Question

x/4+2y/3=7,x/6+3y/5=11
solve by both substitution and elimination...
$\frac{x}{4} + \frac{2y}{3} = 7 \\ \frac{x}{6} + \frac{3y}{5} = 11$
find x and y

plz answer correctly with explanation
10 points for each who answers the question... ​

Answer 1

hey sis your answer is in the given attachment

HOPE IT IS HELPFUL

Answer 2

Answer:

The solution of the equations is

$x=-\frac{564}{7},y=\frac{285}{7}$

Step-by-step explanation:

The given linear equation in two variables are

$\frac{x}{4} +\frac{2y}{3} =7--- > (1)\\\frac{x}{6}+ \frac{3y}{5} =11--- > (2)$

1.Substitution method

In this method,we shall find the value of one variable in terms of other and substitute it in other equation as shown.

$\frac{x}{4} +\frac{2y}{3} =7= > x=4(7-\frac{2y}{3})$

Substituting this value of x in equation(2),we get

$\frac{4(7-\frac{2y}{3}) }{6}+ \frac{3y}{5} =11--- > (2)\\= > \frac{14}{3} -\frac{4y}{9} +\frac{3y}{5}=11= > -\frac{19}{3}+\frac{7y}{45} =0\\= > y=\frac{19\times15}{7} =\frac{285}{7}$

Hence the value of x becomes

$x=4(7-\frac{2y}{3})=4(7-\frac{2}{3} \times \frac{285}{7} )=-\frac{664}{7}$

Hence the solution of the equations is

$x=-\frac{564}{7},y=\frac{285}{7}$

2.Elimination method

Multiplying equation(1) with 4 and equation(2) with 6,we get

$x+\frac{8y}{3}=28-- > (3)\\x+\frac{18y}{5}=66-- > (4)$

Subtracting equation(3) from equation (4),we get

$\frac{18y}{5}- \frac{8y}{3} =66-28=38\\= > \frac{14y}{15} =38= > y=\frac{15\times19}{7} =\frac{285}{7}$

Substituting the value of y in equation(3),we get

$x+\frac{8(\frac{285}{7})}{3}=28= > x=28-\frac{760}{7} =\frac{-564}{7}$

Hence the solution of the equations is

$x=-\frac{564}{7},y=\frac{285}{7}$

#SPJ2