Question

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Algebraically solve the following system of equations.


(x+2)2+(y-3)2=16
x+y-1=0

Answer 1

`(-2-2√(2),3+2√(2))\,,\,(-2+2√(2),3-2√(2))`

Explanation:

Given:

`(x+2)^2+(y-3)^2=16\\x+y-1=0`

To find: value of
`x,y`

Solution:

`(x+2)^2+(y-3)^2=16\,\,\,...(i)\\x+y-1=0\,\,\,(ii)`

From (ii),

`x=1-y`

Put this value of
`x` in (i)

`(1-y+2)^2+(y-3)^2=16\\(3-y)^2+(y-3)^2=16\\(y-3)^2+(y-3)^2=16\\2(y-3)^2=16\\(y-3)^2=8`

`y-3=` ±
`√(8)` = ±
`2√(2)`

`y=3` ±
`2√(2)`

At
`y=3` +
`2√(2)` ,

`x=1-(3+2√(2))=1-3-2√(2)=-2-2√(2)`

At
`y=3-2√(2)` ,

`x=1-(3-2√(2))=1-3+2√(2)=-2+2√(2)`

Solutions are
`(-2-2√(2),3+2√(2))\,,\,(-2+2√(2),3-2√(2))`