Linear Equations in One Variable - 8th math
ncert exercise 2.5 solution-Part-2
Question (6) 
Solution
Given,
After transposing `-(m-2)/3` to LHS, we get
After cross multiplication, we get
5m – 1 = 6 × 1
⇒ 5m – 1 = 6
After transposing –1 to RHS, we get
5m = 6 + 1 =7
⇒ 5m = 7
After transposing 5 to RHS, we get
⇒ m = 7/5 Answer
Checking of Result
Given,
Now, LHS = `m-(m-1)/2` to LHS, we get
After substituting the value of m = 7/5, we get
Thus, LHS = `6/5`
Now, RHS = `1-(m-2)/3`
After substituting the value of m = 7/5, we get
Thus, RHS `=6/5`
Thus, LHS = RHS Proved
Simplify and solve the following linear equations.
Question (7) 3(t – 3) = 5(2t + 1)
Solution
Given, 3(t – 3) = 5(2t + 1)
⇒ 3t – 3 × 3 = 5 × 2t + 1× 5
⇒ 3t – 9 = 10t + 5
After transposing –9 to RHS, we get
⇒ 3t = 10t + 5 + 9
⇒ 3t = 10t + 14
After transposing 10t to LHS, we get
⇒ 3t – 10t = 14
⇒ – 7t = 14
After transposing 7 to RHS, we get
⇒ –t = 14/7 = 2
⇒ t = –2 Answer
Checking of Result
Given, 3(t – 3) = 5(2t + 1)
Now, LHS = 3(t – 3)
After substituting the value of t = –2, we get
3(–2 – 3)
= 3 ( –5)
Thus, LHS = –15
Now, RHS = 5(2t + 1)
After substituting the value of t = –2, we get
5 [2×(–2) + 1]
= 5(–4 + 1)
= 5(–3) = –15
Thus, RHS = –15
Thus, LHS = RHS Proved
Question (8) 15(y –4) – 2(y – 9) + 5(y + 6) = 0
Solution
Given, 15(y –4) – 2(y – 9) + 5(y + 6) = 0
After opening of brackets
⇒ 15y – 60 –2y + 18 + 5y + 30 = 0
After rearranging the above expression
⇒ 15y – 2y + 5 y – 60 + 18 + 30 = 0
⇒ 13y + 5 y – 60 + 48 = 0
⇒ 18y – 12 = 0
After transposing –12 to RHS, we get
18y = 12
After transposing 18 to RHS, we get
Answer
Checking of Result
Given, 15(y –4) – 2(y – 9) + 5(y + 6) = 0
Now, LHS = 15(y –4) – 2(y – 9) + 5(y + 6)
Thus, after substituting the value of y = 2/3, we get
Thus, LHS = RHS Proved
Question (9) 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
Solution
Given, 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
⇒ 15z – 21 – 18z + 22 = 32z – 52 – 17
After rearranging the above expression, we get
⇒ 15z – 18z – 21 + 22 = 32z – 69
⇒ –3z + 1 = 32z – 69
Now after transposing 32z to RHS, we get
–3z – 32z = –69 – 1
⇒ –35z = –70
After transposing –35 to RHS, we get
`z=(-70)/(-35)`
⇒ z = 2 Answer
Checking of Result
Given, 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
Now, LHS = 3(5z – 7) – 2(9z – 11)
After substituting the value of z = 2, we get
3(5 × 2 – 7) – 2(9 × 2 –11)
= 3(10 – 7) – 2(18 – 11)
= 3 (3) – 2(7)
= 9 – 14
⇒ LHS = – 5
Now, RHS = 4(8z – 13) – 17
After substituting the value of z = 2, we get
4(8 × 2 – 13) – 17
= 4(16 – 13) – 17
= 4(3) – 17
= 12 – 17
⇒ RHS = –5
Thus, LHS = RHS Proved
Question (10) 0.25(4f – 3) = 0.05(10f – 9)
Solution
Given, 0.25(4f – 3) = 0.05(10f – 9)
⇒ 1 × f – 0.75 = 0.5f – 0.45
⇒ f – 0.75 = 0.5f – 0.45
Now after transfosing 0.5f to LHS, we get
⇒ f – 0.75 – 0.5f = – 0.45
Again after transposing –0.75 to RHS, we get
⇒ f – 0.5f = –0.45 + 0.75
⇒ 0.5f = 0.3
Now, after transposing 0.5 to RHS, we get
⇒ f = 0.3/0.5
⇒ f = 3/5 = 0.6
⇒ f = 0.6 Answer
Checking of Result
Given, 0.25(4f – 3) = 0.05(10f – 9)
Now, LHS = 0.25(4f – 3)
After substituting the value of f = 0.6, we get
0.25 (4 × 0.6 – 3)
= 0.25(2.4 – 3)
= 0.25 (–0.6)
= – 0.150
⇒ LHS = – 0.15
Now, RHS = 0.05(10f – 9)
After substituting the value of f = 0.6, we get
0.05 (10 × 0.6 – 9)
= 0.05 (6 – 9)
= 0.05 (–3)
⇒ RHS = – 0.15
Thus, LHS = RHS Proved
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