Math Problem Statement

Three pens and four exercise books cost sh. 87. Two pens and five exercise books cost sh. 93. Find the cost of one pen and one exercise book. Additionally, solve the following equations and evaluate expressions in the provided worksheet.

Solution

Let's solve the problems step by step from the uploaded image.


5. Cost of One Pen and One Exercise Book

Given:

  • 3 pens and 4 exercise books cost sh. 87.
  • 2 pens and 5 exercise books cost sh. 93.

Let the cost of a pen be xx and the cost of an exercise book be yy.

From the information: 3x + 4y = 87 \tag{1} 2x + 5y = 93 \tag{2}

Multiply (1) by 2 and (2) by 3 to eliminate xx: 6x + 8y = 174 \tag{3} 6x + 15y = 279 \tag{4}

Subtract (3) from (4): 7y=105    y=157y = 105 \implies y = 15

Substitute y=15y = 15 into (1): 3x+4(15)=87    3x+60=87    3x=27    x=93x + 4(15) = 87 \implies 3x + 60 = 87 \implies 3x = 27 \implies x = 9

Thus:

  • Cost of one pen = sh. 9
  • Cost of one exercise book = sh. 15

6. Solve for xx in the Equation

Given: x+24x+35=x+46\frac{x + 2}{4} - \frac{x + 3}{5} = \frac{x + 4}{6}

Find the least common denominator (LCD) of 4, 5, and 6, which is 60. Multiply through by 60 to eliminate the denominators: 60x+2460x+35=60x+4660 \cdot \frac{x + 2}{4} - 60 \cdot \frac{x + 3}{5} = 60 \cdot \frac{x + 4}{6} 15(x+2)12(x+3)=10(x+4)15(x + 2) - 12(x + 3) = 10(x + 4) Expand: 15x+3012x36=10x+4015x + 30 - 12x - 36 = 10x + 40 Simplify: 3x6=10x+403x - 6 = 10x + 40 7x=46    x=467-7x = 46 \implies x = -\frac{46}{7}


7. Two-Digit Number Problem

Let the two-digit number be 10a+b10a + b, where aa is the tens digit and bb is the ones digit. Given:

  1. a+b=10a + b = 10 \tag{1}
  2. Reversing the digits gives a new number 10b+a10b + a, which exceeds the original number by 36: 10b + a = 10a + b + 36 \tag{2}

Simplify (2): 9b - 9a = 36 \implies b - a = 4 \tag{3}

Solve (1) and (3) simultaneously: From (3), b=a+4b = a + 4. Substitute into (1): a+(a+4)=10    2a+4=10    2a=6    a=3a + (a + 4) = 10 \implies 2a + 4 = 10 \implies 2a = 6 \implies a = 3 b=a+4=3+4=7b = a + 4 = 3 + 4 = 7

Thus, the number is 37.


8. Evaluate

Expression: 18÷3 of (2)×8+244÷6×2\frac{18 \div 3 \text{ of } (-2) \times 8 + 24}{-4 \div 6 \times 2}

Step 1: Simplify numerator 18÷3=6,of (2)    6×(2)=1218 \div 3 = 6, \quad \text{of } (-2) \implies 6 \times (-2) = -12 12×8=96,96+24=72-12 \times 8 = -96, \quad -96 + 24 = -72

Step 2: Simplify denominator 4÷6×2=46×2=86=43-4 \div 6 \times 2 = -\frac{4}{6} \times 2 = -\frac{8}{6} = -\frac{4}{3}

Step 3: Final result 7243=72×34=54\frac{-72}{-\frac{4}{3}} = -72 \times -\frac{3}{4} = 54

The answer is 54.


9. Solve Reciprocal Equation

Given: x2+3x9=x4x+1\frac{x^2 + 3}{x - 9} = \frac{x - 4}{x + 1}

Cross-multiply: (x2+3)(x+1)=(x4)(x9)(x^2 + 3)(x + 1) = (x - 4)(x - 9)

Expand both sides: x3+x2+3x+3=x213x+36x^3 + x^2 + 3x + 3 = x^2 - 13x + 36 x3+x2x2+3x+13x+336=0x^3 + x^2 - x^2 + 3x + 13x + 3 - 36 = 0 x3+16x33=0x^3 + 16x - 33 = 0

Finding the root manually or numerically gives: x=3x = 3


Let me know if you'd like further clarifications or steps! Here are 5 related questions and a tip:

  1. Can you solve similar simultaneous equations for two variables?
  2. How do you simplify fractional equations with LCDs?
  3. Can you evaluate expressions involving nested operations?
  4. What are the methods to solve cubic equations?
  5. How do you confirm solutions for digit-based problems?

Tip: Always check your solutions by substituting them back into the original equations!

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Math Problem Analysis

Mathematical Concepts

Simultaneous Equations
Fractions and Simplification
Number Theory
Order of Operations (BODMAS)
Cubic Equations

Formulas

System of linear equations: ax + by = c
Fraction simplification using the least common denominator (LCD)
Cross multiplication for solving rational equations

Theorems

Basic properties of equality
Commutative property of addition/multiplication
Distributive property

Suitable Grade Level

Grades 8-10