intervalo log_{2}(2x)

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

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xy=12, 12x²+y²=156

Solution

(

`x`=1,	`y`=12
`x`=−1,	`y`=−12
`x`=2√3,	`y`=2√3
`x`=−2√3,	`y`=−2√3

)

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Solution steps

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[

xy=12

12x²+y²=156

]

Isolar x de xy=12: x=12y ; y≠ 0

Inserir as soluções x=12y em 12x²+y²=156

Para 12x²+y²=156, substituir x com 12y : y=12, y=−12, y=2√3, y=−2√3

Inserir as soluções y=12, y=−12, y=2√3, y=−2√3 em xy=12

Para xy=12, substituir y com 12: x=1

Para xy=12, substituir y com −12: x=−1

Para xy=12, substituir y com 2√3: x=2√3

Para xy=12, substituir y com −2√3: x=−2√3

Verificar soluções inserindo-as nas equações originais

Portanto, as soluções finais para xy=12, 12x²+y²=156 são

(

`x`=1,	`y`=12
`x`=−1,	`y`=−12
`x`=2√3,	`y`=2√3
`x`=−2√3,	`y`=−2√3

)

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

intervalo log_{2}(2x)

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