Exercise 2.2
Solve the following linear equations.
1. x 2 1 5 x 3 1 4
Now, let's solve for x. The denominators are 2,5,3,4. The LCM of these numbers is
Multiplying every term in the equation by 60: 60 x 2 − 1 5 60 x 3 + 1 4 30 x − 12 = 20 x + 15
Subtracting 20x from both sides: 30 x − 20 x − 12 = 15 which simplifies to 15
Adding 12 to both sides of the equation with x : 10 x − 12 + 12 = 15 + 12 which gives us 10 x = 27
Finally dividing both sides by 10: x =
Substituting x in the equation: x 2 − 1 5 x 3 1 4
⇒ 27 10 × 2 1 5 27 10 × 3 1 4
2. n 2 3 n 4 5 n 6
Simplifying the equation as LCM of 2, 4 and 6 is
Substituting n in the equation: n 2 − 3 n 4 + 5 n 6
⇒ 36 2 3 × 36 4 5 × 36 6 18 − 27 + 30 =
3. x + 7 - 8 x 3 17 6 5 x 2
Simplifying the equation: We get LCM of 2, 3 and 6 is
Multiplying every term in the equation : 6 × x + 7 − 8 x 3 6 × 17 6 − 5 x 2
Expanding the equation : 6 x + 42 - 2 x 8 x 117 − 3 5 x
Simplifying each term : 6 x − 16 x + − 15 x
Isolating the variable x: 42 = 17 − 15 x which gives us − 10 x + 15 x + 42 = 17 ⇒ 42 = 17
Subtracting 42 from both sides: 5x =
Dividing both sides by 5: x = − 25 5
Substituting x in equation: − 5 + 7 − 8 − 5 3 5 − 5 2
⇒ − 5 + 7 + 40 3 17 6 + 25 2 2 + 40 3
Further simplfying: 6 3 + 40 3 92 6 92 6
We get: 92 6 46 3
4. x − 5 3 x − 3 5
The LCM of 3 and 5 is
Multiplying every term in the equation by 15 to simplify: 15 x − 5 3 15 x − 3 5 x − 5 x − 3
Expanding: 25 = 3 x -
Isolating by subtracting 3 x from both sides: 5 x − 3 x − 25 = -9
Adding 25 to both sides: 2 x − 25 + 25 = − 9 + 25 ⇒ 2 x =
Dividing both sides by 2: x = 16 2
Substituting x in the equation: (
Thus, LHS is equal to RHS.
5. 3 t − 2 4 2 t + 3 3 2 3 − t
The LCM of 4 and 3 is
Multiplying every term in the equation by 12 : 12 3 t − 2 4 12 2 t + 3 3 12 2 3 − 12 t .
Equation simplifies to: 3 3 t − 2 − 4 2 t + 3 8 − 12 t
Expanding, we get: 9 t − 6 − 8 t − 12 = 8 − 12 t .
Separating out like terms : t − 18 = 8 − 12 t .
Isolating 't': t + 12 t − 18 = 8 ⇒ 18 =
Adding 18 to both sides : 13 t − 18 + 18 = 8 + 18 ⇒ 13 t =
Dividing both sides by 13: t = 26 13
Substituting 't' in equation: 3 2 − 2 4 − 2 2 + 3 3 2 3 − 2
⇒ 3 − 7 3 2 − 6 3
6. m − m − 1 2 1 − m − 2 3
Simplify the Equation : 2 m − m − 1 2 1 − m − 2 3 2 m − m + 1 2 1 − m − 2 3
⇒ m + 1 2 1 − m − 2 3 m + 1 2 3 − m − 2 3 m + 1 2 3 − m − 2 3
⇒ m + 1 2 5 − m 3 3 m + 1 2 5 − m 3 m + 3 = 2 5 − m 3 m + 3 = 10 − 2 m .
⇒ 5 m + 3 = 10 ⇒ 5 m = 10 − 3 ⇒
Substituting m in equation: 7 5 7 5 − 1 2 7 5 − 2 3
⇒
Simplify and solve the following linear equations.
7. 3 t − 3 = 5 2 t + 1
Simplifying the equation: 3 t − 9 = 5 2 t + 1 3 t − 9 = 10 t + 5 ⇒ 3 t − 10 t − 9 = 5
Substituting t in equation: 3 − 2 − 3 5 2 − 2 + 1
⇒ 3 − 5 5 − 3
8. 15 y − 4 − 2 y − 9 + 5 y + 6 = 0
Simplifying the equation : 15 − 6 − 2 y + 18 + 5 y + 30 = 0 ⇒ 15 y − 2 y + 5 y − 60 + 18 + 30 = 0 ⇒
⇒ 18y = 12 ⇒ y =
Substituting y in equation: 15 2 3 − 4 2 2 3 − 9 5 2 3 + 6
⇒ 15 × − 10 3 2 − 25 3 5 20 3 − 150 3 50 3 100 3
Thus, − 150 + 50 + 100 3
9. 3 5 z − 7 − 2 9 z − 11 = 4 8 z − 13 − 17
Simplifying the equation : 15 z − 21 − 18 z + 22 = 4 8 z − 13 − 17 ⇒ − 3 z + 1 = 4 8 z − 13 − 17 ⇒ − 3 z +1 =
⇒ − 3 z = 32 z − 70 ⇒ − 70
So, z = − 70 − 35
Substituting z in equation: 3 5 × 2 − 7 − 2 9 × 2 − 11 4 8 × 2 − 13 − 17
⇒ 3 × 3 − 2 × 7 = 4 × 3 − 17 ⇒ LHS =
10. 0.25 4 f − 3 = 0.05 10 f − 9
Simplifying the equation : 25 100 4 f − 3 5 100 10 f − 9 1 4 4 f − 3 1 20 10 f − 9
⇒ 4 f − 3 4 10 f − 9 20 20 4 f − 3 4 10 f − 9
⇒ 80 f − 40 f − 60 = − 36 ⇒ − 36 + 60 ⇒ f =
Substituting f in equation: 0.25 4 × 0.6 − 3 0.05 10 × 0.6 − 9
⇒ LHS