Question

If $\frac{x+1}{4}$ , $\frac{y+2}{3}=1$ and $\frac{1-x}{5}+\frac{3-y}{4} =-\frac{13}{20}$, find the value of $\frac{xy+3}{y}$

Answer 1

Given, pair of linear equations:

x/3 + y/4 = 11 ………..(1)

5x/6 - y/3 = - 7 …………(2)

Substitution method is used to solve this Linear pair of Equations.

This equations can be written as :

From eq 1 :

(4x + 3y)/12 = 11

4x + 3y = 11 × 12

4x + 3y = 132 ………(3)

From eq 2 :

(5x - 2y)/6 = - 7

5x - 2y = -7 × 6

5x - 2y = - 42 ………(4)

From eq (3) :

4x = 132 - 3y

x = (132 - 3y)/4 ………..(5)

On Substituting the value of x in equation (4) we obtain :

5x - 2y = - 42

5(132 - 3y)/4 - 2y = - 42

(660 - 15y)/4 - 2y = - 42

660 - 15y - 8y = - 42 × 4

660 - 23y = - 168

-23y = - 168 - 660

- 23y = - 828

y = 828/23

y = 36

On putting y = 36 in eq (5) we get,

x = (132 - 3y)/4

x = (132 - 3 × 36)/4

x = (132 - 108)/4

x = 24/4

x = 6

Hence the solution of the given system of equation is x = 6 and y = 36.

Hope this answer will help you…

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