solve for x 3x - (1/2) = 2x + (7/4)

Question

Asked 9/12/2012 6:44:21 PM

Updated 12/9/2014 12:48:17 AM

1 Answer/Comment

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javedsarkar|Points 1015|

feurem|Points 1965|

x = 1/2 (3-sqrt(37))

x = 1/2 (3+sqrt(37))

feurem|Points 1965|

mediatrix|Points 10|

stan_r|Points 1332|

Question

Asked 9/12/2012 6:44:21 PM

Updated 12/9/2014 12:48:17 AM

1 Answer/Comment

This conversation has been flagged as incorrect.

Rating

3

One number is 8 more than another number. Let x be one number, the other is x + 8. If the sum of the smaller number and twice the larger number is 46, then: x + 2(x + 8) = 46, x + 2x + 16 = 46, 3x = 46 - 16 = 30, x = 30/3 = 10. The two numbers are 10 and 18 respectively.

Added 12/9/2014 12:31:13 AM

This answer has been confirmed as correct and helpful.

Confirmed by andrewpallarca [12/9/2014 6:45:33 AM]

3

3x - (1/2) = 2x + (7/4)

3x - 2x = (7/4) + (1/2)

x = 9/4 = 2 1/4

3x - 2x = (7/4) + (1/2)

x = 9/4 = 2 1/4

Added 12/9/2014 12:32:30 AM

This answer has been confirmed as correct and helpful.

Confirmed by andrewpallarca [12/9/2014 6:46:40 AM]

3

2x + 3(4 - x) = 3x + 2

2x + 12 - 3x = 3x + 2

-x + 12 = 3x + 2

-x - 3x = 2 - 12

-4x = -10

x = -10/-4 = 2.5

2x + 12 - 3x = 3x + 2

-x + 12 = 3x + 2

-x - 3x = 2 - 12

-4x = -10

x = -10/-4 = 2.5

Added 12/9/2014 12:33:58 AM

This answer has been confirmed as correct and helpful.

Confirmed by andrewpallarca [12/9/2014 6:47:06 AM]

3

x^2 - 3x - 4 = 0

(x - 4)(x + 1) = 0

x = 4 or x = -1

(x - 4)(x + 1) = 0

x = 4 or x = -1

Added 12/9/2014 12:35:26 AM

This answer has been confirmed as correct and helpful.

Confirmed by andrewpallarca [12/9/2014 6:47:39 AM]

3

(9x^4 - 3x^2) + (4x^2 + 5)

= 9x^4 - 3x^2 + 4x^2 + 5

= 9x^4 + x^2 + 5

= 9x^4 - 3x^2 + 4x^2 + 5

= 9x^4 + x^2 + 5

Added 12/9/2014 12:39:38 AM

This answer has been confirmed as correct and helpful.

Confirmed by andrewpallarca [12/9/2014 6:48:29 AM]

3

3x^2 - 4x - 4 = (x - 2)(3x + 2)

Added 12/9/2014 12:47:30 AM

This answer has been confirmed as correct and helpful.

Confirmed by andrewpallarca [12/9/2014 6:49:12 AM]

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