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# Solve the system: x^2 - y = 1 2x^2 + y^2 = 17

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Solve the system:

x^2 - y = 1

2x^2 + y^2 = 17

Posted by kristenmariebieber on July 12, 2013 at 12:04 AM via web and tagged with algebra2, math

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

`2x^2 + y^2 = 17` .................(ii)

From eq. (i) we get:

`x^2=y+1`

Substituting the value of `x^2` in eq. (ii)we get:

`2(y+1)+y^2=17`

`rArr 2y+2+y^2-17=0`

`rArr y^2+2y-15=0`

Now factorizing by splitting the middle term:

`y^2+5y-3y-15=0`

`rArr y(y+5)-3(y+5)=0`

`rArr (y+5)(y-3)=0`

Using the zero product property:

For (y+5)=0 we get y=-5

For (y-3)=0 we get y=3

Now, substitute each value of y in eq. (i) to get the corresponding values of x. So,

when y=-5  ` x^2=-5+1=-4`     `rArr x=sqrt(-1)*sqrt4=2i`    `(sqrt(-1)=i)`

when y=3      ` x^2=3+1=4`        `rArr x=sqrt4=+-2`

Therefore, the soluton set is `{(2i,-5),(+-2,3)}` .

Posted by llltkl on July 12, 2013 at 12:54 AM (Answer #1)