Solve the system:

x^2 + (y + 3)^2 = 25

x - y = 4

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`x^2+(y+3)^2=25` (i)

`x-y=4`

`x=y+4` (ii)

substitute value of x fro (ii) in (i)

`(y+4)^2+(y+3)^2=25`

`y^2+16+8y+y^2+6y+9=25`

`2y^2+14y=0`

`2y(y+7)=0`

`y=0`

`or`

`y=-7`

substitute y=0,-7 in (ii)

if y=0 then x=4

y=-7 then x=-7+4=-3

**Thus solutions are (4,0),(-3,-7).**

`x^2 + (y + 3)^2 = 25` --- (i)

and `x-y=4` --- (ii)

From (ii), `x=y+4`

Substitute this value of x in (i),

`(y+4)^2 + (y + 3)^2 = 25`

`rArr y^2+8y+16+ y^2+6y+9=25`

`rArr 2y^2+14y=25-16-9=0`

`rArr y^2+7y=0`

`rArr y(y+7)=0`

Put each of the factors equal to zero.

`y=0 ` and

y+7=0, i.e. `y=-7`

Agein from (ii), when `y=0, x=4 `

and when` y=-7, x=-7+4=-3`

So, the possible solutions of the set of equations are:

`x=4, y=0` and

`x=-3, y=-7` .

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