Linear Equations in one Variable

Q1: Multiple Choice Type

i. For \(\frac{x+0.5}{6x-1}=\frac{1}{2}\), the value of x is:

Step 1: Cross multiply:
2(x + 0.5) = 1(6x – 1)
Step 2: Expand both sides:
2x + 1 = 6x – 1
Step 3: Bring variables to one side and constants to the other:
2x – 6x = -1 – 1
-4x = -2
Step 4: Divide both sides by -4:
x = (-2)/(-4) = 1/2
Answer: a. x = 1/2

ii. For \(3\left(x+1\right)-4x=0\), the value of x is:

Step 1: Expand the bracket:
3x + 3 – 4x = 0
Step 2: Combine like terms:
(3x – 4x) + 3 = 0 → -x + 3 = 0
Step 3: Solve for x:
-x = -3 → x = 3
Answer: c. x = 3

iii. For \(7x-8=5x+2\), the value of x is:

Step 1: Bring variables to one side and constants to the other:
7x – 5x = 2 + 8
2x = 10
Step 2: Divide by 2:
x = 10 / 2 = 5
Answer: c. x = 5

iv. For \(\frac{5}{x}=\frac{7}{x-4}\), the value of x is:

Step 1: Cross multiply:
5(x – 4) = 7x
Step 2: Expand:
5x – 20 = 7x
Step 3: Bring variables to one side:
5x – 7x – 20 = 0 → -2x – 20 = 0
Step 4: Solve for x:
-2x = 20 → x = -10
Answer: a. x = -10

v. If \(\frac{x-2}{4}-\frac{x-3}{5}=1\), the value of x is:

Step 1: Find LCM of 4 and 5, which is 20:
(5(x-2) – 4(x-3))/20 = 1
Step 2: Expand numerator:
(5x – 10 – 4x + 12)/20 = 1 → (x + 2)/20 = 1
Step 3: Multiply both sides by 20:
x + 2 = 20 → x = 18
Answer: d. x = 18

Q2: Solve \(\frac{3x+2}{x-6} = -7\)

Step 1: Cross multiply:
3x + 2 = -7(x – 6)
Step 2: Expand the right-hand side:
3x + 2 = -7x + 42
Step 3: Bring variables to one side and constants to the other:
3x + 7x + 2 = 42 → 10x + 2 = 42
Step 4: Subtract 2 from both sides:
10x = 42 – 2 → 10x = 40
Step 5: Divide both sides by 10:
x = 40 / 10 → x = 4
Answer: x = 4

Q3: Solve \(3a – 4 = 2(4 – a)\)

Step 1: Expand the bracket on the right-hand side:
3a – 4 = 8 – 2a
Step 2: Bring all variable terms to one side and constants to the other:
3a + 2a – 4 = 8 → 5a – 4 = 8
Step 3: Add 4 to both sides:
5a = 8 + 4 → 5a = 12
Step 4: Divide both sides by 5:
a = 12 / 5 = 2.4
Answer: a = 2.4

Q4: Solve \(3(b – 4) = 2(4 – b)\)

Step 1: Expand both brackets:
3b – 12 = 8 – 2b
Step 2: Bring all variable terms to one side and constants to the other:
3b + 2b – 12 = 8 → 5b – 12 = 8
Step 3: Add 12 to both sides:
5b = 8 + 12 → 5b = 20
Step 4: Divide both sides by 5:
b = 20 / 5 → b = 4
Answer: b = 4

Q5: Solve \(\frac{x+2}{9} = \frac{x+4}{11}\)

Step 1: Cross multiply:
11(x + 2) = 9(x + 4)
Step 2: Expand both sides:
11x + 22 = 9x + 36
Step 3: Bring variables to one side and constants to the other:
11x – 9x = 36 – 22 → 2x = 14
Step 4: Divide both sides by 2:
x = 14 / 2 → x = 7
Answer: x = 7

Q6: Solve \(\frac{x-8}{5} = \frac{x-12}{9}\)

Step 1: Cross multiply:
9(x – 8) = 5(x – 12)
Step 2: Expand both sides:
9x – 72 = 5x – 60
Step 3: Bring variables to one side and constants to the other:
9x – 5x = -60 + 72 → 4x = 12
Step 4: Divide both sides by 4:
x = 12 / 4 → x = 3
Answer: x = 3

Q7: Solve \(5(8x + 3) = 9(4x + 7)\)

Step 1: Expand both brackets:
5 × 8x + 5 × 3 = 9 × 4x + 9 × 7 → 40x + 15 = 36x + 63
Step 2: Bring variables to one side and constants to the other:
40x – 36x = 63 – 15 → 4x = 48
Step 3: Divide both sides by 4:
x = 48 / 4 → x = 12
Answer: x = 12

Q8: Solve \(3(x+1) = 12 + 4(x-1)\)

Step 1: Expand both brackets:
3x + 3 = 12 + 4x – 4 → 3x + 3 = 4x + 8
Step 2: Bring variables to one side and constants to the other:
3x – 4x = 8 – 3 → -x = 5
Step 3: Multiply both sides by -1:
x = -5
Answer: x = -5

Q9: Solve \(\frac{3x}{4} – \frac{1}{4}(x – 20) = \frac{x}{4} + 32\)

Step 1: Expand the bracket on the left-hand side:
\(\frac{3x}{4} – \frac{x}{4} + \frac{20}{4} = \frac{x}{4} + 32\)
\(\frac{3x – x}{4} + 5 = \frac{x}{4} + 32\)
\(\frac{2x}{4} + 5 = \frac{x}{4} + 32\)
Step 2: Simplify \(\frac{2x}{4}\) to \(\frac{x}{2}\):
\(\frac{x}{2} + 5 = \frac{x}{4} + 32\)
Step 3: Bring all terms with x to one side and constants to the other:
\(\frac{x}{2} – \frac{x}{4} = 32 – 5\)
\(\frac{2x – x}{4} = 27\) → \(\frac{x}{4} = 27\)
Step 4: Multiply both sides by 4:
x = 27 × 4 → x = 108
Answer: x = 108

Q10: Solve \(3a – \frac{1}{5} = \frac{a}{5} + 5\frac{2}{5}\)

Step 1: Convert the mixed fraction \(5\frac{2}{5}\) to improper fraction:
\(5\frac{2}{5} = \frac{25 + 2}{5} = \frac{27}{5}\)
Equation becomes: \(3a – \frac{1}{5} = \frac{a}{5} + \frac{27}{5}\)
Step 2: Eliminate denominators by multiplying every term by 5:
5 × (3a) – 5 × (1/5) = 5 × (a/5) + 5 × (27/5)
15a – 1 = a + 27
Step 3: Bring variables to one side and constants to the other:
15a – a = 27 + 1 → 14a = 28
Step 4: Divide both sides by 14:
a = 28 / 14 → a = 2
Answer: a = 2

Q11: Solve \(\frac{x}{3} – 2\frac{1}{2} = \frac{4x}{9} – \frac{2x}{3}\)

Step 1: Convert the mixed fraction \(2\frac{1}{2}\) to improper fraction:
\(2\frac{1}{2} = \frac{5}{2}\)
Equation becomes: \(\frac{x}{3} – \frac{5}{2} = \frac{4x}{9} – \frac{2x}{3}\)
Step 2: Convert \(-\frac{2x}{3}\) to denominator 9:
\(-\frac{2x}{3} = -\frac{6x}{9}\)
Equation becomes: \(\frac{x}{3} – \frac{5}{2} = \frac{4x}{9} – \frac{6x}{9} → \frac{x}{3} – \frac{5}{2} = -\frac{2x}{9}\)
Step 3: Convert \(\frac{x}{3}\) to denominator 9:
\(\frac{x}{3} = \frac{3x}{9}\)
Equation becomes: \(\frac{3x}{9} – \frac{5}{2} = -\frac{2x}{9}\)
Step 4: Bring variables to one side:
\(\frac{3x}{9} + \frac{2x}{9} – \frac{5}{2} = 0 → \frac{5x}{9} – \frac{5}{2} = 0\)
Step 5: Add \(\frac{5}{2}\) to both sides:
\(\frac{5x}{9} = \frac{5}{2}\)
Step 6: Multiply both sides by 9/5:
x = \(\frac{5}{2} × \frac{9}{5} = \frac{9}{2} = 4.5\)
Answer: x = 4.5

Q12: Solve \(\frac{4(y+2)}{5} = 7 + \frac{5y}{13}\)

Step 1: Expand the bracket on the left-hand side:
\(\frac{4y + 8}{5} = 7 + \frac{5y}{13}\)
Step 2: Eliminate denominators by multiplying through by LCM of 5 and 13, which is 65:
65 × \(\frac{4y + 8}{5}\) = 65 × 7 + 65 × \(\frac{5y}{13}\)
13(4y + 8) = 455 + 25y
52y + 104 = 455 + 25y
Step 3: Bring variables to one side and constants to the other:
52y – 25y = 455 – 104 → 27y = 351
Step 4: Divide both sides by 27:
y = 351 / 27 → y = 13
Answer: y = 13

Q13: Solve \(\frac{a+5}{6} – \frac{a+1}{9} = \frac{a+3}{4}\)

Step 1: Find LCM of denominators 6, 9, 4, which is 36, and multiply through by 36 to eliminate denominators:
36 × \(\frac{a+5}{6}\) – 36 × \(\frac{a+1}{9}\) = 36 × \(\frac{a+3}{4}\)
Step 2: Simplify each term:
6(a + 5) – 4(a + 1) = 9(a + 3)
6a + 30 – 4a – 4 = 9a + 27
(6a – 4a) + (30 – 4) = 9a + 27 → 2a + 26 = 9a + 27
Step 3: Bring variables to one side and constants to the other:
2a – 9a = 27 – 26 → -7a = 1
Step 4: Divide both sides by -7:
a = \(-\frac{1}{7}\)
Answer: a = \(-\frac{1}{7}\)

Q14: Solve \(\frac{2x-13}{5} – \frac{x-3}{11} = \frac{x-9}{5} + 1\)

Step 1: Bring like terms together. Subtract \(\frac{x-9}{5}\) from both sides:
\(\frac{2x-13}{5} – \frac{x-9}{5} – \frac{x-3}{11} = 1\)
Step 2: Simplify the terms with denominator 5:
\(\frac{(2x – 13) – (x – 9)}{5} – \frac{x-3}{11} = 1\)
\(\frac{2x – 13 – x + 9}{5} – \frac{x-3}{11} = 1\)
\(\frac{x – 4}{5} – \frac{x-3}{11} = 1\)
Step 3: Eliminate denominators by multiplying both sides by LCM of 5 and 11, which is 55:
55 × \(\frac{x-4}{5}\) – 55 × \(\frac{x-3}{11}\) = 55 × 1
11(x – 4) – 5(x – 3) = 55
Step 4: Expand both brackets:
11x – 44 – 5x + 15 = 55
6x – 29 = 55
Step 5: Add 29 to both sides:
6x = 55 + 29 → 6x = 84
Step 6: Divide both sides by 6:
x = \(\frac{84}{6} = 14\)
Answer: x = 14

Q15: Solve \(6(6x – 5) – 5(7x – 8) = 12(4 – x) + 1\)

Step 1: Expand all brackets:
\(6 \cdot (6x – 5) – 5 \cdot (7x – 8) = 12 \cdot (4 – x) + 1\)
\(36x – 30 – 35x + 40 = 48 – 12x + 1\)
Step 2: Simplify both sides:
\((36x – 35x) + (-30 + 40) = -12x + 49\)
\(x + 10 = -12x + 49\)
Step 3: Bring variables to one side and constants to the other:
\(x + 12x = 49 – 10 → 13x = 39\)
Step 4: Divide both sides by 13:
\(x = \frac{39}{13} → x = 3\)
Answer: \(x = \frac{39}{13} = 3\)

Q16: Solve \((x-5)(x+3) = (x-7)(x+4)\)

Step 1: Expand both sides using distributive property:
LHS: \((x-5)(x+3) = x^2 + 3x – 5x – 15 = x^2 – 2x – 15\)
RHS: \((x-7)(x+4) = x^2 + 4x – 7x – 28 = x^2 – 3x – 28\)
Step 2: Set the equation:
\(x^2 – 2x – 15 = x^2 – 3x – 28\)
Step 3: Subtract \(x^2\) from both sides:
\(-2x – 15 = -3x – 28\)
Step 4: Bring variables to one side and constants to the other:
\(-2x + 3x = -28 + 15 → x = -13\)
Answer: x = -13

Q17: Solve \((x-5)^2 – (x+2)^2 = -2\)

Step 1: Apply the identity \(a^2 – b^2 = (a – b)(a + b)\):
\((x-5)^2 – (x+2)^2 = ((x-5) – (x+2))((x-5) + (x+2))\)
\((x-5) – (x+2) = x – 5 – x – 2 = -7\)
\((x-5) + (x+2) = x – 5 + x + 2 = 2x – 3\)
Equation becomes: \(-7(2x – 3) = -2\)
Step 2: Simplify:
\(-14x + 21 = -2\)
Step 3: Bring constants to the other side:
\(-14x = -2 – 21 → -14x = -23\)
Step 4: Divide both sides by -14:
\(x = \frac{-23}{-14} = \frac{23}{14}\)
Answer: \(x = \frac{23}{14}\)

Q18: Solve \((x-1)(x+6) – (x-2)(x-3) = 3\)

Step 1: Expand both brackets using distributive property:
\((x-1)(x+6) = x^2 + 6x – x – 6 = x^2 + 5x – 6\)
\((x-2)(x-3) = x^2 – 3x – 2x + 6 = x^2 – 5x + 6\)
Step 2: Substitute back into the equation:
\((x^2 + 5x – 6) – (x^2 – 5x + 6) = 3\)
\(x^2 + 5x – 6 – x^2 + 5x – 6 = 3\)
\(10x – 12 = 3\)
Step 3: Bring constants to the other side:
\(10x = 3 + 12 → 10x = 15\)
Step 4: Divide both sides by 10:
\(x = \frac{15}{10} = \frac{3}{2}\)
Answer: \(x = \frac{3}{2}\)

Q19: Solve \(\frac{3x}{x+6} – \frac{x}{x+5} = 2\)

Step 1: Find LCM of the denominators \((x+6)\) and \((x+5)\), which is \((x+6)(x+5)\), and combine the fractions:
\(\frac{3x(x+5) – x(x+6)}{(x+6)(x+5)} = 2\)
Step 2: Expand the numerators:
\(3x(x+5) = 3x^2 + 15x\)
\(x(x+6) = x^2 + 6x\)
So numerator: \(3x^2 + 15x – (x^2 + 6x) = 2x^2 + 9x\)
Step 3: Write the equation:
\(\frac{2x^2 + 9x}{(x+6)(x+5)} = 2\)
Step 4: Multiply both sides by \((x+6)(x+5)\):
\(2x^2 + 9x = 2(x+6)(x+5)\)
\(2x^2 + 9x = 2(x^2 + 11x + 30)\)
\(2x^2 + 9x = 2x^2 + 22x + 60\)
Step 5: Subtract \(2x^2\) from both sides:
\(9x = 22x + 60\)
Step 6: Bring variables to one side:
\(9x – 22x = 60 → -13x = 60\)
Step 7: Divide both sides by -13:
\(x = \frac{-60}{13}\)
Answer: \(x = \frac{-60}{13}\)

Q20: Solve \(\frac{1}{x-1} + \frac{2}{x-2} = \frac{3}{x-3}\)

Step 1: Find LCM of denominators \((x-1)(x-2)(x-3)\) and write each fraction with this denominator:
\(\frac{(x-2)(x-3) + 2(x-1)(x-3)}{(x-1)(x-2)(x-3)} = \frac{3(x-1)(x-2)}{(x-1)(x-2)(x-3)}\)
Step 2: Remove denominators (multiply both sides by \((x-1)(x-2)(x-3)\)):
\((x-2)(x-3) + 2(x-1)(x-3) = 3(x-1)(x-2)\)
Step 3: Expand all brackets:
\((x-2)(x-3) = x^2 – 5x + 6\)
\(2(x-1)(x-3) = 2(x^2 – 4x + 3) = 2x^2 – 8x + 6\)
LHS: \(x^2 – 5x + 6 + 2x^2 – 8x + 6 = 3x^2 – 13x + 12\)
RHS: \(3(x-1)(x-2) = 3(x^2 – 3x + 2) = 3x^2 – 9x + 6\)
Step 4: Write the equation:
\(3x^2 – 13x + 12 = 3x^2 – 9x + 6\)
Step 5: Subtract \(3x^2\) from both sides:
\(-13x + 12 = -9x + 6\)
Step 6: Bring variables to one side and constants to the other:
\(-13x + 9x = 6 – 12 → -4x = -6\)
Step 7: Divide both sides by -4:
\(x = \frac{-6}{-4} = \frac{3}{2}\)
Answer: \(x = \frac{3}{2}\)

Q21: Solve \(\frac{x-1}{7x-14} = \frac{x-3}{7x-26}\)

Step 1: Factor where possible:
\(7x – 14 = 7(x – 2)\)
\(7x – 26\) cannot be factored simply.
Equation becomes: \(\frac{x-1}{7(x-2)} = \frac{x-3}{7x – 26}\)
Step 2: Cross-multiply:
\((x-1)(7x – 26) = (x-3) \cdot 7(x-2)\)
Step 3: Expand both sides:
LHS: \((x-1)(7x – 26) = 7x^2 – 26x – 7x + 26 = 7x^2 – 33x + 26\)
RHS: \((x-3) \cdot 7(x-2) = 7(x-3)(x-2) = 7(x^2 – 5x + 6) = 7x^2 – 35x + 42\)
Step 4: Set equation:
\(7x^2 – 33x + 26 = 7x^2 – 35x + 42\)
Step 5: Subtract \(7x^2\) from both sides:
\(-33x + 26 = -35x + 42\)
Step 6: Bring variables to one side and constants to the other:
\(-33x + 35x = 42 – 26 → 2x = 16\)
Step 7: Divide both sides by 2:
\(x = \frac{16}{2} = 8\)
Answer: x = 8

Q22: Solve \(\frac{1}{x-1} – \frac{1}{x} = \frac{1}{x+3} – \frac{1}{x+4}\)

Step 1: Write each side with a single fraction:
\(\frac{1}{x-1} – \frac{1}{x} = \frac{x – (x-1)}{x(x-1)} = \frac{1}{x(x-1)}\)
\(\frac{1}{x+3} – \frac{1}{x+4} = \frac{x+4 – (x+3)}{(x+3)(x+4)} = \frac{1}{(x+3)(x+4)}\)
Step 2: Now the equation becomes:
\(\frac{1}{x(x-1)} = \frac{1}{(x+3)(x+4)}\)
Step 3: Cross multiply:
\((x)(x-1) = (x+3)(x+4)\)
\(x^2 – x = x^2 + 7x + 12\)
Step 4: Bring all terms to one side:
\(x^2 – x – x^2 – 7x – 12 = 0 → -8x – 12 = 0\)
Step 5: Solve for x:
\(-8x = 12 → x = -\frac{12}{8} = -\frac{3}{2}\)
Answer: \(x = -\frac{3}{2}\)

Q23: Solve: \(\frac{2x}{3}-\frac{x-1}{6}+\frac{7x-1}{4}=2\frac{1}{6}\). Hence, find the value of ‘a’. If \(\frac{1}{a}+5x=8\).

Step 1: Convert mixed fraction to improper fraction:
\(2\frac{1}{6} = \frac{13}{6}\)
Step 2: Find LCM of denominators 3, 6, 4 → LCM = 12. Rewrite all fractions with denominator 12:
\(\frac{2x}{3} = \frac{8x}{12}, \quad \frac{x-1}{6} = \frac{2(x-1)}{12} = \frac{2x-2}{12}, \quad \frac{7x-1}{4} = \frac{3(7x-1)}{12} = \frac{21x-3}{12}\)
Step 3: Combine the fractions on LHS:
\(\frac{8x}{12} – \frac{2x-2}{12} + \frac{21x-3}{12} = \frac{8x – 2x + 2 + 21x – 3}{12} = \frac{27x – 1}{12}\)
Step 4: Set equal to RHS:
\(\frac{27x – 1}{12} = \frac{13}{6}\)
Step 5: Multiply both sides by 12:
\(27x – 1 = 12 \cdot \frac{13}{6} = 26\)
Step 6: Solve for \(x\):
\(27x = 26 + 1 → 27x = 27 → x = 1\)
Step 7: Use second equation \(\frac{1}{a} + 5x = 8\):
\(\frac{1}{a} + 5(1) = 8 → \frac{1}{a} + 5 = 8 → \frac{1}{a} = 3 → a = \frac{1}{3}\)
Answer: \(x = 1, a = \frac{1}{3}\)

Q24: Solve: \(\frac{4-3x}{5}+\frac{7-x}{3}+4\frac{1}{3}=0\). Hence, find the value of ‘p’, if \(3p-2x+1=0\).

Step 1: Convert mixed fraction to improper fraction:
\(4\frac{1}{3} = \frac{13}{3}\)
Step 2: Find LCM of denominators 5 and 3 → LCM = 15. Rewrite fractions with denominator 15:
\(\frac{4-3x}{5} = \frac{3(4-3x)}{15} = \frac{12-9x}{15}\)
\(\frac{7-x}{3} = \frac{5(7-x)}{15} = \frac{35-5x}{15}\)
Convert \(\frac{13}{3} = \frac{65}{15}\)
Step 3: Combine all fractions:
\(\frac{12-9x + 35-5x + 65}{15} = \frac{112 – 14x}{15}\)
Step 4: Set equal to 0:
\(\frac{112 – 14x}{15} = 0 → 112 – 14x = 0\)
Step 5: Solve for x:
\(-14x = -112 → x = 8\)
Step 6: Use second equation \(3p – 2x + 1 = 0\):
\(3p – 2(8) + 1 = 0 → 3p -16 + 1 = 0 → 3p – 15 = 0\)
\(3p = 15 → p = 5\)
Answer: x = 8, p = 5

Q25: Solve \(0.25 + \frac{1.95}{x} = 0.9\)

Step 1: Subtract 0.25 from both sides:
\(\frac{1.95}{x} = 0.9 – 0.25 = 0.65\)
Step 2: Solve for \(x\) by cross multiplication:
\(1.95 = 0.65x → x = \frac{1.95}{0.65}\)
Step 3: Divide 1.95 by 0.65:
\(\frac{1.95}{0.65} = 3\)
Answer: x = 3

Q26: Solve \(5x – \left(4x + \frac{5x-4}{7}\right) = \frac{4x-14}{3}\)

Step 1: Simplify LHS:
\(5x – \left(4x + \frac{5x-4}{7}\right) = 5x – 4x – \frac{5x-4}{7} = x – \frac{5x-4}{7}\)
Step 2: Write \(x\) as \(\frac{7x}{7}\) to combine fractions:
\(x – \frac{5x-4}{7} = \frac{7x – (5x-4)}{7} = \frac{7x – 5x + 4}{7} = \frac{2x + 4}{7}\)
Step 3: Now equation becomes:
\(\frac{2x + 4}{7} = \frac{4x – 14}{3}\)
Step 4: Cross multiply:
\(3(2x + 4) = 7(4x – 14)\)
\(6x + 12 = 28x – 98\)
Step 5: Bring variables to one side and constants to the other:
\(6x – 28x = -98 – 12 → -22x = -110\)
Step 6: Divide both sides by -22:
\(x = \frac{-110}{-22} = 5\)
Answer: x = 5