Question

7/2x+1+13/y+2=27 ; 13/2x+1+7/y+2=33 Solve the simultaneous equation

Answer 1

Solution:-

given by simultaneous equations:-
$7 \times ( \frac{1}{2x + 1} ) + 13 \times( \frac{1}{y + 2} ) = 27 \\ let \: assume \: ( \frac{1}{2x + 1} )as \: m \: and \: ( \frac{1}{y + 2} ) \: as \: n \: we \: get \\ 7m + 13n = 27..............(1) \\ 13 \times ( \frac{1}{2x + 1} + 7 \times ( \frac{1}{y + 2} ) = 33 \\ 13m + 7n = 33............(2) \\ multipy \: in \: eq(1)with \: 13 \: and \: in \: eq(2)with \: 7 \\ 91m + 169n = 351........(3) \\ 91m + 49n = 231.........(4) \\ substract \: eq \: (4)in \: eq \: (3) \: we \: get \\ 120n = 120 \\ n = 1 \\ put \: n = 1 \: in \: eq(1) \\ 7m + 13 \times 1 = 27 \\ 7m = 14 \\ m = 2 \\ now \\ \: m \: = 2 =( \frac{1}{2x + 1} ) \\ 4x + 2 = 1 \\ x = \frac{ - 1}{4} \\ n = 1 = ( \frac{1}{y + 2} ) \\ y + 2 = 1 \\ y = - 1$
here (x , y) = (-1/4 ,-1)

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Answer 2

7/(2x + 1) + 13/(y + 2) = 27
13/(2x + 1) + 7/(y + 2) = 33


$\left[\begin{array}{cc}p = \frac{1}{2x + 1} , q = \frac{1}{y + 2}\end{array}\right$

7p + 13q = 27-------------( 1 )
13p + 7q = 33-------------( 2 )

adding-----------( 1 ) & -----------( 2 )

7p + 13q = 27
13p + 7q = 33
–····················–
20p + 20q = 60

p + q = 3-------------( 3 )

Subtracting-------( 1 ) & ----------( 2 )

7p + 13q = 27
13p + 7 q = 33
–······················–
-6p + 6q = -6

p - q = 1------------( 4 )


from---------( 3 ) & -----------( 4 )

p + q = 3
p - q = 1
–··········–
2p = 4

p = 2 [ put in -----( 1 ) ]

2 + q = 3

q = 1 , p = 2

p = 2 = 1/(2x + 1)

4x + 2 = 1
x = -1/4

q = 1 = 1/(y + 2)

y + 2 = 1

y = -1 , x = -1/4