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If 2 x^2 - 2x - 12 = 0 and y^2 -5y + 6 = 0 when x = -

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If 2 x^2 - 2x - 12 = 0 and y^2 -5y + 6 = 0 when x = -  [#permalink]

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New post 09 Apr 2018, 09:28
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If \(2 x^2 - 2x - 12 = 0\) and \(y^2 - 5y + 6 = 0\) when \(x = -y\), then what is the value of \(x\) ?

a. -3
b. -2
c. 0
d. 2
e. 3

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If 2 x^2 - 2x - 12 = 0 and y^2 -5y + 6 = 0 when x = -  [#permalink]

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New post 09 Apr 2018, 13:37
QZ wrote:
If \(2 x^2 - 2x - 12 = 0\) and \(y^2 - 5y + 6 = 0\) when \(x = -y\), then what is the value of \(x\) ?

a. -3
b. -2
c. 0
d. 2
e. 3

• Simplify and factor the first quadratic equation
\(2 x^2 - 2x - 12 = 0\)
\(x^2 - x - 6 = 0\)
\((x - 3)(x + 2) = 0\)

\(x = 3\) OR
\(x = -2\)


• Factor the second quadratic equation
\(y^2 - 5y + 6 = 0\)
\((y - 2)(y - 3) = 0\)

\(y = 2\) OR
\(y = 3\)

• Heed the condition:
when \(x = -y\)

For values of x and y, only one pair yields

\(x = -y\)
when
\(x = -2\)and \(y = 2\)

When \(x = -y\):

\(x = -2\)

Answer B
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Re: If 2 x^2 - 2x - 12 = 0 and y^2 -5y + 6 = 0 when x = -  [#permalink]

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New post 09 Apr 2018, 20:31
1
QZ wrote:
If \(2 x^2 - 2x - 12 = 0\) and \(y^2 - 5y + 6 = 0\) when \(x = -y\), then what is the value of \(x\) ?

a. -3
b. -2
c. 0
d. 2
e. 3


We can get the value of x without solving quadratics.

Substitute y = -x into \(y^2 - 5y + 6 = 0\): \(x^2 + 5x + 6 = 0\).

Multiply by 2: \(2x^2 + 10x + 12 = 0\).

Subtract the first equation: \(2x^2 + 10x + 12 -(2 x^2 - 2x - 12)= 0\);

\(12x+24=0\);

\(x=-2\).

Answer: B.
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Re: If 2 x^2 - 2x - 12 = 0 and y^2 -5y + 6 = 0 when x = -  [#permalink]

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New post 10 Apr 2018, 03:22
Ans B, solving for Y it gives value for Y = 2 or 3. Using equation X = -Y and solving equation 2 X^2 - 2X - 12 = 0 for X = -2, satisfies the equation hence Ans B

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Re: If 2 x^2 - 2x - 12 = 0 and y^2 -5y + 6 = 0 when x = -  [#permalink]

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New post 11 Apr 2018, 15:19
QZ wrote:
If \(2 x^2 - 2x - 12 = 0\) and \(y^2 - 5y + 6 = 0\) when \(x = -y\), then what is the value of \(x\) ?

a. -3
b. -2
c. 0
d. 2
e. 3


Dividing the first equation by 2 and solving for x, we have:

x^2 - x - 6 = 0

(x - 3)(x + 2) = 0

x = 3 or x = -2

Solving for y in the second equation, we have:

(y - 3)(y - 2) = 0

y = 3 or y = 2

Since x = -y, we see that x must be -2 (and y must be 2).
Alternate Solution:

Note that 2x^2 can also be expressed as 2x^2 = 2x + 12 using the first equation.

Since x = -y, let’s substitute y = -x in the second equation:

(-x)^2 -5(-x) + 6 = 0

x^2 + 5x + 6 = 0

Let’s multiply each side of this equality by 2:

2x^2 + 10x + 12 = 0

2x^2 = -10x - 12

Since we have two expressions for 2x^2, let’s set them equal to each other:

2x + 12 = -10x -12

12x = -24

x = -2

Answer: B
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Re: If 2 x^2 - 2x - 12 = 0 and y^2 -5y + 6 = 0 when x = - &nbs [#permalink] 11 Apr 2018, 15:19
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