Question

$5 + \sqrt{3} \div 7 - 4 \sqrt{3} = 47a + \sqrt{3} b$
if a and b are both rational numbers find values of a and b​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Step-by-step explanation:

$\rm 5 + \sqrt{3} \div 7 - 4 \sqrt{3} = 47a + \sqrt{3} b$

$\frac{5 + \sqrt{3} }{7 - 4 \sqrt{3} }$

On rationalise the denominator

$\frac{5 + \sqrt{3} }{7 - 4 \sqrt{3} } \times \frac{7 + 4 \sqrt{3} }{7 + 4 \sqrt{3} }$

$\frac{(5 + \sqrt{3} )(7 + 4\sqrt{3} )}{(7 - 4 \sqrt{3})(7 + 4 \sqrt{3} ) }$

$\frac{5(7 + 4 \sqrt{3}) + \sqrt{3} (7 + 4 \sqrt{3} )}{ {(7})^{2} - {(4 \sqrt{3}) }^{2} }$

$\frac{35 + 20 \sqrt{3} + 7 \sqrt{3} + 4 \sqrt{9} }{49 - 16 \times \sqrt{9} }$

$\frac{35 + 27 \sqrt{3} + 4 \times {3} }{49 - 16 \times 3 }$

$\frac{35 + 27 \sqrt{3} +12}{49 - 48 }$

$\frac{35 +12+ 27 \sqrt{3} }{1 }$

$47 + 27 \sqrt{3}$

$\therefore \: \: \rm a = 1 \: \: ad \: \: b = 27$

I hope it is helpful