3/x+y+2/x-y=2 9/x+y-4/x-y=1 Solve from elimination method
Answers
Answer:
To find the value of x & y if,
\frac{3}{x+y} + \frac{2}{x-y} = 2
x+y
3
+
x−y
2
=2 ..(i)
&
\frac{9}{x+y} - \frac{4}{x-y} = 1
x+y
9
−
x−y
4
=1 ..(ii)
Solution :
Adding both the equations,.
⇒ (i) × 2 + (ii)
⇒ (\frac{3}{x+y} + \frac{2}{x-y})(2) + (\frac{9}{x+y} - \frac{4}{x-y}) = 2(2) + 1(
x+y
3
+
x−y
2
)(2)+(
x+y
9
−
x−y
4
)=2(2)+1
⇒ \frac{6}{x+y} + \frac{4}{x-y} + (\frac{9}{x+y} - \frac{4}{x-y}) = 5
x+y
6
+
x−y
4
+(
x+y
9
−
x−y
4
)=5
⇒ \frac{15}{x+y} = 5
x+y
15
=5
⇒ \frac{x + y}{15} =\frac{1}{5}
15
x+y
=
5
1
⇒ x + y =\frac{15}{5}x+y=
5
15
⇒ x + y =3x+y=3 ..(iii)
By substituting value of (iii) in (i),
We get,
⇒ \frac{3}{x+y} + \frac{2}{x-y} = 2
x+y
3
+
x−y
2
=2
⇒ \frac{3}{3} + \frac{2}{x-y} = 2
3
3
+
x−y
2
=2
⇒ 1 + \frac{2}{x-y} = 21+
x−y
2
=2
⇒ \frac{2}{x-y} = 2 - 1
x−y
2
=2−1
⇒ \frac{2}{x-y} = 1
x−y
2
=1
⇒ x-y = 2x−y=2 ...(iv)
By adding (iii) & (iv),
We get,
⇒ (iii) + (iv) ⇒ (x + y) + (x - y) = 3 + 2(x+y)+(x−y)=3+2
⇒ 2x= 52x=5
⇒ x = \frac{5}{2}x=
2
5
By subtracting (iv) from (iii),
We get,
⇒ (iii) - (iv) ⇒ (x + y) - (x - y) = 3 - 2(x+y)−(x−y)=3−2
⇒ 2y = 12y=1
⇒ y =\frac{1}{2}y=
2
1
∴ x = \frac{5}{2}x=
2
5
∴ y =\frac{1}{2}y=
2
1