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Elimu school bought 25 textbooks and 35 exercise books for sh. 13500. Soma school bought 21 textbooks and 38 exercise books for sh. 12200. Find the cost of one textbook and one exercise book.

Mathematics

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Here's the solution to question 3:

Let xx be the cost of one textbook and yy be the cost of one exercise book.

  1. Formulate equations based on the given information.

    For Elimu school:
    25 textbooks and 35 exercise books cost sh. 13500.

    25x+35y=13500(Equation 1)25x + 35y = 13500 \quad \text{(Equation 1)}

    For Soma school:
    Soma school bought 21 textbooks and 38 exercise books.
    They spent sh. 1300 less than Elimu school.
    Elimu school spent sh. 13500, so Soma school spent 13500−1300=1220013500 - 1300 = 12200 sh.

    21x+38y=12200(Equation 2)21x + 38y = 12200 \quad \text{(Equation 2)}
  2. Solve the system of linear equations using the elimination method.

    Multiply Equation 1 by 21 and Equation 2 by 25 to eliminate xx.
    Equation 1 ×\times 21:

    21(25x)+21(35y)=21(13500)21(25x) + 21(35y) = 21(13500) 525x+735y=283500(Equation 3)525x + 735y = 283500 \quad \text{(Equation 3)}

    Equation 2 ×\times 25:

    25(21x)+25(38y)=25(12200)25(21x) + 25(38y) = 25(12200) 525x+950y=305000(Equation 4)525x + 950y = 305000 \quad \text{(Equation 4)}
  3. Subtract Equation 3 from Equation 4 to solve for yy.

    (525x+950y)−(525x+735y)=305000−283500(525x + 950y) - (525x + 735y) = 305000 - 283500 215y=21500215y = 21500 y=21500215y = \frac{21500}{215} y=100y = 100

    The cost of one exercise book is sh. 100.

  4. Substitute the value of yy into Equation 1 to solve for xx.

    25x+35(100)=1350025x + 35(100) = 13500 25x+3500=1350025x + 3500 = 13500 25x=13500−350025x = 13500 - 3500 25x=1000025x = 10000 x=1000025x = \frac{10000}{25} x=400x = 400

    The cost of one textbook is sh. 400.

    The cost of one textbook is sh. 400\boxed{\text{sh. } 400}.
    The cost of one exercise book is sh. 100\boxed{\text{sh. } 100}.

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