Math, asked by raih87261, 4 days ago

√2x+3y=5 and x-√3y=11 solve by substitution method.
 \sqrt{2}  + 3y = 5 \\ x -  \sqrt{3}y = 11

Answers

Answered by jahnavisuthar7
9

I have done half rest you should do

Attachments:
Answered by mathdude500
31

Answer:

 \sf \:\qquad \:\boxed{\begin{aligned}& \qquad \:\bf \: x  = \dfrac{33 + 5 \sqrt{3}}{3 +  \sqrt{6} }\qquad \: \\ \\& \qquad \:\bf \:  y = \dfrac{5 - 11\sqrt{2}}{3 +  \sqrt{6}} \end{aligned}} \qquad  \\  \\

Step-by-step explanation:

Consider,

\sf \:  \sqrt{2}x + 3y = 5 \\  \\

\sf \: 3y = 5  -  \sqrt{2}x \\  \\

\sf\implies y = \dfrac{5 -  \sqrt{2} x}{3}   \:  \:  \:  \: -  -  - (1) \\  \\

Now, Consider

\sf \: x -  \sqrt{3}y = 11 \\  \\

On substituting the value of y from equation (1), we get

\sf \: x -  \sqrt{3}\bigg(\dfrac{5 -  \sqrt{2}x}{3} \bigg)  = 11 \\  \\

\sf \: x -\dfrac{5 \sqrt{3}  -  \sqrt{6}x}{3}  = 11 \\  \\

\sf \: \dfrac{3x - 5 \sqrt{3} +\sqrt{6}x}{3}  = 11 \\  \\

\sf \: 3x - 5 \sqrt{3} +  \sqrt{6}x= 33 \\  \\

\sf \: 3x +\sqrt{6}x= 33 + 5 \sqrt{3}  \\  \\

\sf \: (3 +\sqrt{6})x= 33 + 5 \sqrt{3}  \\  \\

\sf\implies \sf \: x=  \dfrac{33 + 5 \sqrt{3}}{3 +  \sqrt{6} } \\  \\

On substituting the value of x in equation (1), we get

\sf \:  y = \dfrac{5 -  \sqrt{2}  \times \dfrac{33 + 5 \sqrt{3}}{3 +  \sqrt{6} }}{3} \\  \\

\sf \:  y = \dfrac{5 -   \dfrac{33 \sqrt{2}  + 5 \sqrt{6}}{3 +  \sqrt{6} }}{3} \\  \\

\sf \:  y = \dfrac{\dfrac{15 + 5 \sqrt{6} -  33 \sqrt{2} -  5 \sqrt{6}}{3 +  \sqrt{6} }}{3} \\  \\

\sf \:  y = \dfrac{15 - 33 \sqrt{2} }{3(3 +  \sqrt{6} )} \\  \\

\sf \:  y = \dfrac{3(5 - 11\sqrt{2} )}{3(3 +  \sqrt{6} )} \\  \\

\sf\implies \sf \:  y = \dfrac{5 - 11\sqrt{2}}{3 +  \sqrt{6}} \\  \\

Hence,

\sf\implies \sf \:\qquad \:\boxed{\begin{aligned}& \qquad \:\bf \: x  = \dfrac{33 + 5 \sqrt{3}}{3 +  \sqrt{6} }\qquad \: \\ \\& \qquad \:\bf \:  y = \dfrac{5 - 11\sqrt{2}}{3 +  \sqrt{6}} \end{aligned}} \qquad  \\  \\

Verification

Consider,

\sf \:  \sqrt{2}x + 3y = 5 \\  \\

On substituting the value of x and y, we get

\sf \:  \sqrt{2}\bigg(\dfrac{33 + 5 \sqrt{3}}{3 +  \sqrt{6} }\bigg) + 3\bigg( \dfrac{5 - 11\sqrt{2}}{3 +  \sqrt{6}} \bigg)  = 5 \\  \\

\sf \:  \dfrac{33 \sqrt{2}  + 5 \sqrt{6}}{3 +  \sqrt{6} } +  \dfrac{15 - 33\sqrt{2}}{3 +  \sqrt{6}}   = 5 \\  \\

\sf \:  \dfrac{33 \sqrt{2}  + 5 \sqrt{6} + 15 - 33 \sqrt{2} }{3 +  \sqrt{6} }   = 5 \\  \\

\sf \:  \dfrac{ 5 \sqrt{6} + 15 }{3 +  \sqrt{6} }   = 5 \\  \\

\sf \:  \dfrac{ 5 (\sqrt{6} + 3) }{3 +  \sqrt{6} }   = 5 \\  \\

\sf\implies  \: 5 = 5 \\  \\

Hence, Verified

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