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Solve the system: x^2 + (y + 3)^2 = 25 x - y = 4
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High School Teacher
substitute value of x fro (ii) in (i)
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).
Posted by aruv on July 13, 2013 at 4:29 AM (Answer #1)
`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`
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` .
Posted by llltkl on September 27, 2013 at 3:54 AM (Answer #1)
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