 4x = -21 + 16 – 11
⇒ 28x – 32x = -32 + 16
⇒ -4x = -16
⇒ x = 4
Verification:
L.H.S. = (3x – 8) (3x + 2) – (4x – 11) (2x + 1)
= (3 x 4 – 8) (3 x 4 + 2) – (4 x 4 – 11) (2 x 4 + 1)
= (12 – 8) (12 + 2) – (16 – 11) (8 + 1)
= 4 x 14 – 5 x 9 = 56 – 45 = 11
R.H.S. = (x – 3) (x + 7) = (4 – 3) (4 + 7) = 1 x 11 = 11
L.H.S. = R.H.S.

Question 25.
[(2x + 3) + (x + 5)]² + [(2x + 3) – (x + 5)]² = 10x² + 92
Solution:
[(2x + 3) + (x + 5)]² + [(2x + 3) – (x + 5)]² = 10x² + 92
⇒ (2x + 3 + x + 5)² + (2x + 3 – x – 5)² = 10x² + 92
⇒ (3x + 8)² + (x – 2)² = 10x² + 92
⇒ 9x² + 2 x 3x x 8 + 64 + x² – 2 x x x 2 + 4 = 10x² + 92
⇒ 9x² + 48x + 64 + x² – 4x + 4 = 10x² + 92
⇒ 9x² + x² – 10x² + 48x – 4x = 92 – 64 – 4
⇒ 44x = 24

Hope given RD Sharma Class 8 Solutions Linear Equations in One Variable Ex 9.2 are helpful to complete your math homework.

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