x^2+y=5, x^2+y^2=7

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

☐²

x^☐

√☐

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□□

log_□

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∫

ddx

≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

simplify solve for inverse tangent line

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x²+y=5, x²+y²=7

Solution

(

`x`=√6,	`y`=−1
`x`=−√6,	`y`=−1
`x`=√3,	`y`=2
`x`=−√3,	`y`=2

)

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Solution steps

One step at a time

[

x²+y=5

x²+y²=7

]

Subtract the equations

x²+y=5

−

x²+y²=7

x²+y−(x²+y²)=5−7

Simplify

−y²+y=−2

Solve −y²+y=−2: y=−1, y=2

Plug the solutions y=−1, y=2 into x²+y=5

For x²+y=5, subsitute y with −1: x=√6, x=−√6

For x²+y=5, subsitute y with 2: x=√3, x=−√3

Verify solutions by plugging them into the original equations

Therefore, the final solutions for x²+y=5, x²+y²=7 are

(

`x`=√6,	`y`=−1
`x`=−√6,	`y`=−1
`x`=√3,	`y`=2
`x`=−√3,	`y`=2

)

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

x^2+y=5, x^2+y^2=7

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