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In a shop, 2 pens and 3 books cost K27. 1 pens and 2 books cost K23. Find the cost of one pen and one book.

Mathematics

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2 parts
b

K3\text{K}3

c

K7\text{K}7

K3\text{K}3

Step-by-step solution

6 steps

Part 5

SIMULTANEOUS EQUATIONS

Part a

Form the two equations.

Let pp be the cost of one pen (in K).
Let bb be the cost of one book (in K).

From the problem statement:

  1. "2 pens and 3 books cost K27"
  2. "3 pens and 2 books cost K23"

The two equations are:

2p+3b=27(Equation 1)2p + 3b = 27 \quad \text{(Equation 1)} 3p+2b=23(Equation 2)3p + 2b = 23 \quad \text{(Equation 2)}

Part b

K3\text{K}3

Use elimination to find one variable.

  1. Make the coefficients of one variable equal.

    Let's make the coefficients of bb equal.
    Multiply Equation 1 by 2:

    2(2p+3b)=2(27)  ⟹  4p+6b=54(Equation 3)2(2p + 3b) = 2(27) \implies 4p + 6b = 54 \quad \text{(Equation 3)}

    Multiply Equation 2 by 3:

    3(3p+2b)=3(23)  ⟹  9p+6b=69(Equation 4)3(3p + 2b) = 3(23) \implies 9p + 6b = 69 \quad \text{(Equation 4)}

  2. Subtract Equation 3 from Equation 4 to eliminate bb.

    (9p+6b)−(4p+6b)=69−54(9p + 6b) - (4p + 6b) = 69 - 54 5p=155p = 15

  3. Solve for pp.

    p=155p = \frac{15}{5} p=3p = 3

    The cost of one pen is K3\boxed{\text{K}3}.

Part c

K7\text{K}7

Find the second variable and verify.

  1. Substitute the value of p=3p=3 into Equation 1.

    2(3)+3b=272(3) + 3b = 27 6+3b=276 + 3b = 27

  2. Solve for bb.

    3b=27−63b = 27 - 6 3b=213b = 21 b=213b = \frac{21}{3} b=7b = 7

    The cost of one book is K7\boxed{\text{K}7}.

  3. Verify the solution using Equation 2.

    Substitute p=3p=3 and b=7b=7 into Equation 2:

    3p+2b=3(3)+2(7)3p + 2b = 3(3) + 2(7) 9+14=239 + 14 = 23

    Since 23=2323 = 23, the solution is correct.

    The cost of one pen is K3\boxed{\text{K}3} and the cost of one book is K7\boxed{\text{K}7}.

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