Question

answer this question

Answer 1

${a}^{2} + {b}^{2} = 117$
Adding 2ab to the above equation,
$\: \: \: \: \: \: \: {a}^{2} + {b}^{2} + 2ab = 117 + 2ab \\ = > {a}^{2} + {b}^{2} + 2ab = 117 + 2 \times 54 \\ = > {(a + b)}^{2} = 225 \\ = > (a + b) = 15 \: \: or\: \: ( - 15) - - - - - > 1$



Now subtracting 2ab from the very first equation,
$\: \: \: \: \: \: \: {a}^{2} + {b}^{2} - 2ab = 117 +( - 2ab) \\ = > {a}^{2} + {b}^{2} - 2ab = 117 - 2 \times 54 \\ = > {(a - b)}^{2} = 9 \\ = > (a - b) = 3 \: \: \: or \: \: \: \: ( - 3) - - - - - > 2$



Now from eq. 1 & 2,
$\frac{( a+b )}{(a - b)} = \frac{15}{3} \: \: or \: \: \frac{ - 15}{ - 3} = 5.$


So,
5 is the answer.




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Answer 2

Hay mate here is your answer ✌❤☺

a²+b²= 117

ab = 54

a²+b²= 117

by adding and subtracting 2ab

a²+b²+2ab-2ab= 117

(a+b)² -2ab = 117

by putting the value of an = 54

(a+b)² - 2x54 = 117

(a+b)² = 117 + 108

(a+b)² = 225

a+b = √225

a+b = 25

like this

a-b = 3

than,

(a+b)/(a-b) = 25/3

= 5. Ans

Hope it will help you ☺❤