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Solve the equations: x + y = 1 x^2 + 4y = 18

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xetaalpha2 | Student | (Level 2) Honors

Posted April 19, 2011 at 2:38 PM via web

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Solve the equations:

x + y = 1

x^2 + 4y = 18

2 Answers | Add Yours

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted April 19, 2011 at 3:00 PM (Answer #1)

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We have to solve x + y = 1 and x^2 + 4y = 18

x + y = 1

=> x = 1 - y

substitute in x^2 + 4y = 18

=> (1 - y)^2 + 4y = 18

=> 1 + y^2 - 2y + 4y = 18

=> y^2 + 2y - 17 = 0

y1 = -2/2 + sqrt (4 + 68)/2

=> -1 + 3*sqrt 2

y2 = -1 - 3*sqrt 2

x1 = 2 - 3*sqrt 2

x2 = 2 + 3*sqrt 2

The solution of the equations are (2 - 3*sqrt 2, -1 + 3*sqrt 2) and (2 + 3*sqrt 2, -1 - 3*sqrt 2)

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atyourservice | Student, Grade 10 | (Level 3) Valedictorian

Posted March 22, 2014 at 8:19 PM (Answer #2)

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x + y = 1

y=x+1

`x^2 + 4(x+1) = 18    `

`x^2+4x+1=18 `

`x^2+4x+1-18=18-18 `

`x^2+4x-17=0 `

`x^2+4x=17 `

`(-b/2)^2 `

`4/2=2^2=4 `

`x^2+4x+4=17-4 `

`x^2+4x+4=13 `

`(x+2)^2=13 `

`x+2=sqrt(13) `

`x=-2+-sqrt(13) `

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