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a^2 + b ^2 = 290 ...............................................(2)

we will use the algebraic identity formula

(a + b) ^2 = a ^2 + b^2 + 2ab.............................(3)

18^2 = 290 = 2ab

324 - 290 = 2ab

34 = 2ab

ab = 34 /2

ab = 17,..............................................................(4)

a = 17 / b............................................................ (5 )

apply (5) in (1) we get,

a + b = 18

17/ b + b = 18

17+ b^2 = 18b

b^2 - 18b + 17

(b - 17) (b -1)

b = 17,1 ..............................................................(6)

if b= 17 apply in 5 we get

a = 17/17

a = 1

if b= 1

then a = 17,

(a,b) = ( 17,1) or ( a,b) = (1,17)

a^2 + b ^2 = 290

(a + b) ^2 = a ^2 + b^2 + 2ab

18^2 = 290 = 2ab

324 - 290 = 2ab 34 = 2ab

ab = 34 /2 =17

a = 17 / b

when we put 5 in 1

a + b = 18

17/ b + b = 18

17+ b^2 = 18b

b^2 - 18b + 17

(b - 17) (b -1)

b = 17,1

if b= 17 applied in 5 comes 1

a = 17/17

a = 1

b= 1

a = 17,

b=1