To solve the equation \( 5 \frac{3}{8} = k + 3 \frac{1}{5} \), we first convert the mixed numbers into improper fractions.
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Convert \( 5 \frac{3}{8} \): \[ 5 \frac{3}{8} = \frac{5 \times 8 + 3}{8} = \frac{40 + 3}{8} = \frac{43}{8} \]
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Convert \( 3 \frac{1}{5} \): \[ 3 \frac{1}{5} = \frac{3 \times 5 + 1}{5} = \frac{15 + 1}{5} = \frac{16}{5} \]
Now, substituting back into the equation: \[ \frac{43}{8} = k + \frac{16}{5} \]
To isolate \( k \), we can subtract \( \frac{16}{5} \) from both sides: \[ k = \frac{43}{8} - \frac{16}{5} \]
To carry out the subtraction, we need a common denominator. The least common multiple of 8 and 5 is 40.
- Convert both fractions: \[ \frac{43}{8} = \frac{43 \times 5}{8 \times 5} = \frac{215}{40} \] \[ \frac{16}{5} = \frac{16 \times 8}{5 \times 8} = \frac{128}{40} \]
Now substituting these values back in: \[ k = \frac{215}{40} - \frac{128}{40} = \frac{215 - 128}{40} = \frac{87}{40} \]
Thus, the final answer is: \[ k = \frac{87}{40} \]
If you prefer to convert it back into a mixed number: \[ \frac{87}{40} = 2 \frac{7}{40} \]
So, \( k = 2 \frac{7}{40} \) or \( k = \frac{87}{40} \).