Gujarat Board GSEB Textbook Solutions Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5 Textbook Questions and Answers.
Gujarat Board Textbook Solutions Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5
Question 1.
Solve the following linear equations.
(i) \(\frac { x }{ 2 }\) – \(\frac { 1 }{ 5 }\) = \(\frac { x }{ 3 }\) + \(\frac { 1 }{ 4 }\)
(ii) \(\frac { n }{ 2 }\) – \(\frac { 3n }{ 4 }\) + \(\frac { 5n }{ 6 }\) = 21
(iii) x + 7 – \(\frac { 3 }{ 2 }\)
(iv) \(\frac { x-5 }{ 3 }\) = \(\frac { x – 3 }{ 5 }\)
(v) \(\frac { 3t – 2 }{ 4 }\) – \(\frac { 2t+3 }{ 3 }\) = \(\frac { 2 }{ 3 }\) – t
(vi) m – \(\frac { m – 1 }{ 2 }\) = 1 – \(\frac { m – 2 }{ 3 }\)
Solution:
(i) \(\frac { x }{ 2 }\) – \(\frac { 1 }{ 5 }\) = \(\frac { x }{ 3 }\) + \(\frac { 1 }{ 4 }\)
Solution:
Transposing (-1) to RHS and \(\frac { x }{ 3 }\) to LHS, we have
\(\frac { x }{ 2 }\) – \(\frac { x }{ 3 }\) = \(\frac { 1 }{ 4 }\) + \(\frac { 1 }{ 5 }\) or \(\frac { 3x – 2x }{ 6 }\) = \(\frac { 5 + 4 }{ 20 }\) or \(\frac { x }{ 6 }\) = \(\frac { 9 }{ 20 }\) × \(\frac { 9 }{ 20 }\) or x = \(\frac { 9 }{ 20 }\) × 6
(Multiplyinng both sides by 6)
= \(\frac { 27 }{ 10 }\) = x = \(\frac { 27 }{ 10 }\)
(ii) \(\frac { n }{ 2 }\) – \(\frac { 3n }{ 4 }\) + \(\frac { 5n }{ 6 }\) = 21
Solution:
∵LCM of 2, 4 and 6 = 12
∴Multiplyinng both sides by 12, we have
12 × \(\frac { n }{ 2 }\) – 12 × \(\frac { 3n }{ 4 }\) + 12 × \(\frac { 5n }{ 6 }\) = 12 × 21
or 6n – 9n + 10n = 252 or 7n = 252
or n = \(\frac { 252 }{ 7 }\) = 36
∴ n = 36
(iii) x + 7 – \(\frac { 8x }{ 3 }\) = \(\frac { 8x }{ 3 }\) = \(\frac { 17 }{ 6 }\) – \(\frac { 5x }{ 2 }\)
∵LCM of 3, 6 and 2 = 6
∴Multiplyinng both sides by 6, we have
6 × x + 6 × 7 – 6 × \(\frac { 8x }{ 3 }\) = 6 × \(\frac { 17 }{ 6 }\) – 6 × \(\frac { 5x }{ 2 }\)
or 6x + 42 – 16x = 17 – 15x
or (6 -16)x + 42 = 17 – 15x
or -10x + 42 = 17 – 15x
Transposing 42 to RHS and -15x to LHS, we have
-10x + 15x = 17 – 42 or 5x = -25
or x = \(\frac { -25 }{ 5 }\) = -5
(Dividing both sides by 5)
∴ x = -5
(iv) \(\frac { x-5 }{ 3 }\) = \(\frac { x – 3 }{ 5 }\)
∵LCM of 3, 6 and 5 is 15
∴Multiplyinng both sides by 15, we have
15 × \(\frac { x-5 }{ 3 }\) = \(\frac { x-3 }{ 5 }\) or 5(x-5) = 3(x-3)
or 5x – 25 = 3x – 9
Transposing (-25) to RHS and 3x to LHS, we have
5x – 3x = -9 + 25
or 2x = 16 or x = \(\frac { 16 }{ 2 }\)
(Dividing both sides by 2)
∴ x = 8
(v) \(\frac { 3t – 2 }{ 4 }\) – \(\frac { 2t+3 }{ 3 }\) = \(\frac { 2 }{ 3 }\) – t
∵LCM of 4 and 3 is 12
∴Multiplyinng both sides by 12, we have
12 × \(\frac { 3t – 2 }{ 4 }\) – 12 × \(\frac { 2t + 3 }{ 3 }\) = 12 × \(\frac { 2 }{ 3 }\) – 12 × the
or 3(3t – 2) -4(2t +3) -4(2t + 3) = (4 ×2) – 12t
or 9t – 6 – 8t – 12 = 8 – 12t
or (9 – 8)t – (6+12) = 8 – 12t
or t – 18 = 8 – 12t
Transposing -18 to RHS and -12t to LHS, we have
t + 12t = 8 + 18
or 133t = 26 or t = \(\frac { 26 }{ 13 }\)
(Dividing both sides by 2)
∴ t = 2
(vi) m – \(\frac { m – 1 }{ 2 }\) = 1 – \(\frac { m – 2 }{ 3 }\)
Since, LCM of 2 and 3 is 6
∴ Multiplyinng both sides by 6, we have
6 × m – 6 × \(\frac { m-1 }{ 2 }\) = 6 × 1 – 6 × \(\frac { m-2 }{ 3 }\)
or 6m – 3(m – 1) = 6 – 2(m – 2)
or 6m – 3m + 3 = 6 – 2m + 4
or (6 – 3)m + 3 = (6+4) – 2m
or 3m + 3 = 10 – 2m
Transposing 3 to RHS and -2m to LHS , we have
3m =2m + 10 – 3
or 5m = 7 m = \(\frac { 7 }{ 5 }\)
(Dividing both sides by 5)
Question 2.
Simplify and solve the following linear equations.
(vii) 3(t – 3) = 5(2t + 1)
(viii) 15(y – 4) – 2(y – 9) + 5(y + 6) = 0
(ix) 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
(x) 0.25(4f – 3) = 0.05(10f – 9)
Solutions:
(vii) 3(t-3) = 5(2t+ 1) or 3t – 9 = 10t + 5
Transposing (-9) to RHS and 10t to LHS,
we have
3t – 10t = 5 + 9 or -7t = 14
or t = \(\frac { 14 }{ -7 }\) = -2(Dividing both sides by -7)
∴ t = -2
(viii) 15(y – 4) – 2(y – 9) + 5(y + 6) = 0
Opening the brackets, we have
15y – 60 – 2y + 18 + Sy + 30 = 0
Collecting the like terms,
(15 – 2 + 5)y + (-60 + 18 + 30) = 0
or 18y + (-12) = 0
Transposing (-12) to RHS, we have
18y = 12 or y = \(\frac { 12 }{ 18 }\) = \(\frac { 2 }{ 3 }\)
(Dividing both sides by 18)
∴ y = \(\frac { 2 }{ 3 }\)
(ix) 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
Opening the brackets, we have
15z – 21 – 18z + 22 = 32z – 52 – 17
Collecting the like terms on both sides,
(15 – 18)z + (-21 + 22)
= 32z + (-52 – 17)
or -3z + 1 = 32z – 69
Transposing 1 to RHS and 32z to LHS,
we have
-3z – 32z = -69 – 1 or -35z = -70
or z = \(\frac { -70 }{ -35 }\) = 2
(Dividing both sides by -35)
∴ z = 2
(x) 0.25(4f – 3) = 0.05(10f – 9)
Opening the brackets, we have
0.25 x 4f – 3 x 0.25
= 0.05 x 10f – 0.05 x 9
or \(\frac { 25 }{ 100 }\) × 4f – 3 × \(\frac { 25 }{ 100 }\) × 10f = \(\frac { 5 }{ 100 }\) × 9
or f – \(\frac { 3 }{ 4 }\) = \(\frac { 1 }{ 2 }\) f – \(\frac { 9 }{ 20 }\)
Transposing (\(\frac { -3 }{ 4 }\)) to RHS and (\(\frac { 1 }{ 2 }\)f) to LHS, we have
