Question

If x+y, y+z, z+x, is in the ratio of 6:7:8 and x+y+z=14 then find the value of x

Answer 1

Answer:

x = 14/3

y = 10/3

z = 6

Step-by-step explanation:

Assume:

x + y = 6k

y + z = 7k

x+ z = 8k

=> 2(x+y+z) = 21k = 2* 14 = 28

=> k = 28/21 = 4/3

x + y + x + z = 14k

=> x + 14 = 14k = 56/3

=> x = 56/3 - 14 = 14/3

Answer 2

$\displaystyle\large\boxed {\sf {x=\dfrac {14}{3}}}$

Let,

• $\displaystyle\sf {x+y=6k\quad\quad\dots (1)}$

• $\displaystyle\sf {y+z=7k\quad\quad\dots (2)}$

• $\displaystyle\sf {z+x=8k\quad\quad\dots (3)}$

On adding these (1), (2) and (3),

$\displaystyle\longrightarrow\sf {(x+y)+(y+z)+(z+x)=6k+7k+8k}$

$\displaystyle\longrightarrow\sf {2(x+y+z)=21k\quad\quad\dots (4)}$

But given that,

• $\displaystyle\sf {x+y+z=14\quad\quad\dots (5)}$

Thus (4) becomes,

$\displaystyle\longrightarrow\sf {21k=2\times 14}$

$\displaystyle\longrightarrow\sf {21k=28}$

$\displaystyle\longrightarrow\sf {k=\dfrac {4}{3}\quad\quad\dots (6)}$

Substituting (6) in (2),

$\displaystyle\longrightarrow\sf {y+z=7\times\dfrac {4}{3}}$

$\displaystyle\longrightarrow\sf {y+z=\dfrac {28}{3}\quad\quad\dots(7)}$

Hence, on subtracting (7) from (5),

$\displaystyle\longrightarrow\sf {(x+y+z)-(y+z)=14-\dfrac {28}{3}}$

$\displaystyle\longrightarrow\sf {\underline {\underline {x=\dfrac {14}{3}}}}$