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WAEC Mathematics 2024 Paper

Question : 27

Total: 50

Solution:

Given that $6244_x - 1621_x = 4323_x$ , find the value of $x$ .

{Step 1: Convert to decimal:}

6244_x = 6x^3 + 2x^2 + 4x + 4

1621_x = 1x^3 + 6x^2 + 2x + 1

4323_x = 4x^3 + 3x^2 + 2x + 3

f{Step 2: Set up the equation:}

(6x^3 + 2x^2 + 4x + 4) - (1x^3 + 6x^2 + 2x + 1) = 4x^3 + 3x^2 + 2x + 3

{Step 3: Simplify the left side:}

(6x^3 - 1x^3) + (2x^2 - 6x^2) + (4x - 2x) + (4 - 1) = 5x^3 - 4x^2 + 2x + 3

{Step 4: Set the equation:}

5x^3 - 4x^2 + 2x + 3 = 4x^3 + 3x^2 + 2x + 3

{Step 5: Combine like terms:}

5x^3 - 4x^3 - 4x^2 - 3x^2 + 2x - 2x + 3 - 3 = 0

x^3 - 7x^2 = 0

{Step 6: Factor the equation:}

x^2(x - 7) = 0

{Step 7: Solutions:}

x = 7 \quad (\text{valid}), \quad x = 0 \quad (\text{not valid})

\Thus, the base $x$ is 7

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