Question

(z+5)/6 - (z+1)/9 = (z+3)/4 linear equation..
plz solve this question.....​

Answer 1

Hope this given attachment helps you

Answer 2

★Question given :

$\mapsto\tt\dfrac{z+5}{6}-\dfrac{z+1}{9}=\dfrac{(z+3)}{4}$

★Solution :

$\mapsto\tt\dfrac{3(z+5)-2(z+1)}{18}=\dfrac{(z+3)}{4}$(Taking the lcm of 6 & 9=18)

$\mapsto\tt\dfrac{3z+15-2z-2}{18}=\dfrac{(z+3)}{4}$

$\mapsto\tt\dfrac{3z-2z+15-2}{18}=\dfrac{(z+3)}{4}$

$\mapsto\tt\dfrac{z+13}{18}=\dfrac{(z+3)}{4}$

$\mapsto\sf{18(z+3)=4(z+13)}$

[By cross multiplication]

$\mapsto\sf{18z+54=4z+52}$

$\mapsto\sf{18z-4z=52-54}$

$\mapsto\sf{14z=-2}$

$\mapsto\sf{z=}\cancel\dfrac{-2}{14^7}^{-1}$

$\mapsto\sf{z=}\dfrac{-1}{7}$

Thus,the value of $\sf\red{z=}\dfrac{-1}{7}$