第 8 类 NCERT 解决方案 – 第 2 章一个变量中的线性方程 – 练习 2.5
求解下列线性方程。
问题 1. x/2 – 1/5 = x/3 + 1/4
解决方案：
(5x – 2)/10 = (4x + 3)/12 …(Taking LCM on both the sides)
12(5x – 2) = 10 (4x + 3) …(Cross multiplying)
60x – 24 = 40x + 30 …(Solving the brackets)
60x – 40x = 30 + 24 …(Transposing terms of x to LHS and others to RHS)
20x = 54
x = 54/20 or 27/10 … (Solution)
Verification:
Putting value of “x” in the equation to check if our answer is correct
27/20 – 1/5 = 27/30 + 1/4
(27 – 4)/20 = (108 + 30)/120
23/20 = 138/120
23/20 = 23/20
LHS = RHS (Hence Proved that solution is correct)
问题 2. n/2 – 3n/4 + 5n/6 = 21
解决方案：
(6n – 9n + 10n)/12 = 21 …(Taking LCM and solving LHS)
7n/12 = 21 (Solving LHS)
7n = 21 × 12
n = 36 …(Solution)
Verification:
Putting value of “n” in the equation to check if our answer is correct
36/2 – 108/4 + 180/6 = 21
18 – 27 + 30 = 21
21 = 21
LHS = RHS (Hence Proved that solution is correct)
问题 3. x + 7 – 8x/3 = 17/6 – 5x/2
解决方案：
x – 8x/3 + 5x/2 = 17/6 – 7 …(Transposing terms of x to LHS and others to RHS)
(6x – 16x + 15x)/6 = (17 – 42)/6 …(Taking LCM and solving)
5x/6 = -25/6
x = -5 …(Solution)
Verification –
Putting value of “x” in the equation to check if our answer is correct
-5 + 7 – (-40)/3 = 17/6 – (-25)/2
2 + 40/3 = 17/6 + 25/2
46/3 = (17 + 75)/6
46/3 = 92/6
46/3 = 46/3
LHS = RHS (Hence Proved that solution is correct)
问题 4. (x – 5)/3 = (x – 3)/5
解决方案：
5(x – 5) = 3(x – 3) …(Cross multiply)
5x – 25 = 3x – 9
2x = 16
x = 8 …(Solution)
Verification –
Putting value of “x” in the equation to check if our answer is correct
(8 – 5)/3 = (8 – 3)/5
3/3 = 5/5
1 = 1
LHS = RHS (Hence Proved that solution is correct)
问题 5. (3t – 2)/4 – (2t + 3)/3 = 2/3 – t
解决方案：
3t/4 – 1/2 – 2t/3 -1 = 2/3 – t …(Solving brackets)
3t/4 – 2t/3 + t = 2/3 + 1 + 1/2 …(Transposing terms of x to LHS and others to RHS)
(9t – 8t + 12t)/12 = (4 + 6 + 3)/6 …(Taking LCM both sides)
13t/12 = 13/6
t = 2 …(Solution)
Verification –
Putting value of “t” in the equation to check if our answer is correct
(3 × 2 – 2)/4 – (2 × 2 + 3)/3 = 2/3 – 2
4/4 – 7/3 = 2/3 – 2
(12 – 28)/12 = (2 – 6)/3
-16/12 = -4/3
-4/3 = -4/3
LHS = RHS (Hence Proved that solution is correct)
问题 6. m – (m – 1)/2 = 1 – (m – 2)/3
解决方案：
(2m – m + 1)/2 = (3 – m + 2)/3 …(Taking LCM both sides)
(m + 1)/2 = (5 – m)/3
3(m + 1) = 2(5 – m) …(Cross multiplying)
3m + 3 = 10 – 2m
5m = 7
m = 7/5 …(Solution)
Verification –
Putting value of “m” in the equation to check if our answer is correct
7/5 – (7/5 – 1)/2 = 1 – (7/5 – 2)/3
7/5 – 1/5 = 1 – (-3)/15
6/5 = 1 + 1/5
6/5 = 6/5
LHS = RHS (Hence Proved that solution is correct)
问题 7. 3(t – 3) = 5(2t + 1)
解决方案：
3t – 9 = 10t + 5 …(Opening brackets)
3t – 10t = 9 + 5
-7t = 14
t = -2 …(Solution)
Verification –
Putting value of “t” in the equation to check if our answer is correct
3(-2 – 3) = 5(2(-2) + 1)
3(-5) = 5(-4 +1)
-15 = -15
LHS = RHS (Hence Proved that solution is correct)
问题 8. 15(y – 4) – 2(y – 9) + 5(y + 6) = 0
解决方案：
15y – 60 – 2y + 18 + 5y + 30 = 0
18y – 12 = 0
y = 12/18 or 2/3 …(Solution)
Verification –
Putting value of “y” in the equation to check if our answer is correct
15(2/3 – 4) – 2(2/3 – 9) + 5(2/3 + 6) = 0
10 – 60 – 4/3 +18 + 10/3 + 30 = 0
-50 -4/3 + 48 + 10/3 = 0
-2 + 6/3 = 0
-2 + 2 = 0
0 = 0
LHS = RHS (Hence Proved that solution is correct)
问题 9. 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
解决方案：
15z – 21 – 18z + 22 = 32z – 52 – 17 …(Solving the brackets)
-3z + 1 = 32z – 69
-35z = -70
z = 2 …(Solution)
Verification –
Putting value of “z” in the equation to check if our answer is correct
3(5(2) – 7) – 2(9(2) – 11) = 4(8(2) – 13) – 17
3(3) – 2(7) = 4(3) – 17
9 – 14 = 12 – 17
-5 = -5
LHS = RHS (Hence Proved that solution is correct)
问题10。0.25(4f – 3) = 0.05(10f – 9)
解决方案：
f – 0.25(3) = 0.5f – 0.05(9)
f – 0.75 = 0.5f – 0.45
0.5f = 0.75 – 0.45
f = 3/5 or 0.6 (Solution)
Verification –
Putting value of “f” in the equation to check if our answer is correct
0.25(4(0.6) – 3) = 0.05(10(0.6) – 9)
0.25(2.4 – 3) = 0.05(6 – 9)
0.25 × (-0.6) = 0.05 × (-3)
-0.15 = -0.15
LHS = RHS (Hence Proved that solution is correct)