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8th class > Linear Equations in One Variable > Exercise 2.2

Exercise 2.2

Solve the following linear equations.

Instructions

1. x2 - 15 = x3 + 14
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: 60x215 = 60x3+14 gives us 30x12 = 20x+15
Subtracting 20x from both sides: 30x20x12 = 15 which simplifies to - = 15
Adding 12 to both sides of the equation with x : 10x12+12 = 15+12 which gives us 10x=27
Finally dividing both sides by 10: x =
Substituting x in the equation: x215 = x3 + 14
2710×2 - 15= 2710×3 + 14
2720 - 15 = 2730 + 14 =
2. n2 - 3n4 + 5n6 = 21
Simplifying the equation as LCM of 2, 4 and 6 is .
6n33n+25n12 = 21 ⇒ 6n9n+10n12 = 21 ⇒ 6n+n12 = 21
7n12 = 21 ⇒ 7n = 21×12 ⇒ n = 21×127 ⇒ n =
Substituting n in the equation: n23n4+5n6
362 - 3×364 + 5×3661827+30 =
3. x+7- 8x3 = 176 - 5x2
Simplifying the equation: We get LCM of 2, 3 and 6 is .
Multiplying every term in the equation : 6×x+78x3 = 6×1765x2
Expanding the equation : 6x+42 - 2x8x= 11735x
Simplifying each term : 6x16x + = 15x
Isolating the variable x: + 42 = 1715x which gives us 10x+15x+42=17 + 42 = 17
Subtracting 42 from both sides: 5x =
Dividing both sides by 5: x = 255 ⇒ x =
Substituting x in equation: 5+7853 = - 552
5+7+403 = 176+2522+403 =
Further simplfying: 63+403 = 926 = 926
We get: 926 = 463 i.e. LHS equal to RHS for x = .
4. x53 = x35
The LCM of 3 and 5 is .
Multiplying every term in the equation by 15 to simplify: 15x53 = 15x35 x5 = x3
Expanding: - 25 = 3x -
Isolating by subtracting 3x from both sides: 5x3x25 = -9
Adding 25 to both sides: 2x25+25 = 9+252x =
Dividing both sides by 2: x=162 which gives us x =
Substituting x in the equation: (-5)/3 = (-3)/3 ⇒ /3 = /5 where both further reduce to .
Thus, LHS is equal to RHS.
5. 3t24 - 2t+33 = 23t
The LCM of 4 and 3 is .
Multiplying every term in the equation by 12 : 123t24 - 122t+33 = 122312t.
Equation simplifies to: 33t242t+3 = 812t
Expanding, we get: 9t68t12 = 812t.
Separating out like terms : t18 = 812t.
Isolating 't': t+12t18 = 8 ⇒ - 18 =
Adding 18 to both sides : 13t18+18 = 8+1813t =
Dividing both sides by 13: t = 2613 ⇒ t =
Substituting 't' in equation: 322422+33 = 232
373 = 263 =
6. mm12 = 1m23
Simplify the Equation : 2mm12 = 1m232mm+12 = 1m23.
m+12 = 1m23m+12 = 3m23m+12 = 3m23.
m+12 = 5m33m+1 = 25m3m+3 = 25m3m+3 = 102m.
5m+3 = 105m = 103 = ⇒ m = .
Substituting m in equation: 75 - 7512 = 1 - 7523
= (LHS = RHS)

Simplify and solve the following linear equations.

7. 3t3=52t+1
Simplifying the equation: 3t9 = 52t+13t9 = 10t+53t10t9 = 5
7t9= 5 ⇒ -7t = 5+9 ⇒ -7t = 14 ⇒ t =
Substituting t in equation: 323 = 522+1
35 = 53 ⇒ LHS = RHS
8. 15y42y9+5y+6=0
Simplifying the equation : 1562y+18+5y+30 = 0 ⇒ 15y2y+5y60+18+30 = 0 ⇒ = 0
⇒ 18y = 12 ⇒ y =
Substituting y in equation: 15234 - 2239 + 523+6 = 0
15×103- 2253+ 5203 = 0 ⇒ 1503 + 503 + 1003 = 0
Thus, 150+50+1003 = i.e. LHS is equal to RHS = .
9. 35z729z11=48z1317
Simplifying the equation : 15z2118z+22 = 48z13173z+1 = 48z13173z+1 = -
3z = 32z70 = 70
So, z = 7035 ⇒ z =
Substituting z in equation: 35×2729×211 = 48×21317
3×32×7 = 4×317 ⇒ LHS = and RHS =
10. 0.254f3=0.0510f9
Simplifying the equation : 251004f3 = 510010f9144f3 = 12010f9.
4f34 = 10f920204f3 = 410f9
80f40f60 = 36 f = 36+60 ⇒ f = ⇒ f =
Substituting f in equation: 0.254×0.63 = 0.0510×0.69
⇒ LHS = RHS