Question

If 4x-3y-2z = 0 and 2x + 5y-3z = 0 then what is the value of x: y: z​

Answer 1

Answer:

Step-by-step explanation:

0=0

4x-3y-2z = 2x+5y-3z

2x-8y+z =0

2x+z = 8y -----(1)

2(4x-3y-2z) -4(2x+5y-3z)=0

=>8x -6y -4z -8x -20y +12z=0

=>8z -26y=0

8z = 26y

4z = 13y

z =13y/4 -------(2)

Putting value of z (2) in eq (1)

2x +13y/4 = 8y

2x = 32y -13y /4

2x = 19y/4

x = 19y/8 ---------(3)

Puttinf value of xyz in

4x-3y-2z = 0

19y/2 - 3y - 13y/2 = 0

19y -6y -13y =0

First set of values is

x=y=z=0

Answer 2

Given,

$4x-3y-2z=0 ---- > (1)\\2x+5y-3z =0---- > (2)$

by solving $(1)$$-2*(2)$ we get

$4x-3y-2z-[2*(2x+5y-3z)]=0\\4x-3y-2z-4x-10y+6z=0\\-13y+4z=0\\13y=4z\\ \\y = \frac{4z}{13}$

by substituting the value of 'y' in equation($1$) we get

$4x-(3*\frac{4z}{13})-2z=0\\ \\4x-\frac{12z}{13}-2z=0\\ \\4x-[\frac{12z}{13}+2z]=0\\ \\4x-[\frac{12z+26z}{13}]=0\\ \\4x-[\frac{38z}{13}]=0\\ \\ 4x=[\frac{38z}{13}]\\ \\x=[\frac{38z}{13*4}] \\ \\\\x=\frac{38z}{52}$

Now, the ratio between x,y,z is

$x:y:z = \frac{38z}{52}:\frac{4z}{13}:z$

by taking the LCM we get

$x:y:z=\frac{38}{52}:\frac{4}{13}:1\\ \\x:y:z = \frac{38}{4}:4:13\\ \\x:y:z = 38:16:52$

∴The value of $x:y:z = 38:16:52$

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