Q1: Solve \(4x – 9 = 2x + 7\)Step 1: Write the given equation:
\[
4x – 9 = 2x + 7
\]Step 2: Transpose \(2x\) to the left-hand side and \(-9\) to the right-hand side:
\[
4x – 2x = 7 + 9
\]Step 3: Simplify both sides:
\[
2x = 16
\]Step 4: Divide both sides by 2:
\[
x = \frac{16}{2} = 8
\]Answer:x = 8

Q2: Solve \(5y + 18 = 11 – 2y\)Step 1: Write the given equation:
\[
5y + 18 = 11 – 2y
\]Step 2: Transpose \(-2y\) to the left-hand side and 18 to the right-hand side:
\[
5y + 2y = 11 – 18
\]Step 3: Simplify both sides:
\[
7y = -7
\]Step 4: Divide both sides by 7:
\[
y = \frac{-7}{7} = -1
\]Answer:y = -1

Q3: Solve \(21 – 3(x – 7) = x + 20\)Step 1: Write the given equation:
\[
21 – 3(x – 7) = x + 20
\]Step 2: Expand the bracket:
\[
21 – 3x + 21 = x + 20 \\
42 – 3x = x + 20
\]Step 3: Transpose \(x\) to the left-hand side and 42 to the right-hand side:
\[
-3x – x = 20 – 42 \\
-4x = -22
\]Step 4: Divide both sides by -4:
\[
x = \frac{-22}{-4} = \frac{22}{4} = \frac{11}{2}
\]Answer:x = \(\frac{11}{2}\)

Q4: Solve \(3(y-7)-2(3y-4) = (2-5y)\)Step 1: Write the given equation:
\[
3(y – 7) – 2(3y – 4) = 2 – 5y
\]Step 2: Expand the brackets:
\[
3y – 21 – 6y + 8 = 2 – 5y \\
-3y – 13 = 2 – 5y
\]Step 3: Transpose \(-5y\) to the left-hand side and \(-13\) to the right-hand side:
\[
-3y + 5y = 2 + 13 \\
2y = 15
\]Step 4: Divide both sides by 2:
\[
y = \frac{15}{2}
\]Answer:y = \(\frac{15}{2}\)

Q5: Solve \(3(t-5)-16t = 12 – 2(t-3)\)Step 1: Write the given equation:
\[
3(t – 5) – 16t = 12 – 2(t – 3)
\]Step 2: Expand the brackets:
\[
3t – 15 – 16t = 12 – 2t + 6 \\
-13t – 15 = 18 – 2t
\]Step 3: Transpose \(-2t\) to the left-hand side and \(-15\) to the right-hand side:
\[
-13t + 2t = 18 + 15 \\
-11t = 33
\]Step 4: Divide both sides by -11:
\[
t = \frac{33}{-11} = -3
\]Answer:t = -3

Q6: Solve \(\frac{3x}{4} – \frac{(x-4)}{3} = \frac{5}{3}\)Step 1: Write the given equation:
\[
\frac{3x}{4} – \frac{(x-4)}{3} = \frac{5}{3}
\]Step 2: Take LCM of denominators 4 and 3 for the left-hand side: LCM = 12
\[
\frac{9x}{12} – \frac{4(x-4)}{12} = \frac{5}{3} \\
\frac{9x – 4(x-4)}{12} = \frac{5}{3}
\]Step 3: Simplify numerator:
\[
9x – 4x + 16 = 5x + 16 \\
\frac{5x + 16}{12} = \frac{5}{3}
\]Step 4: Cross multiply:
\[
3(5x + 16) = 12 \cdot 5 \\
15x + 48 = 60
\]Step 5: Simplify:
\[
15x = 60 – 48 \\
15x = 12 \\
x = \frac{12}{15} = \frac{4}{5}
\]Answer:x = \(\frac{4}{5}\)

Q7: Solve \(\frac{4x+1}{3} + \frac{2x-1}{2} – \frac{3x-7}{5} = 6\)Step 1: Write the given equation:
\[
\frac{4x+1}{3} + \frac{2x-1}{2} – \frac{3x-7}{5} = 6
\]Step 2: Take LCM of denominators 3, 2, 5 for the left-hand side: LCM = 30
\[
\frac{10(4x+1)}{30} + \frac{15(2x-1)}{30} – \frac{6(3x-7)}{30} = 6 \\
\frac{10(4x+1) + 15(2x-1) – 6(3x-7)}{30} = 6
\]Step 3: Expand numerators:
\[
\frac{40x + 10 + 30x – 15 – 18x + 42}{30} = 6 \\
\frac{(40x + 30x – 18x) + (10 – 15 + 42)}{30} = 6 \\
\frac{52x + 37}{30} = 6
\]Step 4: Multiply both sides by 30:
\[
52x + 37 = 180
\]Step 5: Simplify:
\[
52x = 180 – 37 \\
52x = 143 \\
x = \frac{143}{52} = \frac{143 \div 13}{52 \div 13} = \frac {11}{4}
\]Answer:x = \(\frac{11}{4}\)

Q8: Solve \(\frac{z+5}{6} – \frac{z+1}{9} = \frac{z+3}{4}\)Step 1: Write the given equation:
\[
\frac{z+5}{6} – \frac{z+1}{9} = \frac{z+3}{4}
\]Step 2: Take LCM of denominators 6, 9, 4 for the left-hand side: LCM = 36
\[
\frac{6(z+5) – 4(z+1)}{36} = \frac{z+3}{4}
\]Step 3: Simplify numerator:
\[
\frac{6z + 30 – 4z – 4}{36} = \frac{2z + 26}{36} = \frac{z+13}{18} \\
\frac{z+13}{18} = \frac{z+3}{4}
\]Step 4: Cross multiply:
\[
4(z+13) = 18(z+3) \\
4z + 52 = 18z + 54
\]Step 5: Simplify:
\[
4z – 18z = 54 – 52 \\
-14z = 2 \\
z = -\frac{2}{14} = -\frac{1}{7}
\]Answer:z = -\(\frac{1}{7}\)

Q9: Solve \(\frac{2-9z}{17-4z} = \frac{4}{5}\)Step 1: Write the given equation:
\[
\frac{2-9z}{17-4z} = \frac{4}{5}
\]Step 2: Cross multiply:
\[
5(2-9z) = 4(17-4z)
\]Step 3: Expand both sides:
\[
10 – 45z = 68 – 16z
\]Step 4: Transpose terms:
\[
-45z + 16z = 68 – 10 \\
-29z = 58
\]Step 5: Divide both sides by -29:
\[
z = \frac{58}{-29} = -2
\]Answer:z = -2

Q10: Solve \(\frac{2x-3}{3x-1} = \frac{2x+3}{3x+4}\)Step 1: Write the given equation:
\[
\frac{2x-3}{3x-1} = \frac{2x+3}{3x+4}
\]Step 2: Cross multiply:
\[
(2x-3)(3x+4) = (2x+3)(3x-1)
\]Step 3: Expand both sides:
\[
6x^2 + 8x – 9x – 12 = 6x^2 – 2x + 9x – 3 \\
6x^2 – x – 12 = 6x^2 + 7x – 3
\]Step 4: Transpose terms to bring all variables to one side:
\[
6x^2 – x – 12 – 6x^2 – 7x + 3 = 0 \\
-8x – 9 = 0
\]Step 5: Solve for x:
\[
-8x = 9 \\
x = -\frac{9}{8}
\]Answer:x = -\(\frac{9}{8}\)

Q11: Solve \(\frac{2-9x}{16+5x}=0\)Step 1: Write the given equation:
\[
\frac{2-9x}{16+5x} = 0
\]Step 2: A fraction is zero when its numerator is zero:
\[
2 – 9x = 0
\]Step 3: Solve for x:
\[
-9x = -2 \\
x = \frac{2}{9}
\]Answer:x = \(\frac{2}{9}\)

Q12: Solve \(\frac{0.5z+4}{1.2z+6} = \frac{5}{3}\)Step 1: Write the given equation:
\[
\frac{0.5z + 4}{1.2z + 6} = \frac{5}{3}
\]Step 2: Cross multiply:
\[
3(0.5z + 4) = 5(1.2z + 6)
\]Step 3: Expand both sides:
\[
1.5z + 12 = 6z + 30
\]Step 4: Transpose terms:
\[
1.5z – 6z = 30 – 12 \\
-4.5z = 18
\]Step 5: Divide both sides by -4.5:
\[
z = \frac{18}{-4.5} = -4
\]Answer:z = -4

Q13: Solve \(\frac{3}{2x-1} + \frac{4}{2x+1} = \frac{7}{2x}\)Step 1: Write the given equation:
\[
\frac{3}{2x-1} + \frac{4}{2x+1} = \frac{7}{2x}
\]Step 2: Take LCM of the denominators \((2x-1)(2x+1)(2x)\) to eliminate fractions:
\[
3 \cdot 2x \cdot (2x+1) + 4 \cdot 2x \cdot (2x-1) = 7 \cdot (2x-1)(2x+1)
\]Step 3: Expand each term:
\[
3(2x)(2x+1) = 12x^2 + 6x \\
4(2x)(2x-1) = 16x^2 – 8x \\
7(2x-1)(2x+1) = 7(4x^2 – 1) = 28x^2 – 7
\]Step 4: Combine like terms on the left-hand side:
\[
(12x^2 + 6x) + (16x^2 – 8x) = 28x^2 – 2x \\
28x^2 – 2x = 28x^2 – 7
\]Step 5: Transpose terms:
\[
-2x = -7 \\
x = \frac{7}{2}
\]Answer:x = \(\frac{7}{2}\)

Q14: Solve \(\frac{3}{x-2} – \frac{2}{x-3} = \frac{4}{x-3} – \frac{3}{x-1}\)Step 1: Write the given equation:
\[
\frac{3}{x-2} – \frac{2}{x-3} = \frac{4}{x-3} – \frac{3}{x-1}
\]Step 2: Bring like terms together:
\[
\frac{3}{x-2} + \frac{3}{x-1} = \frac{4}{x-3} + \frac{2}{x-3}
\]Step 3: Simplify both sides:
\[
\frac{3}{x-2} + \frac{3}{x-1} = \frac{6}{x-3}
\]Step 4: Take LCM of the left-hand side \((x-2)(x-1)\):
\[
\frac{3(x-1) + 3(x-2)}{(x-2)(x-1)} = \frac{6}{x-3}
\]Step 5: Simplify the numerator:
\[
3(x-1) + 3(x-2) = 3x -3 + 3x -6 = 6x -9 \\
\frac{6x – 9}{(x-2)(x-1)} = \frac{6}{x-3}
\]Step 6: Cross multiply:
\[
(6x – 9)(x-3) = 6(x-2)(x-1)
\]Step 7: Expand both sides:

LHS: \(6x(x-3) – 9(x-3) = 6x^2 -18x -9x +27 = 6x^2 -27x +27\)

RHS: \(6(x^2 -3x +2) = 6x^2 -18x +12\)
Step 8: Subtract RHS from LHS:
\[
6x^2 -27x +27 – (6x^2 -18x +12) = 0 \\
-9x +15 = 0
\]Step 9: Solve for x:
\[
-9x = -15 \\
x = \frac{15}{9} = \frac{5}{3}
\]Answer:x = \(\frac{5}{3}\)

Q15: Solve \((z+3)(z-3) – z(z+5) = 6\)Step 1: Write the given equation:
\[
(z+3)(z-3) – z(z+5) = 6
\]Step 2: Expand both terms using distributive property:
\[
(z+3)(z-3) = z^2 – 9 \\
z(z+5) = z^2 + 5z
\]Step 3: Substitute back into the equation:
\[
(z^2 – 9) – (z^2 + 5z) = 6
\]Step 4: Simplify the left-hand side:
\[
z^2 – 9 – z^2 – 5z = 6 \\
-5z – 9 = 6
\]Step 5: Solve for z:
\[
-5z = 6 + 9 \\
-5z = 15 \\
z = -3
\]Answer:z = -3

Q16: Solve \(y(2y+3) – 2y(y-5) = 26\)Step 1: Write the given equation:
\[
y(2y+3) – 2y(y-5) = 26
\]Step 2: Expand the brackets:
\[
2y^2 + 3y – 2y^2 + 10y = 26
\]Step 3: Combine like terms:
\[
(2y^2 – 2y^2) + (3y + 10y) = 26 \\
13y = 26
\]Step 4: Solve for y:
\[
y = \frac{26}{13} = 2
\]Answer:y = 2

Q17: Solve \(\frac{x+6}{4} – \frac{5x-4}{8} + \frac{x-3}{5} = 0\)Step 1: Write the given equation:
\[
\frac{x+6}{4} – \frac{5x-4}{8} + \frac{x-3}{5} = 0
\]Step 2: Take LCM of 4 and 8 for the first two terms: LCM = 8
\[
\frac{2(x+6) – (5x-4)}{8} + \frac{x-3}{5} = 0
\]Step 3: Simplify the numerator:
\[
2x + 12 – 5x + 4 = -3x + 16 \\
\frac{-3x + 16}{8} + \frac{x-3}{5} = 0
\]Step 4: Take LCM of 8 and 5 = 40
\[
\frac{5(-3x + 16) + 8(x-3)}{40} = 0
\]Step 5: Simplify the numerator:
\[
-15x + 80 + 8x – 24 = -7x + 56 \\
\frac{-7x + 56}{40} = 0
\]Step 6: Set the numerator = 0:
\[
-7x + 56 = 0 \\
-7x = -56 \\
x = 8
\]Answer:x = 8

Q18: Solve \(\frac{3}{4}(7x-1) – \left(2x – \frac{1-x}{2}\right) = x + \frac{3}{2}\)Step 1: Expand the first term:
\[
\frac{3}{4} \cdot 7x – \frac{3}{4} \cdot 1 = \frac{21x}{4} – \frac{3}{4}
\]Step 2: Simplify the second term:
\[
2x – \frac{1-x}{2} = 2x – \frac{1}{2} + \frac{x}{2} = 2x + \frac{x}{2} – \frac{1}{2} = \frac{5x}{2} – \frac{1}{2}
\]Step 3: Substitute back into the equation:
\[
\frac{21x}{4} – \frac{3}{4} – \left(\frac{5x}{2} – \frac{1}{2}\right) = x + \frac{3}{2}
\]Step 4: Distribute the negative sign:
\[
\frac{21x}{4} – \frac{3}{4} – \frac{5x}{2} + \frac{1}{2} = x + \frac{3}{2}
\]Step 5: Convert \(\frac{5x}{2}\) to \(\frac{10x}{4}\) and \(\frac{1}{2}\) to \(\frac{2}{4}\) for LCM:
\[
\frac{21x}{4} – \frac{10x}{4} – \frac{3}{4} + \frac{2}{4} = x + \frac{3}{2} \\
\frac{11x}{4} – \frac{1}{4} = x + \frac{3}{2}
\]Step 6: Multiply both sides by 4 to eliminate denominators:
\[
11x – 1 = 4x + 6
\]Step 7: Solve for x:
\[
11x – 4x – 1 = 6 \\
7x – 1 = 6 \\
7x = 7 \\
x = 1
\]Answer:x = 1