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Question Number 100404 by PengagumRahasiamu last updated on 26/Jun/20

Commented by PengagumRahasiamu last updated on 26/Jun/20
Thank you
Commented by Rasheed.Sindhi last updated on 26/Jun/20

$(i) \Rightarrow x = \frac{28 - y}{y + 2}, (i i) \Rightarrow x = \frac{37 - z}{z + 3}$ $\frac{28 - y}{y + 2} = \frac{37 - z}{z + 3} . . . . . . . . (iv)$ $(iii) \Rightarrow y = \frac{42 - 2 z}{z + 3}$ $(iii) & (iv) \Rightarrow$ $\frac{28 - \frac{42 - 2 z}{z + 3}}{\frac{42 - 2 z}{z + 3} + 2} = \frac{37 - z}{z + 3}$ $\frac{\frac{28 z + 84 - 42 + 2 z}{z + 3}}{\frac{42 - 2 z + 2 z + 6}{z + 3}} = \frac{37 - z}{z + 3}$ $\frac{30 z + 42}{48} = \frac{37 - z}{z + 3}$ $\frac{5 z + 7}{8} = \frac{37 - z}{z + 3}$ ${5 z}^{2} + 22 z + 21 = 296 - 8 z$ ${5 z}^{2} + 30 z - 275 = 0$ $z^{2} + 6 z - 55 = 0$ $(z + 11) (z - 5) = 0$ $z = - 11 (d e s c a r d a b l e), 5$ $x = \frac{37 - z}{z + 3} = \frac{37 - 5}{5 + 3} = \frac{32}{8} = 4$ $y = \frac{42 - 2 z}{z + 3} = \frac{42 - 2 (5)}{5 + 3} = \frac{32}{8} = 4$ $x = 4, y = 4, z = 5$ $x + \sqrt{y} + z = 4 + \sqrt{4} + 5 = 11$
Commented by bobhans last updated on 27/Jun/20
$x= ((28−y)/(y+2)) = ((37−z)/(z+3)) ∧ y= ((42−2z)/(z+3)) ((28−(((42−2z)/(z+3))))/(((42−2z)/(z+3)) +2)) = ((37−z)/(z+3)) ⇒ ((30z+42)/((z+3)(48))) = ((37−z)/(z+3)) ((30z+42)/(48)) = ((37−z)/(z+3)) ⇒(5z+7)(z+3) = 296−8z 5z^2 +22z+21=296−8z 5z^2 +30z−275=0, z^2 +6z−55=0 (z+11)(z−5)=0 ⇒ { ((z=5)),((y=((42−10)/8)= 4)),((x=((28−4)/6)=4)) :} ∴ x+(√y) +z = 4+2+5 = 11 ■$
$x = \frac{28 - y}{y + 2} = \frac{37 - z}{z + 3} \land y = \frac{42 - 2 z}{z + 3}$ $\frac{28 - (\frac{42 - 2 z}{z + 3})}{\frac{42 - 2 z}{z + 3} + 2} = \frac{37 - z}{z + 3} \Rightarrow \frac{30 z + 42}{(z + 3) (48)} = \frac{37 - z}{z + 3}$ $\frac{30 z + 42}{48} = \frac{37 - z}{z + 3} \Rightarrow (5 z + 7) (z + 3) = 296 - 8 z$ ${5 z}^{2} + 22 z + 21 = 296 - 8 z$ ${5 z}^{2} + 30 z - 275 = 0, z^{2} + 6 z - 55 = 0$ $(z + 11) (z - 5) = 0 \Rightarrow {\begin{cases} z = 5 \\ y = \frac{42 - 10}{8} = 4 \\ x = \frac{28 - 4}{6} = 4 \end{cases}$ $∴ x + \sqrt{y} + z = 4 + 2 + 5 = 11 ◼$
Commented by Rasheed.Sindhi last updated on 26/Jun/20

$S i r b o b h a n s x, y, z \in Z^{+}$ $T h e n h o w z = \frac{303}{13} ?$
Commented by bramlex last updated on 26/Jun/20

$i t s h o u l d b e \Leftrightarrow \frac{30 z + 42}{48} = \frac{37 - z}{z + 3}$ $\frac{5 z + 7}{8} = \frac{37 - z}{z + 3} \Leftrightarrow (5 z + 7) (z + 3) = 8 (37 - z)$
Commented by Dwaipayan Shikari last updated on 26/Jun/20

$x = \frac{28 - y}{y + 2}$ $y = \frac{42 - 2 z}{z + 3}$ $x = \frac{37 - z}{z + 3}$ $So, \frac{37 - z}{z + 3} = \frac{28 - \frac{42 - 2 z}{z + 3}}{\frac{42 - 2 z}{z + 3} + 2} \Rightarrow \frac{37 - z}{z + 3} = \frac{30 z + 42}{48}$ $\frac{296 - 8 z}{z + 3} = 5 z + 7 \Rightarrow 296 - 8 z = {5 z}^{2} + 22 z + 21$ ${5 z}^{2} + 30 z - 275 = 0 \Rightarrow z^{2} + 6 z - 55 \Rightarrow z = 5 or - 11 but z \neq - 11$ $So; x = 4$ $y = 4$ $so x + \sqrt{y} + z = 11$
Commented by Dwaipayan Shikari last updated on 26/Jun/20

$Sir answer will be 11$
Commented by Dwaipayan Shikari last updated on 26/Jun/20

$s i r a n s w e r i s 11 . i t h i n k y o u h a v e d o n e s o m e t h i n g m i s t a k e$
Commented by Dwaipayan Shikari last updated on 26/Jun/20

$O h h h s i r y o u p u t 2 i n s t e a d o f 5 w h i l e f i n d i n g t h e v a l u e o f y$
Commented by Rasheed.Sindhi last updated on 26/Jun/20

$T h a n k y o u s i r f o r m e n t i o n i n g$ $m i s t a k e . N o w I c o r r e c t e d i t . Y o u r$ $a n s w e r i s a l s o c o r r e t .$
Commented by bobhans last updated on 27/Jun/20

$sorry sir Rashed . miscalculated$
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