solve x and y :- 2/x+y + 3/x-y=1 , 8/x+y - 7/x-y = 5/6 explain step by step
Answers
\begin{gathered} \frac{2}{x + y} + \frac{3}{x - y} = 1 \\ \frac{8}{x + y} - \frac{7}{x - y} = \frac{5}{6} \\ \\ Let, \: \: \frac{1}{x + y} \: \: be \: \: u \: \: and \: \: \frac{1}{x - y} \: \: be \: \: v \\ \\ 2u + 3v = 1 \: \: \: \: \: \: (1) \\ 8u - 7v = \frac{5}{6} \: \: \: \: \: (2) \\ \\ 2u + 3v = 1 \\ 2u = 1 - 3v \\ u = \frac{1 - 3v}{2} \\ \\ 8u - 7v = \frac{5}{6} \\ 8( \frac{1 - 3v}{2} ) - 7v = \frac{5}{6} \\ \frac{8 - 24v}{2} - 7v = \frac{5}{6} \\ \frac{8 - 24v - 14v}{2} = \frac{5}{6} \\ \frac{8 - 38v}{2} = \frac{5}{6} \\ 6(8 - 38v) = 2 \times 5 \\ 48 - 228v = 10 \\ 48 - 10 = 228v \\ \frac{38}{228} = v \\ \boxed{ \underline {\underline{ \frac{1}{6}}} = v }\\ \\ 2u + 3v = 1 \\ 2u + 3( \frac{1}{6} ) = 1 \\ 2u + \frac{1}{2} = 1 \\ 2u = 1 - \frac{1}{2} \\ 2u = \frac{2 - 1}{2} \\ u = \frac{1}{2} \div 2 \\ u = \frac{1}{2} \times \frac{1}{2} \\ \boxed{u = \underline {\underline{ \frac{1}{4} }}} \\ \\ \frac{1}{x + y} = u \\ \frac{1}{x + y} = \frac{1}{4} \\ \boxed{ \boxed{4 = x + y}} \\ \\ \frac{1}{x - y} = v \\ \frac{1}{x - y} = \frac{1}{6} \\ \boxed{ \boxed{6 = x - y}} \\ 6 + y = x \\ \\ x + y = 4 \\ (6 + y) + y = 4 \\ 6 + y + y = 4 \\ 6 + 2y = 4 \\ 2y = 4 - 6 \\ y = \frac{ - 2}{2} \\ \boxed{ \boxed{y = \underline{ \underline{- 1}}}} \\ \\ x - y = 6 \\ x - ( - 1) = 6 \\ x + 1 = 6 \\ x = 6 - 1 \\ \boxed{ \boxed{x = \underline{ \underline{5}}}}\end{gathered}
x+y
2
+
x−y
3
=1
x+y
8
−
x−y
7
=
6
5
Let,
x+y
1
beuand
x−y
1
bev
2u+3v=1(1)
8u−7v=
6
5
(2)
2u+3v=1
2u=1−3v
u=
2
1−3v
8u−7v=
6
5
8(
2
1−3v
)−7v=
6
5
2
8−24v
−7v=
6
5
2
8−24v−14v
=
6
5
2
8−38v
=
6
5
6(8−38v)=2×5
48−228v=10
48−10=228v
228
38
=v
6
1
=v
2u+3v=1
2u+3(
6
1
)=1
2u+
2
1
=1
2u=1−
2
1
2u=
2
2−1
u=
2
1
÷2
u=
2
1
×
2
1
u=
4
1
x+y
1
=u
x+y
1
=
4
1
4=x+y
x−y
1
=v
x−y
1
=
6
1
6=x−y
6+y=x
x+y=4
(6+y)+y=4
6+y+y=4
6+2y=4
2y=4−6
y=
2
−2
y=
−1
x−y=6
x−(−1)=6
x+1=6
x=6−1
x=
5
Step-by-step explanation:
x = 5
y = -1