Question

plesae solve the questions attachment is qiven of linear equatoin

Answer 1

Hi friend, ashish here.

Here is your answer:

Given that,

\frac{\sqrt{11}- \sqrt{7} }{\sqrt{11}+\sqrt{7}} = a- b \sqrt{77}11​+7​11​−7​​=a−b77​ 

To find,  

The value of a & b.

Solution:

First we must rationalize the denominator of the given number by multiplying and dividing with it's conjugate.

Conjugate of denominator is √11 - √7.

Then,

 \frac{\sqrt{11}- \sqrt{7} }{\sqrt{11}+ \sqrt{7} } \times \frac{\sqrt{11}- \sqrt{7} }{\sqrt{11}- \sqrt{7} } = \frac{(\sqrt{11}- \sqrt{7} )^{2} }{(\sqrt{11}- \sqrt{7})(\sqrt{11}+ \sqrt{7}) }11​+7​11​−7​​×11​−7​11​−7​​=(11​−7​)(11​+7​)(11​−7​)2​ 

Here in the denominator to multiply we can use the identity:

(x+y)(x-y) = x² - y².

Then,

(√11 + √7) (√11 - √7) = (√11)² - (√7)² = 11 - 7 = 4

So,

\frac{(\sqrt{11}- \sqrt{7} )^{2} }{(\sqrt{11}- \sqrt{7})(\sqrt{11}+ \sqrt{7}) } = \frac{11+7-2( \sqrt{11})( \sqrt{7}) }{4} = \frac{18 - 2 \sqrt{77} }{4}(11​−7​)(11​+7​)(11​−7​)2​=411+7−2(11​)(7​)​=418−277​​

⇒  \frac{18 - 2 \sqrt{77} }{4} = a-b \sqrt{77}418−277​​=a−b77​ 

⇒  \frac{18}{4} - \frac{2 \sqrt{77} }{4} = a-b \sqrt{77}418​−4277​​=a−b77​ 

Now by comparing the above equation , We get:

a= \frac{18}{4} = \frac{9}{2} : b= \frac{2}{4} = \frac{1}{2}a=418​=29​:b=42​=21​ 

Hence these are the values of a & b.
__________________________________________________

Hope my answer is helpful to you

Answer 2

Step-by-step explanation:

Refer To Attachment

hope it's help you