x , y , z ∈R: If { (((1/x) +(1/(y+z)) =(1/2))),(((1/y) +(1/(x+z)) = (1/3))),(((1/z


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 183668 by mnjuly1970 last updated on 28/Dec/22
$x , y , z ∈R: If { (((1/x) +(1/(y+z)) =(1/2))),(((1/y) +(1/(x+z)) = (1/3))),(((1/z_ ) +(1/(x+y)) =(1/4))) :} ⇒ x , y , z =?$
$x, y, z \in R :$ $If {\begin{cases} \frac{1}{x} + \frac{1}{y + z} = \frac{1}{2} \\ \frac{1}{y} + \frac{1}{x + z} = \frac{1}{3} \\ \frac{1}{z_{}} + \frac{1}{x + y} = \frac{1}{4} \end{cases}$ $\Rightarrow x, y, z = ?$
Commented by Frix last updated on 28/Dec/22

$This is easy to solve with$ $y = a x \land z = b x \land a, b \neq 0 \land a + b + 1 \neq 0$ $W e get$ $a = \frac{5}{3} \land b = 5 \Rightarrow x = \frac{23}{10} \land y = \frac{23}{6} \land z = \frac{23}{2}$
	Answered by Rasheed.Sindhi last updated on 29/Dec/22
	${ (((1/x) +(1/(y+z)) =(1/2))),(((1/y) +(1/(x+z)) = (1/3))),(((1/z_ ) +(1/(x+y)) =(1/4))) :} { ((((x+y+z)/(x(y+z))) =(1/2)⇒x+y+z=((x(y+z))/2))),((((x+y+z)/(y(x+z))) = (1/3)⇒x+y+z=((y(x+z))/3))),((((x+y+z)/(z(x+y))) =(1/4)⇒x+y+z=((z(x+y))/4))) :} ⇒ ((xy+xz)/2)=((xy+yz)/3)=((xz+yz)/4) =((xy+xz+xy+yz+xz+yz)/(2+3+4)) =((2(xy+yz+xz))/9) { ((9(xy+xz)=4(xy+yz+xz))),((9(xy+yz)=6(xy+yz+xz))),((9(xz+yz)=8(xy+yz+xz))) :} { ((5xy−4yz+5xz=0)),((3xy+3yz−6xz=0)),((-8xy+yz+xz=0)) :} Dividing by xyz to both sides: { (((5/z)−(4/x)+(5/y)=0....(i))),(((3/z)+(3/x)−(6/y)=0....(ii))),((((-8)/z)+(1/x)+(1/y)=0....(iii))) :} (i)×3: ((15)/z)−((12)/x)+((15)/y)=0...(iv) (ii)×5: ((15)/z)+((15)/x)−((30)/y)=0...(v) (iv)−(v) : ((-27)/x)+((45)/y)=0 ((27)/x)=((45)/y) (3/x)=(5/y)....(viii) (ii)×1:(3/z)+(3/x)−(6/y)=0...(vi) (iii)×3:((-24)/z)+(3/x)+(3/y)=0...(vii) (vi)−(vii): ((27)/z)−(9/y)=0 (3/z)=(1/y) ((15)/z)=(5/y) (3/x)=(5/y).......(ix) (viii)&(ix): (3/x)=(5/y)=((15)/z) x=3k ,y=5k , z=15k (1/x) +(1/(y+z)) =(1/2) (first given eqn) ⇒(1/(3k))+(1/(5k+15k))=(1/2) 60k∙(1/(3k))+60k∙(1/(20k))=60k∙(1/2) 20+3=30k⇒k=((23)/(30)) x=3k=3(((23)/(30)))=((23)/(10)) y=5k=5(((23)/(30)))=((23)/6) z=15k=15(((23)/(30)))=((23)/2)$
	${\begin{cases} \frac{1}{x} + \frac{1}{y + z} = \frac{1}{2} \\ \frac{1}{y} + \frac{1}{x + z} = \frac{1}{3} \\ \frac{1}{z_{}} + \frac{1}{x + y} = \frac{1}{4} \end{cases}$ ${\begin{cases} \frac{x + y + z}{x (y + z)} = \frac{1}{2} \Rightarrow x + y + z = \frac{x (y + z)}{2} \\ \frac{x + y + z}{y (x + z)} = \frac{1}{3} \Rightarrow x + y + z = \frac{y (x + z)}{3} \\ \frac{x + y + z}{z (x + y)} = \frac{1}{4} \Rightarrow x + y + z = \frac{z (x + y)}{4} \end{cases}$ $\Rightarrow \frac{x y + x z}{2} = \frac{x y + y z}{3} = \frac{x z + y z}{4}$ $= \frac{x y + x z + x y + y z + x z + y z}{2 + 3 + 4}$ $= \frac{2 (x y + y z + x z)}{9}$ ${\begin{cases} 9 (x y + x z) = 4 (x y + y z + x z) \\ 9 (x y + y z) = 6 (x y + y z + x z) \\ 9 (x z + y z) = 8 (x y + y z + x z) \end{cases}$ ${\begin{cases} 5 x y - 4 y z + 5 x z = 0 \\ 3 x y + 3 y z - 6 x z = 0 \\ - 8 x y + y z + x z = 0 \end{cases}$ $D i v i d i n g b y x y z t o b o t h s i d e s :$ ${\begin{cases} \frac{5}{z} - \frac{4}{x} + \frac{5}{y} = 0 . . . . (i) \\ \frac{3}{z} + \frac{3}{x} - \frac{6}{y} = 0 . . . . (i i) \\ \frac{- 8}{z} + \frac{1}{x} + \frac{1}{y} = 0 . . . . (i i i) \end{cases}$ $(i) \times 3 : \frac{15}{z} - \frac{12}{x} + \frac{15}{y} = 0 . . . (i v)$ $(i i) \times 5 : \frac{15}{z} + \frac{15}{x} - \frac{30}{y} = 0 . . . (v)$ $(i v) - (v) : \frac{- 27}{x} + \frac{45}{y} = 0$ $\frac{27}{x} = \frac{45}{y}$ $\frac{3}{x} = \frac{5}{y} . . . . (v i i i)$ $(i i) \times 1 : \frac{3}{z} + \frac{3}{x} - \frac{6}{y} = 0 . . . (v i)$ $(i i i) \times 3 : \frac{- 24}{z} + \frac{3}{x} + \frac{3}{y} = 0 . . . (v i i)$ $(v i) - (v i i) : \frac{27}{z} - \frac{9}{y} = 0$ $\frac{3}{z} = \frac{1}{y}$ $\frac{15}{z} = \frac{5}{y}$ $\frac{3}{x} = \frac{5}{y} . . . . . . . (i x)$ $(v i i i) & (i x) : \frac{3}{x} = \frac{5}{y} = \frac{15}{z}$ $x = 3 k, y = 5 k, z = 15 k$ $\frac{1}{x} + \frac{1}{y + z} = \frac{1}{2} (f i r s t g i v e n e q n)$ $\Rightarrow \frac{1}{3 k} + \frac{1}{5 k + 15 k} = \frac{1}{2}$ $60 k \cdot \frac{1}{3 k} + 60 k \cdot \frac{1}{20 k} = 60 k \cdot \frac{1}{2}$ $20 + 3 = 30 k \Rightarrow k = \frac{23}{30}$ $x = 3 k = 3 (\frac{23}{30}) = \frac{23}{10}$ $y = 5 k = 5 (\frac{23}{30}) = \frac{23}{6}$ $z = 15 k = 15 (\frac{23}{30}) = \frac{23}{2}$
	Commented by mnjuly1970 last updated on 29/Dec/22

	$t a n k s a l o t a l i . . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum