Question

x/3+y/4=11,5x/6-y/3=-7 solve in substitution method

Answer 1

→ refer to attachment

now,

put the value of x in equation (1)

y= 132-132 - 4{6})/3

= (132 - 24)/3

= 108/3

= 36

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Answer 2

Answer:

Solution of given system of equation is $x=\frac{138}{11}\:\:and\:\:y=\frac{300}{11}$

Step-by-step explanation:

Given System of equation are $\frac{x}{3}+\frac{y}{4}=11\:\:and\:\:\frac{x}{6}-\frac{y}{3}=-7$

We have to find solution by substitution method.

Consider,

$\frac{x}{3}+\frac{y}{4}=11$

$\frac{4x}{12}+\frac{3y}{12}=11$

$4x+3y=11\times12$

$4x+3y=132$  .................(1)

$\frac{x}{6}-\frac{y}{3}=-7$

$\frac{x}{6}-\frac{2y}{6}=-7$

$x-2y=-7\times6$

$x-2y=-42$ ...........(2)

from (1)

$4x=132-3y$

$x=\frac{132-3y}{4}$ .......(3)

Put this value of x in (2)

$\frac{132-3y}{4}-2y=-42$

$132-3y-8y=-42\times4$

$132-11y=-168$

$11y=132+168$

$y=\frac{300}{11}$

Now put this value of y in (3)

$x=\frac{132-3\times\frac{300}{11}}{4}$

$x=\frac{132\times11-3\times300}{4\times11}$

$x=\frac{1452-900}{44}$

$x=\frac{552}{44}$

$x=\frac{138}{11}$

Therefore, Solution of given system of equation is $x=\frac{138}{11}\:\:and\:\:y=\frac{300}{11}$