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Solve the system:  (x-1)^2 + y^2 = 25 x-y^2 = -4

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kristenmarieb... | Student, Grade 10 | Valedictorian

Posted July 11, 2013 at 4:47 AM via web

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Solve the system: 

(x-1)^2 + y^2 = 25

x-y^2 = -4

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llltkl | College Teacher | Valedictorian

Posted July 11, 2013 at 5:42 AM (Answer #2)

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Let `(x-1)^2 + y^2 = 25` be equation (i)

`x-y^2 = -4`   .........equation (ii)

From (ii) we get:

`y^2=x+4`

Substituting the value of `y^2` in eq. (i) we get:

`(x-1)^2 + x+4 = 25`

`rArr x^2-2x+1+x+4=25`

`rArr x^2-x-20=0`

Now factorize by splitting the middle term,

`x^2-5x+4x-20=0`

`rArr x(x-5)+4(x-5)=0`

`rArr (x-5)(x+4)=0`

Using zero product property:

For (x-5)=0 we get x=5

For (x+4)=0 we get x=-4

Now substitute each value of x in eq.(ii) to obtain the corresponding values of y:

For x=5,   `y^2=5+4=9`     ` rArr y=+-3`

For x=-4   `y^2=-4+4=0`      ` rArr y=0`

Therefore, the solution set is `{(5,+-3),(-4,0)}` .

 

Sources:

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msnavalta | eNotes Newbie

Posted July 11, 2013 at 4:56 AM (Answer #1)

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Isolate and substitute.

Isolate y^2, so you get y^2 = x+4

Substitute y^2 into the other equation, so you get:  (x-1)^2 + x + 4 = 25

Simplify:

x^2 - x - x + 1 + x + 4 = 25

x^2 - x + 5 = 25

x^2 - x - 20 = 0

Factor:  (x-5) (x+4) = 0

x-5=0, so x=5

x+4=0, so x= -4

Solutions: x=5, x=-4

 

 

 

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