2^x+y=2^x-y=root8
find the value of y=?
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HEY MATE HERE IS YOUR ANSWER
2(x + y) = \sqrt{8} \\ 2x + 2y = 2 \sqrt{2} = = = > (1)eq
2(x - y) = \sqrt{8} \\ 2x - 2y = 2 \sqrt{2} = = = > (2)eq
by adding 1 eq and 2 eq
2x + 2y + 2x - 2y= 2 \sqrt{2} + 2 \sqrt{2} \\ 4x = 4 \sqrt{2} \\ x = \sqrt{2}
substitute the value of x in 1 or 2 eq
2( \sqrt{2} ) + 2y = 2 \sqrt{2} \\ 2y = 2 \sqrt{2} - 2 \sqrt{2} \\ y = 0
Step-by-step explanation:
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