Question

If x^2+y^2+z^2-xy-yz-zx=0 find the value of 2x + y:3x+4y​

Answer 1

answer 3:7

Step-by-step explanation:

x^2+y^2+z^2-xy-yz-zx=0

or,2x^2+2y^2+2z^2-2xy-2yz-2zx=0

or,x^2+y^2-2xy+y^2+z^2-2yz+z^2+x^2-2zx=0

or,(x-y)^2+(y-z)^2+(z-x)^2=0

so x-y=0,y-z=0,z-x=0

therefore x=y,y=z,z=x

so x=y=z=k(,say)

now,(2x+y):(3x+4y)=(2k+k):(3k+4k)

=3k:7k

=3:7

Answer 2

Given :

${x}^{2} + {y}^{2} + {z}^{2} - xy - yz - zx = 0 \: \: \: \: \: \: \: \: - - - (1)$

To find :

$2x + y</u><u> </u><u>:</u><u> </u><u>3x + 4y = </u><u>?</u><u> </u><u>$

Solution :

Multiply both side of equation ( 1 ) by 2 .

$2( {x}^{2} + {y}^{2} + {z}^{2} - xy - yz - zx) = 2 \times 0 \\ \\ 2 {x}^{2} + 2 {y}^{2} + 2 {z}^{2} - 2xy - 2yz - 2zx = 0$

By rearranging the terms :

$( {x}^{2} + {y}^{2} - 2xy) + ( {y}^{2} + {z}^{2} - 2yz) + ( {x}^{2} + {z}^{2} - 2zx) = 0$

By using identity ;

${a}^{2} + {b}^{2} -2ab ={ (a - b )}^{2}$

We get ,,,,

${(x - y)}^{2} + {(y - z)}^{2} + {(x - z)}^{2} = 0 \\ \\ = = > \: \: \: \: \: {(x - y)}^{2} = 0 \\ \: \:x - y = \sqrt{0} \\ \: \: x - y = 0 \\ \: \: x = y$

${lll}^{ly} .....x = z \: \: \: \: \: and \: \: \: \: \: \: y = z$

==> x = y = z

Let x = y = z = k where k is any real number

$2x + y</strong><strong>:</strong><strong>3x + 4y = \frac{2x + y}{3x + 4y} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{2k + k}{3k + 4k} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{3k}{7k} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{3}{7}$

2x + y : 3x + 4y = 3 : 7

Hope it's helpful .....

Please mark it as Brainliest ..... ❤