((⋰BeMath⋱)/★) Given ((2x)/(2x+6)) = ((5y)/(5y+25)) = ((4z)/(4z+16)) and xy + yz + xz = 188 . Find the solution


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Question Number 108769 by bemath last updated on 19/Aug/20

$\frac{\iddots B e M a t h ⋱}{★}$ $G i v e n \frac{2 x}{2 x + 6} = \frac{5 y}{5 y + 25} = \frac{4 z}{4 z + 16}$ $a n d x y + y z + x z = 188 . F i n d t h e$ $s o l u t i o n$
	Answered by john santu last updated on 19/Aug/20
	$((⊸JS⊸)/♥) ⇒ (x/(x+3)) = (y/(y+5)) = (z/(z+4)) ⇒ ((x+3)/x) = ((y+5)/y) = ((z+4)/z) ⇒ (3/x) = (5/y) = (4/z) → { ((y = 5b)),((x = 3b )),((z = 4b)) :} ⇒15b^2 + 12b^2 + 20b^2 = 188 ⇒47b^2 = 188 ; → { ((b=2)),((b=−2)) :} case(1) b = 2 → { ((x=6)),((y=10)),((z=8)) :} case(2) b=−2 → { ((x=−6)),((y=−10)),((z=−8)) :}$
	$\frac{⊸ J S ⊸}{♡}$ $\Rightarrow \frac{x}{x + 3} = \frac{y}{y + 5} = \frac{z}{z + 4}$ $\Rightarrow \frac{x + 3}{x} = \frac{y + 5}{y} = \frac{z + 4}{z}$ $\Rightarrow \frac{3}{x} = \frac{5}{y} = \frac{4}{z} \to {\begin{cases} y = 5 b \\ x = 3 b \\ z = 4 b \end{cases}$ $\Rightarrow 15 b^{2} + 12 b^{2} + 20 b^{2} = 188$ $\Rightarrow 47 b^{2} = 188; \to {\begin{cases} b = 2 \\ b = - 2 \end{cases}$ $c a s e (1) b = 2 \to {\begin{cases} x = 6 \\ y = 10 \\ z = 8 \end{cases}$ $c a s e (2) b = - 2 \to {\begin{cases} x = - 6 \\ y = - 10 \\ z = - 8 \end{cases}$
	Commented by bemath last updated on 19/Aug/20

	$j o o s s$
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