\[\frac{7}{2x + 1} + \frac{13}{y +2} = 27\]
and
\[\frac{13}{2x + 1} + \frac{7}{y + 2} = 33\]
Answer:
Let ,
\[\frac {1}{2x + 1}= m\] and \[\frac{1}{y + 2}= n\]
now, equations become,
7m + 13n = 27 ——–A
13m + 7n = 33 ——– B
adding A and B,
7m + 13n = 27
13m + 7n = 33
20m + 20n = 60
dividing above equation by 20 on both sides
now, m + n = 3 —————C
subtracting equation B from A
7m + 13n = 27
13m + 7n = 33
-6m + 6n = -6
Diving above equation by (-6 ) on both sides,
m – n = 1 ————-D
Solving equation C and D,
m + n = 3
m – n = 1
2m = 4
m =2
put m =2 in equation C,
m + n = 3
2 + n = 3
n = 3 – 2
n = 1
n = 1
we have
\[\frac {1}{2x + 1}= m\]
\[\frac {1}{2x + 1}= 2\]
2(2x + 1) = 1
4x + 2 = 1
4x = 1-2
4x = -1
\[x= \frac{-1}{4}\]
we have,
\[\frac{1}{y + 2}= n\]
\[\frac{1}{y + 2}= 1\]
y + 2= 1
y= 1 – 2
y = -1
Answer: \[x= \frac{-1}{4} and y= -1 \]