if 5+root3/5-root3=x-root15 y, then x+y=?
Answers
Answer:
The value of x+y = \frac{14-\sqrt{5}}{11}
11
14−
5
.
Given equation is \frac{5+\sqrt{3}}{5-\sqrt{3}}
5−
3
5+
3
= x-y√15
We have to find the value of x+y
We solve the equality first
\frac{5+\sqrt{3}}{5-\sqrt{3}}
5−
3
5+
3
= x-y\sqrt{15}x−y
15
We solve the Left Hand side of the equation,
\frac{5+\sqrt{3}}{5-\sqrt{3}}
5−
3
5+
3
Multiplying numerator and denominator with the conjugate of the denominator,
= \frac{5+\sqrt{3}}{5-\sqrt{3}}
5−
3
5+
3
× \frac{5+\sqrt{3}}{5+\sqrt{3}}
5+
3
5+
3
= \frac{(5+\sqrt{3})^2}{(5-\sqrt{3})(5+\sqrt{3})}
(5−
3
)(5+
3
)
(5+
3
)
2
= \frac{(5+\sqrt{3})^2}{5^2 - {(\sqrt{3})}^2}
5
2
−(
3
)
2
(5+
3
)
2
= \frac{25+10\sqrt{3}+3}{25 - 3}
25−3
25+10
3
+3
= \frac{28+10\sqrt{3}}{22}
22
28+10
3
= \frac{28}{22} + \frac{10\sqrt{3}}{22}
22
28
+
22
10
3
Now equating with the Right hand side of the equation,
\frac{28}{22} + \frac{10\sqrt{3}}{22}
22
28
+
22
10
3
= x-y\sqrt{15}x−y
15
Therefore, x = \frac{28}{22}
22
28
and -y√15 = \frac{10\sqrt{3}}{22}
22
10
3
x = \frac{14}{11}
11
14
and y√15 = - \frac{\sqrt{5}\sqrt{15}}{11}
11
5
15
x = \frac{14}{11}
11
14
and y = - \frac{\sqrt{5}}{11}
11
5
Now, x+y = \frac{14-\sqrt{5}}{11}
11
14−
5