5 + root 3 by 5 -root 3 equal to x minus y root 15 ,then find the value of x+y
Answers
The value of x+y = .
- Given equation is = x-y√15
- We have to find the value of x+y
- We solve the equality first
=
- We solve the Left Hand side of the equation,
- Multiplying numerator and denominator with the conjugate of the denominator,
= ×
=
=
=
=
=
- Now equating with the Right hand side of the equation,
=
- Therefore, x = and -y√15 =
x = and y√15 = -
x = and y = -
- Now, x+y =
Step-by-step explanation:
The value of x+y = \frac{14-\sqrt{5}}{11}1114−5 .
Given equation is \frac{5+\sqrt{3}}{5-\sqrt{3}}5−35+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}}5−35+3 = x-y\sqrt{15}x−y15
We solve the Left Hand side of the equation,
\frac{5+\sqrt{3}}{5-\sqrt{3}}5−35+3
Multiplying numerator and denominator with the conjugate of the denominator,
= \frac{5+\sqrt{3}}{5-\sqrt{3}}5−35+3 × \frac{5+\sqrt{3}}{5+\sqrt{3}}5+35+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}52−(3)2(5+3)2
= \frac{25+10\sqrt{3}+3}{25 - 3}25−325+103+3
= \frac{28+10\sqrt{3}}{22}2228+103
= \frac{28}{22} + \frac{10\sqrt{3}}{22}2228+22103
Now equating with the Right hand side of the equation,
\frac{28}{22} + \frac{10\sqrt{3}}{22}2228+22103 = x-y\sqrt{15}x−y15
Therefore, x = \frac{28}{22}2228 and -y√15 = \frac{10\sqrt{3}}{22}22103
x = \frac{14}{11}1114 and y√15 = - \frac{\sqrt{5}\sqrt{15}}{11}11515
x = \frac{14}{11}1114 and y = - \frac{\sqrt{5}}{11}115
Now, x+y = \frac{14-\sqrt{5}}{11}1114−5