Question

X+y,y+z,z+x In the ratio of 6:7:8 and x+y+z=14 then the valu of x

Answer 1

Answer:

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

Step-by-step explanation:

We have been given the ratio:

$\frac{x+y}{y+z}=\frac{6}{7}------(i) \\\frac{x+y}{z+x}=\frac{6}{8}$------(ii)

Find the equation from the (i)

$\frac{x+y}{y+z}=\frac{6}{7}\\\ 7x+7y=6y+6z\\7x+y=6z$----(iii)

Find the equation from the (ii)

$\frac{x+y}{z+x}=\frac{6}{8} \\ 8x+8y=6z+6x\\2x+8y=6z$------(iv)

According to equation (iii) and (iv) find the value of x by substitution method as shown below:

Firstly subtract equation (iii) from (iv)

$-5x+7y=0\\x=\frac{7y}{5}$

Substitute the value of x in equation (iii)

$7(\frac{7y}{5} )+y=6z\\z=\frac{9y}{5}$

Substitute the value of x and z values in the given equation $x+y+z=14$ so that we calculate the value of x.

$x+y+z=14\\\frac{7y}{5}+y+\frac{9y}{5}=14\\\frac{7y+5y+9y}{5} =14\\\frac{21y}{5}=14\\y=\frac{14*5}{21} \\y=\frac{10}{3}$

Substitute this above value of y in the value of x:

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

Thus the value of x is $\frac{14}{3}$.