Algebra is the branch of mathematics that uses letters and symbols to represent numbers and relationships between them.

Kiswahili: Algebra ni sehemu ya hesabu inayotumia herufi na alama kuwakilisha namba na uhusiano kati ya namba hizo.

Why is it important? Algebra allows us to solve problems without knowing the exact numbers. It is like a language that explains patterns, relationships, and real-life situations.

Before algebra, humans used arithmetic and geometry to solve problems:

• People used numbers to count, add, subtract, multiply, divide.
• They also measured areas, volumes, and lengths to solve practical problems like farming, building, and trade.

Kiswahili: Kabla ya algebra, watu walitumia hesabu rahisi na jiometri kuhesabu na kupima. Hii ilikuwa muhimu kwa kilimo, biashara na ujenzi.

Limitation: Without algebra, you could not solve problems where numbers were unknown or changing.

The word "Algebra" comes from the Arabic word “al-jabr”, meaning "reunion of broken parts". It was introduced by Al-Khwarizmi around 820 AD in Baghdad.

• Al-Khwarizmi wrote the book “Kitab al-Mukhtasar fi Hisab al-Jabr wal-Muqabala” (The Compendious Book on Calculation by Completion and Balancing).
• He created rules for solving equations systematically for the first time.

Kiswahili: Al-Khwarizmi alianzisha algebra kuweza kutatua matatizo bila namba halisi. Alipa njia ya kufuata hatua kwa hatua (systematic).

• Diophantus (Ancient Greece) – Worked on solving equations with symbols. Sometimes called “the father of algebra”.
• Bhaskara (India, 12th century) – Solved quadratic equations and developed methods to find unknowns in practical problems.
• European Mathematicians (16th-17th century) – Introduced modern notation (like x, y, etc.) and symbols (+, −, =) for easier writing.

Kiswahili: Wataalamu wa hesabu walichangia algebra kwa kutatua equations za aina tofauti na kuunda alama rahisi za kufuatilia hesabu.

Before algebra, problems like “I have a number, add 5, then double it, and the result is 30. What is the number?” were very hard to solve systematically.

Algebra allows you to write x + 5 for the unknown number and solve: 2*(x + 5) = 30 → x + 5 = 15 → x = 10.

Kiswahili: Algebra inatupa njia rahisi ya kutatua matatizo ambayo ni magumu kwa arithmetic ya kawaida. Hii ni muhimu katika maisha yetu ya kila siku, biashara, uhandisi, sayansi, na teknolojia.

• Algebra allowed creation of formulas: like distance = speed × time.
• It led to calculus, statistics, and all modern mathematics.
• Today, algebra is essential in computers, engineering, economics, and science.

Kiswahili: Baada ya algebra, tunaweza kuunda formulas, kutatua problems za kweli, na kuendeleza sayansi na teknolojia. Bila algebra, dunia ya kisayansi ingekuwa ngumu kuelewa.

Algebra is not just about numbers; it is about thinking logically and solving real-life problems:

• Calculating finances: budgeting, loans, investments.
• Engineering and architecture: designing buildings, bridges.
• Science experiments: predicting results using formulas.
• Computer programming: using variables and conditions to write programs.

Kiswahili: Algebra inatufundisha kufikiri kimantiki, kutatua matatizo halisi, na ni msingi wa sayansi, uhandisi na teknolojia.

Imagine a world without algebra:

• No formulas to calculate areas or volumes.
• Engineers and architects could not design efficiently.
• Computers would not understand variables.

Learning algebra is like learning a new language that opens doors to understanding the universe.

Kiswahili: Kujifunza algebra ni kama kujifunza lugha mpya ya kuelewa dunia na kutatua matatizo kwa urahisi.

• Algebra began in ancient times, formalized by Al-Khwarizmi.
• It solved the limitations of arithmetic for unknown numbers.
• Other mathematicians improved notation and methods.
• Algebra is essential in real life: science, technology, economy, and daily life.
• Even beginners can understand algebra step by step and see patterns.

Kiswahili: Algebra ni muhimu sana, inatufundisha kufikiri kimantiki, na kutatua matatizo katika maisha halisi. Hii ndio sababu lazima tuipende na tuisome kwa makini.

Definition: Algebra is a branch of mathematics where letters (called variables) represent numbers.

Kiswahili: Algebra ni sehemu ya hesabu ambapo tunatumia herufi (kama x, y, a, b) kuwakilisha namba. Hii inatusaidia kufanya mahesabu bila kutumia namba halisi.

Example / Mfano:

• x + 3 = 7 → solve for x
• x = 7 – 3 = 4

Kiswahili: Hapa tunapata x kwa kuondoa 3 kutoka 7.

Definition: An algebraic expression is a combination of letters and numbers with operations (like +, -, *, /).

Kiswahili: Expression ya algebra ni mchanganyiko wa namba na herufi ukiwa na alama za kujumlisha, kutoa, kuzidisha, au kugawanya.

Example: 3x + 2y – 5

• 3x → 3 multiplied by x
• 2y → 2 multiplied by y
• -5 → constant number (namba isiyobadilika)

Combine like terms: terms with the same letter and exponent.

Kiswahili: Unganisha vipengele vinavyofanana: vile vilivyo na herufi moja na exponent moja.

Example:

• 2a + 3a – 5b = (2+3)a – 5b = 5a – 5b
• x – (x – y) + (2x – 7) → x – x + y + 2x – 7 → 2x + y – 7

Use distributive property: (a + b)(c + d) = a*c + a*d + b*c + b*d

Kiswahili: Tumia distributive property: zingatia kila sehemu iendane na nyingine.

Examples:

• (a – 4)(a + 6) → a² + 6a – 4a – 24 → a² + 2a – 24
• (2x – 3)(3x – 3) → 6x² – 6x – 9x + 9 → 6x² – 15x + 9

Factor out common terms: a*b + a*c = a(b + c)

Kiswahili: Toa kipengele cha kawaida ili kupunguza expression.

Examples:

• 2ax + 3ay – 8bx – 12by → (2x + 3y)(a – 4b)
• r – 1 – r² + r → - (r – 1)²

Examples:

• (a + b)² = a² + 2ab + b²
• (a – b)² = a² – 2ab + b²
• (a + b)(a – b) = a² – b²

The coefficient is the number in front of the variable.

Kiswahili: Coefficient ni namba iliyo mbele ya herufi.

Example:

• 5x² → coefficient of x² = 5
• -3xy → coefficient of xy = -3

The constant term is a number without a variable.

Kiswahili: Constant ni namba isiyo na herufi.

Example: In 3x + 7 → constant = 7

Formula: Area = 1/2 * (sum of parallel sides) * height

Kiswahili: Eneo = 1/2 * jumla ya misingi ya pembeni * urefu wima

Example: Parallel sides = x+4, x+2; Height = 2x-3 → Area = 1/2*(2x+6)*(2x-3) → Expand → 2x²+3x-9

Perimeter = total length → 4 * side = y → side = y/4 → Area = side² = y²/16

Kiswahili: Tunapata side kwa kugawanya perimeter kwa 4, kisha area = side²

Always follow these steps:

1. Combine like terms
2. Expand brackets carefully
3. Factorize by taking common factors
4. Identify coefficients and constants
5. Check your work with numbers if possible

Kiswahili: Kila wakati: Unganisha vipengele vinavyofanana, panua bracket, toa vipengele vya kawaida, tambua coefficient na constant, hakikisha matokeo sahihi.

• Think of letters as boxes holding numbers.
• Always combine numbers first if letters are same.
• Use color-coding if writing on paper to identify like terms.
• Check your answer by substituting numbers for variables.

Kiswahili: Fikiria herufi kama sanduku lenye namba, ungana namba tu ikiwa herufi sawa, tumia rangi kwenye karatasi, hakikisha kwa kuweka namba badala ya herufi.

Solution:

Step 1: Sum first two expressions: (2a – 3b – 4c) + (4b – 3a – 2c)

Combine like terms:

• 2a – 3a = -a
• -3b + 4b = b
• -4c – 2c = -6c

Result: -a + b – 6c

Step 2: Sum next two expressions: (3c – 4b – 5a) + (–a + 2b – 5c)

• -5a – a = -6a
• -4b + 2b = -2b
• 3c – 5c = -2c

Result: -6a – 2b – 2c

Step 3: Sum last two expressions: (2a – 3b – 4c) + (4b – 3a – 2c)

• 2a – 3a = -a
• -3b + 4b = b
• -4c – 2c = -6c

Result: -a + b – 6c

Step 4: Add all three results together:

• (-a + b – 6c) + (-6a – 2b – 2c) + (-a + b – 6c)

• -a – 6a – a = -8a
• b – 2b + b = 0
• -6c – 2c – 6c = -14c

Final Answer: -8a – 14c

Maelezo ya Kiswahili: Tumetoa hesabu kwa hatua kwa hatua kwa kuunganisha vipengele vinavyofanana (like terms), mfano 'a', 'b', na 'c'.

Solution:

a) Algebraic expression for second class:

Total passengers = 3x² – 13x + 4

Passengers in first class = x – y + z

Passengers in third class = 2x – 3y – z

Passengers in second class = Total – (first + third)

• Second = (3x² – 13x + 4) – [(x – y + z) + (2x – 3y – z)]
• Combine terms inside brackets: (x + 2x) + (-y – 3y) + (z – z) = 3x – 4y + 0 = 3x – 4y
• Second = 3x² – 13x + 4 – (3x – 4y)
• Second = 3x² – 13x + 4 – 3x + 4y
• Second = 3x² – 16x + 4 + 4y

Answer: 3x² – 16x + 4 + 4y

b) Numerical calculation: x = 43, y = 38, z = 6

• First class = x – y + z = 43 – 38 + 6 = 11
• Third class = 2x – 3y – z = 86 – 114 – 6 = -34 (impossible, interpret as 0 passengers, example halisi: train inafaa kuwa na negative value inaonyesha data isiyo halisi)
• Second class = 3x² – 16x + 4 + 4y = 3(43²) – 16(43) + 4 + 4(38)
• 43² = 1849 → 3*1849 = 5547
• -16*43 = -688
• 4 + 4*38 = 4 + 152 = 156
• Total second class = 5547 – 688 + 156 = 5015

Answer: Second class = 5015 passengers

Maelezo ya Kiswahili: Tumetumia hatua ya kujua idadi ya abiria wa kila daraja kwa kuondoa wale walioko first na third class kutoka total. Hapa environment ya mifano: treni kubwa yenye daraja tatu.

Solution:

a) a(a – b) – b(b + a)

• Expand a(a – b) = a² – ab
• Expand -b(b + a) = -b² – ab
• Combine: a² – ab – b² – ab = a² – 2ab – b²

Answer: a² – 2ab – b²

b) x – (x – y) + (2x – 7)

• Remove brackets: x – x + y + 2x – 7
• Combine like terms: x – x + 2x = 2x; y remains; -7 remains

Answer: 2x + y – 7

Solution:

a) 2ax + 3ay – 8bx – 12by

• Group terms: (2ax + 3ay) – (8bx + 12by)
• Factor each group: a(2x + 3y) – 4b(2x + 3y)
• Factor out common: (2x + 3y)(a – 4b)

Answer: (2x + 3y)(a – 4b)

b) 3ac + 2ba + ad + 6bc

• Group: (3ac + 6bc) + (2ba + ad)
• Factor each: 3c(a + 2b) + a(2b + d)
• Observe common pattern: rearrange 2ba + ad = a(2b + d)
• Combine: (a + 3c)(?) – actually check carefully: 3c(a + 2b) + a(d + 2b) = a(d + 2b) + 3c(a + 2b) = (a + 3c)(?)
• Final factorization: (a + 3c)(d + 2b)

Answer: (a + 3c)(d + 2b)

c) 6pr + 6qs – 9ps – 4qr

• Group: (6pr – 9ps) + (6qs – 4qr)
• Factor each: 3p(2r – 3s) + 2q(3s – 2r)
• Notice second term negative: 2q(-2r + 3s) = 2q(3s – 2r)
• Factor common: (2r – 3s)(3p – 2q)

Answer: (2r – 3s)(3p – 2q)

d) 8x² + 6xy – 4xy – 3y²

• Combine like terms: 8x² + (6xy – 4xy) – 3y² = 8x² + 2xy – 3y²
• Factor: (4x – 3y)(2x + y)

Answer: (4x – 3y)(2x + y)

e) r – 1 – r² + r

• Combine like terms: r + r – 1 – r² = 2r – 1 – r²
• Rearrange: -r² + 2r – 1 = -(r² – 2r + 1) = -(r – 1)²

Answer: -(r – 1)²

Solution:

a) (a – 4)(a + 6)

• Multiply each: a² + 6a – 4a – 24
• Combine like terms: a² + 2a – 24

Answer: a² + 2a – 24

b) (2x – 3)(3x – 3)

• 2x*3x = 6x², 2x*-3 = -6x, -3*3x = -9x, -3*-3 = 9
• Combine like terms: 6x² – 15x + 9

Answer: 6x² – 15x + 9

c) (x + 2y)(2p + q)

• x*2p = 2px, x*q = xq, 2y*2p = 4py, 2y*q = 2yq
• Combine: 2px + xq + 4py + 2yq

Answer: 2px + xq + 4py + 2yq

d) (2a + 3b)(3a + 2b)

• 2a*3a = 6a², 2a*2b = 4ab, 3b*3a = 9ab, 3b*2b = 6b²
• Combine like terms: 6a² + 13ab + 6b²

Answer: 6a² + 13ab + 6b²

e) (3r – 3s)(r – 4s)

• 3r*r = 3r², 3r*-4s = -12rs, -3s*r = -3rs, -3s*-4s = 12s²
• Combine: 3r² – 15rs + 12s²

Answer: 3r² – 15rs + 12s²

Solution:

• Expand: 4x*x = 4x², 4x*2 = 8x, -3*x = -3x, -3*2 = -6
• Combine like terms: x terms → 8x – 3x = 5x

Coefficient of x: 5

Solution:

• Expand: x*6x = 6x², x*-2 = -2x, -3*6x = -18x, -3*-2 = 6
• Combine: constants = 6

Answer: 6

Solution:

• Expand: 4y*2y = 8y², 4y*3 = 12y, -6*2y = -12y, -6*3 = -18
• Coefficient of y² = 8

Answer: 8

Solution:

• Formula for trapezium area: A = 1/2 * (sum of parallel sides) * height
• Sum of parallel sides = (x + 4) + (x + 2) = 2x + 6
• Height = 2x – 3
• Area = 1/2 * (2x + 6)(2x – 3) = (x + 3)(2x – 3)
• Expand: x*2x = 2x², x*-3 = -3x, 3*2x = 6x, 3*-3 = -9
• Combine like terms: 2x² + 3x – 9

Answer: 2x² + 3x – 9

Solution:

• Expand first: 4a*2a = 8a², 4a*8 = 32a, -6*2a = -12a, -6*8 = -48 → sum = 8a² + 20a – 48
• Expand second: 2a*4a = 8a², 2a*-3 = -6a, 5*4a = 20a, 5*-3 = -15 → sum = 8a² + 14a – 15
• Subtract second from first: (8a² + 20a – 48) – (8a² + 14a – 15) = 6a – 33

Answer: 6a – 33

Solution:

• Perimeter of square = 4 * side = y
• Side = y/4
• Area = side² = (y/4)² = y² / 16

Answer: y² / 16

Kiswahili: Tumetumia formula ya mraba; side = perimeter/4, area = side²

Solution:

• Combine like terms: a² + 6a + a = a² + 7a
• –6x(2 – a) = –12x + 6ax
• Bring +8: final = a² + 7a + 8 –12x + 6ax

Answer: a² + 7a + 8 – 12x + 6ax

Kiswahili: Tukichanganya vipengele vya aina moja na ku-distribute multiplication ya -6x kwenye bracket.

13. Determine whether or not each of the following is an identity:

(a) 1 - b / 1 + b = (b - 1) / (b + 1)

Solution: Check by simplifying left side (LS) and right side (RS):
LS: 1 - b / 1 + b = (1 - b) / (1 + b)
RS: (b - 1) / (b + 1) = -(1 - b)/(b + 1) = -(1 - b)/(1 + b)
LS ≠ RS because LS = (1-b)/(1+b) but RS = -(1-b)/(1+b). Hence, not an identity.
Example: If b = 2, LS = (1-2)/(1+2) = -1/3, RS = (2-1)/(2+1) = 1/3 → not equal.

(b) x*(y + 2) + y*(x + 2) = 2*(x*y + x + y)

Solution: Expand both sides:
LS: x*y + 2x + y*x + 2y = 2xy + 2x + 2y
RS: 2*(xy + x + y) = 2xy + 2x + 2y
LS = RS, so it is an identity.
Example: x=1, y=3 → LS=2*3+2+6? Check: LS=1*(3+2)+3*(1+2)=5+9=14, RS=2*(3+1+3)=2*7=14

(c) (a*d + c*x)/(a*d) = c*x

Solution: Divide term by a*d:
LS: (a*d + c*x)/(a*d) = 1 + (c*x)/(a*d)
RS: c*x
Not equal unless special values (a*d=1). Not an identity.

(d) (a + b)2 = a2 + b2

Solution: Expand LS:
(a+b)² = a² + 2ab + b²
RS: a² + b²
LS ≠ RS, so not an identity.
Example: a=2, b=3 → LS=25, RS=13 → not equal.

(e) a*(a + 1) = a*(1 - a)

Solution: LS: a² + a, RS: a - a²
LS ≠ RS, so not an identity.

(f) 2*(x - 1) - 4*(2 - x) = 2*(3x - 5)

Solution: LS: 2x-2 - 8 + 4x = 6x-10
RS: 2*(3x-5)=6x-10
LS=RS, so it is an identity.

(g) x*(y + z) + x = x*y + x + x*z

Solution: LS: xy + xz + x, RS: xy + x + xz
Both sides = xy + xz + x, so it is an identity.

14. Give the missing term so that each expression becomes an identity:

(a) 6x − 7y − 8a + 9b = (6x − 7y) − (8a − 9b)

Explanation: When we expand the RHS: (6x − 7y) − (8a − 9b) = 6x − 7y − 8a + 9b

(b) 2p − 3q − 4r − 5s = (2p − 3q) − (4r + 5s)

Check: 2p−3q−(4r+5s)=2p−3q−4r−5s

(c) 6x + 6 − 9x − 4 = (x + 1) − (3x − 2)

Check: (x+1)-(3x-2)=x+1-3x+2=-2x+3 Incorrect hmm need adjust:

Correct missing: (6x+6)-(9x+4)= -3x+2 Correct

(d) 6ab + 2a − 9bc + 3c = 2a*(3b + 1) − (9bc − 3c)

Explanation: 2a*(3b+1) = 6ab + 2a, then subtract 9bc-3c = 6ab+2a-9bc+3c

(e) (3a−4b)*(2c−3d) = (6ac−9ad) − (8bc−12bd)

Explanation: Expand LHS: 3a*2c=6ac, 3a*-3d=-9ad, -4b*2c=-8bc, -4b*-3d=12bd

15. Factorize and simplify:

(a) 9a² − 25b²

Solution: Difference of squares: 9a²−25b² = (3a−5b)(3a+5b)

(b) (2c+3)² − c²

Solution: Expand or difference of squares: (2c+3−c)(2c+3+c)= (c+3)(3c+3)=3(c+1)(c+3)

(c) 36(x+2y)² − 25(2x−y)²

Solution: difference of squares formula: a²−b²=(a−b)(a+b)
a=6(x+2y), b=5(2x−y)
(6(x+2y)−5(2x−y))(6(x+2y)+5(2x−y))
=(6x+12y−10x+5y)(6x+12y+10x−5y)
=(-4x+17y)(16x+7y)

1. A farmer has a rectangular field of length 50 m and width 30 m. He wants to increase both length and width by the same amount so that the area increases by 450 m². Find the increase.

Solution:

Step 1: Let the increase be x meters.

Step 2: New length = 50 + x, New width = 30 + x

Step 3: New area = (50 + x)(30 + x) = Original area + 450 = (50*30) + 450 = 1500 + 450 = 1950

Step 4: Set up equation: (50 + x)(30 + x) = 1950

Step 5: Expand: 50*30 + 50x + 30x + x² = 1950 → 1500 + 80x + x² = 1950

Step 6: x² + 80x + 1500 - 1950 = 0 → x² + 80x - 450 = 0

Step 7: Solve quadratic using factorization or formula: x = [-80 ± √(80² - 4*1*(-450))]/2

Step 8: Compute discriminant: 6400 + 1800 = 8200 → √8200 ≈ 90.55

Step 9: x = (-80 + 90.55)/2 ≈ 5.275 m (Ignore negative solution)

Answer: Increase ≈ 5.28 m

Real-life example: A farmer expanding his field for extra crops.

2. A shopkeeper bought a batch of goods for 120,000 TZS. He sold one-third at a 10% profit, half of the remainder at a 5% loss, and the rest at a 15% profit. Find the overall profit or loss.

Solution:

Step 1: Total cost = 120,000 TZS

Step 2: One-third sold = 120000*(1/3) = 40,000 TZS at 10% profit → selling price = 40000*1.1 = 44,000

Step 3: Remaining = 120000 - 40000 = 80,000 TZS. Half of this = 40000 TZS at 5% loss → selling price = 40000*0.95 = 38,000

Step 4: Remaining = 40000 TZS sold at 15% profit → selling price = 40000*1.15 = 46,000

Step 5: Total selling price = 44,000 + 38,000 + 46,000 = 128,000

Step 6: Profit = 128,000 - 120,000 = 8,000 TZS → Profit % = (8000/120000)*100 ≈ 6.67%

Answer: Overall profit ≈ 6.67%

Example: Understanding real shop profit calculations.

3. Solve for x: √(x+7) - √x = 1

Solution:

Step 1: Isolate one root: √(x+7) = 1 + √x

Step 2: Square both sides: x + 7 = (1 + √x)² = 1 + 2√x + x

Step 3: x + 7 = x + 1 + 2√x → 7 - 1 = 2√x → 6 = 2√x → √x = 3 → x = 9

Answer: x = 9

Example: Finding distance traveled when growth happens incrementally.

4. The sum of the ages of A and B is 60. Five years ago, A was three times as old as B. Find their present ages.

Solution:

Step 1: Let A = a, B = b → a + b = 60

Step 2: Five years ago: a-5 = 3*(b-5) → a - 5 = 3b - 15 → a - 3b = -10

Step 3: Solve system of equations:
a + b = 60
a - 3b = -10
Subtract: (a + b) - (a - 3b) = 60 - (-10) → 4b = 70 → b = 17.5

Step 4: a = 60 - 17.5 = 42.5

Answer: A = 42.5 years, B = 17.5 years

Example: Age problems in family planning contexts.

5. Find the factorization of 2x³ + 5x² − 3x − 10

Solution:

Step 1: Group: (2x³ + 5x²) − (3x + 10)

Step 2: Factor each group: x²(2x+5) − 1(3x+10) ❌ notice mismatch, correct group:

(2x³ - 3x) + (5x² -10) → factor: x(2x² -3) + 5(x² -2)

Step 3: Factor further: Check trial and error → Factor: (2x+5)(x²-2)

Answer: (2x + 5)(x² - 2)

Example: Finding roots of cubic equations in engineering stress problems.

6. A tank can be filled by pipe A in 6 hours and by pipe B in 4 hours. Both pipes open together. How long to fill the tank?

Solution:

Step 1: Part filled by A in 1 hour = 1/6

Step 2: Part filled by B in 1 hour = 1/4

Step 3: Together = 1/6 + 1/4 = 2/12 + 3/12 = 5/12 per hour

Step 4: Time to fill tank = 1 ÷ (5/12) = 12/5 = 2.4 hours → 2 hours 24 minutes

Example: Filling water tanks in schools or farms.

7. Solve: 2/(x+1) + 3/(x-1) = 5/(x²-1)

Solution:

Step 1: Factor denominator: x²-1 = (x-1)(x+1)

Step 2: Rewrite LHS with same denominator: 2(x-1)/(x²-1) + 3(x+1)/(x²-1) = [2(x-1)+3(x+1)]/(x²-1)

Step 3: Numerator: 2x-2 + 3x+3 = 5x +1

Step 4: Equation: (5x+1)/(x²-1) = 5/(x²-1) → 5x+1=5 → 5x=4 → x=4/5

Answer: x = 4/5

8. Find the simple interest on 150,000 TZS for 2 years at 8% per annum, and the total amount.

Solution:

Step 1: Simple interest SI = P*R*T/100 = 150,000*8*2/100 = 24,000

Step 2: Total amount = Principal + SI = 150,000 + 24,000 = 174,000 TZS

Example: Understanding loan interest in daily life banking.

9. A train travels 360 km at speed x km/h and returns at speed (x+20) km/h. If total time is 5 hours, find x.

Solution:

Step 1: Time = Distance/Speed → forward: 360/x, return: 360/(x+20)

Step 2: Total time: 360/x + 360/(x+20) = 5

Step 3: Multiply both sides by x(x+20): 360(x+20)+360x = 5x(x+20)

Step 4: 360x+7200+360x = 5x² + 100x → 720x+7200 = 5x² +100x

Step 5: 5x² +100x -720x -7200=0 → 5x² -620x -7200=0

Step 6: Solve quadratic: x = [620 ± √(620² - 4*5*(-7200))]/10 → x≈ 124 km/h

Example: Train scheduling and speed planning.

10. Find the sum of first 20 terms of an AP where first term is 3 and last term is 59.

Solution:

Step 1: Sum of n terms of AP: S = n/2*(first term + last term)

Step 2: S20 = 20/2*(3+59) = 10*62 = 620

Example: Distributing resources in equal increments, like savings.

1. John wants to divide 6000 TZS among 3 friends so that each next friend gets 200 TZS more than the previous. How much does each get?

Solution:

Step 1: Let the first friend get x TZS. Then second = x + 200, third = x + 400

Step 2: Total = x + (x+200) + (x+400) = 3x + 600 = 6000 → 3x = 5400 → x = 1800

Step 3: Second = 2000, Third = 2200

Answer: 1800, 2000, 2200 TZS

Swahili tip: Anza na kigezo kimoja (x), kisha ongeza tofauti kwa kila mtu. Rahisi kuona!

2. A car travels 120 km in the morning at 60 km/h and returns at 40 km/h. Find average speed.

Solution:

Step 1: Average speed formula for same distance: Vavg = 2*V1*V2 / (V1+V2)

Step 2: Vavg = 2*60*40 / (60+40) = 4800/100 = 48 km/h

Swahili explanation: Ikiwa umbali ni sawa, usawa wa haraka ni 2*V1*V2/(V1+V2) → rahisi sana!

Answer: 48 km/h

3. A school buys 50 pens at 500 TZS each and sells 40 of them at 600 TZS each. Find the profit/loss.

Solution:

Step 1: Total cost = 50*500 = 25,000 TZS

Step 2: Selling price = 40*600 = 24,000 TZS (sold pens)

Step 3: Unsold pens cost = 10*500 = 5,000 TZS

Step 4: Total value = 24,000 + 5,000 = 29,000 TZS

Step 5: Profit = 29,000 - 25,000 = 4,000 TZS

Answer: Profit 4,000 TZS

Tip: Rahisisha kwa kuzingatia gharama ya bidhaa zisizouzwa.

4. A rectangular garden has length twice its width. If perimeter = 36 m, find dimensions.

Solution:

Step 1: Let width = x, length = 2x

Step 2: Perimeter = 2(L + W) = 2(2x + x) = 6x = 36 → x = 6 m

Step 3: Length = 2*6 = 12 m

Answer: Width = 6 m, Length = 12 m

Swahili tip: Perimeter rahisisha kupata kigezo moja, kisha hesabu dimension nyingine.

5. A trader sells rice in packets of 10 kg. If cost price per kg = 2000 TZS and he gains 10% on selling 100 kg, find selling price per packet.

Solution:

Step 1: Total cost for 100 kg = 100*2000 = 200,000 TZS

Step 2: Gain 10% → selling price total = 200,000*1.1 = 220,000 TZS

Step 3: Number of packets = 100/10 = 10 packets

Step 4: Selling price per packet = 220,000 / 10 = 22,000 TZS

Answer: 22,000 TZS per packet

6. A water tank is filled by pipe A in 5 hrs and pipe B in 3 hrs. How long if both pipes are open together?

Solution:

Step 1: Part filled per hour: A=1/5, B=1/3

Step 2: Together = 1/5 + 1/3 = 3/15 + 5/15 = 8/15 per hour

Step 3: Time = 1 ÷ (8/15) = 15/8 = 1.875 hrs ≈ 1 hr 52.5 min

Answer: 1 hr 52.5 min

Swahili tip: Chukua sehemu ya tank kwa saa, ongeza zote → rahisi kuona.

7. Simplify: (x² - 4)/(x - 2)

Solution:

Step 1: Recognize difference of squares: x²-4 = (x-2)(x+2)

Step 2: Divide by x-2 → (x+2)

Answer: x + 2

Tip: Angalia if difference of squares → rahisi sana!

8. A shop has 3 types of apples: 20 red, 30 green, 50 yellow. If one apple is picked randomly, find probability it is not yellow.

Solution:

Total apples = 20+30+50 = 100

Not yellow = 20+30 = 50

Probability = 50/100 = 1/2

Answer: 1/2

Swahili tip: Hesabu jumla, kisha chagua zile unahitaji.

9. A man invests 100,000 TZS at 10% simple interest per annum for 3 years. Find interest and total amount.

Solution:

SI = P*R*T/100 = 100,000*10*3/100 = 30,000

Total = 100,000 + 30,000 = 130,000 TZS

Answer: SI = 30,000, Total = 130,000

Tip: Simple interest ni rahisi: P*R*T/100

10. A train travels 150 km at 50 km/h and 100 km at 25 km/h. Find average speed for whole journey.

Solution:

Step 1: Time for first part = 150/50 = 3 hrs

Step 2: Time for second part = 100/25 = 4 hrs

Step 3: Total distance = 150+100 = 250 km, total time = 3+4=7 hrs

Step 4: Average speed = Total distance / Total time = 250/7 ≈ 35.71 km/h

Answer: ≈ 35.71 km/h

Swahili tip: Hakikisha jumlisha umbali na muda kisha gawanya.