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Solve the system: 3x^2 + 2y^2 - 54y - 143 = 0 x-3y -3 = 0

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annaball111 | Student, College Freshman | eNotes Newbie

Posted November 19, 2010 at 12:44 AM via web

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Solve the system:
3x^2 + 2y^2 - 54y - 143 = 0
x-3y -3 = 0

3 Answers | Add Yours

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted November 19, 2010 at 12:45 AM (Answer #1)

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3x^2 + 2y^2 - 54y - 143 = 0............(1)

 x-3y -3 = 0............(2)

Let us rewrite equation (2) :

x - 3y - 3 = 0

==> x= 3y + 3= 3(y+1)

==> x= 3(y+1)

Now we will substitute x value in (1):

3x^2 + 2y^2 - 54y - 143 = 0

3(3(y+1))^2 + 2y^2 - 54y - 143 = 0

3(9(y^2 + 2y+1) + 2y^2 - 54y - 143 = 0

27y^2 + 54y + 27 + 2y^2 - 54y - 143 = 0

Now combine like terms:

29y^2 - 116 = 0

==> 29y^2 = 116

Now divide by 29:

==> y^2 = 4

==> y= +- 2

==> x= 3(y+1)

==> x1= 3(2+1) = 9

==> x2= 3(-2+1) = -3

==> the answer is:

(9, 2) and (-3, -2)

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted November 19, 2010 at 12:46 AM (Answer #2)

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We have to solve for x and y using 3x^2 + 2y^2 - 54y - 143 = 0  and x-3y -3 = 0

Let us write x-3y -3 = 0 as x = 3y + 3

Therefore 3x^2 + 2y^2 - 54y - 143 = 0

=> 3(3y + 3)^2 + 2y^2 - 54y - 143 =0

=> 27y^2 + 54y +27 + 2y^2 - 54y - 143 =0

=> 29y^2 - 116 = 0

=> y^2 = 116/29 = 4

Therefore y can be +2 or -2

As x = 3y + 3, for y = +2, x= 9 and for y=-2, x=-3

So the values of x and y are (-3, -2) and (9, 2).

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kjcdb8er | Teacher | (Level 1) Associate Educator

Posted November 19, 2010 at 12:49 AM (Answer #3)

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y = -(3-x)/3  --> y^2 = (3-x)^2 / 9

3x^2 + 2y^2 - 54y - 143 = 0

3x^2 + 2(3-x)^2 / 9 + 54(3-x)/3 - 143 = 0

Combine terms:

(29 x^2)/9-(58 x)/3-87 = 0

Factor:

29/9 (x^2-6 x-27) = 0
29/9 (x-9) (x+3) = 0

x = 9, -3

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