Question

solve by cross Multiplication method :
3x + 14y=5xy,21y-x=2xy.​

Answer 1

Step-by-step explanation:

3x+14y=5xy......(1)

21y-x=2xy.... Rearranging

-x+21y=2xy .....(2)

Multiplying equ (1) by 21 & equ (2) by 14, we get

63x+294y=105xy.....(3)

-14x+294y=28xy......(4)

Subtracting equ (3) & equ (4),

63x+294y=105xy

- -14x+294y=28xy

77x+0y=77xy

77x=77xy

=77xy

77x

y=1

Substituting the value of y=1 in equ (1),

3x+14y=5xy

3x+14(1)=5x(1)

3x+14=5x

5x - 3x = 14

2x= 14

x = 14

2

x= 7

Therefore, (x,y) = (7,1)

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Answer 2

Step-by-step explanation:

$\huge \boxed{\underbrace{\mathfrak {\underline {\fcolorbox {black}{pink}{Answer}}}}}$

${ 3x+14y=5xy ........(1) }$

${ 21y-x=2xy }$

$\implies { -x+21y=2xy.......(2) }$

multiplying equation1 by 21 and equation2 by 14,We get....

${ 63x+294y=105xy.............(3) }$

${ -14x+294y=28xy...............(4) }$

solving the two equation3 and 4, we got that..

$\implies { y=1 }$

then put the value ${ y=1 }$In equation 1

${ 3x+14y=5xy }$

$\implies {3x+14=5x }$

$\implies { 5x-3x=14 }$

$\implies { 2x=14 }$

$\implies { x=\frac{14}{2} }$

$\implies { x=7 }$

we got the value ${(x,y ) }$=${( 7,1 ) }$