Question

4.
If x+y, yuz, z+x are in the ratio 6:7:8 and x + y + z =14 then the value of x is
(a) 6
(b) 7
(c) 8
(d) 10​

Answer 1

Answer:

x= $\frac{14}{3}$

Step-by-step explanation:

We are given that

$(x+y):(y+z):(z+x)=6:7:8$

$x+y+z=14$

We have to find the value of x.

According to question

$\frac{x+y}{y+z}=\frac{6}{7}$

$7x+7y=6y+6z$

$7x+7y-6y=6z$

$7x+y=6z$  (equation I)

$\frac{x+y}{z+x}=\frac{6}{8}$

$8x+8y=6z+6x$

$8x-6x+8y=6z$

$2x+8y=6z$  (equation II)

Subtract equation I from equation II

$-5x+7y=0$

$7y=5x$

$x=\frac{7}{5}y$

Substitute the value of y in equation I

$\frac{7\times 7y}{5}+y=6z$

$\frac{49y+5y}{5\times 6}=z$

$z=\frac{9y}{5}=\frac{9y}{5}$

Substitute the values

$\frac{7y}{5}+y+\frac{9y}{5}=14$

$\frac{7y+5y+9y}{5}=14$

$\frac{21y}{5}=14$

$y=\frac{14\times 5}{21}=\frac{10}{3}$

Substitute the value

$x=\frac{7\times 10}{5\times 3}=\frac{14}{3}$

Hence, the value of x= $\frac{14}{3}$