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(x+1)/(x+4)=(2x-1)/(x+6)Solve for x

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syedajmain | Student | eNoter

Posted July 20, 2009 at 11:15 PM via web

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(x+1)/(x+4)=(2x-1)/(x+6)

Solve for x

5 Answers | Add Yours

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kjcdb8er | Teacher | (Level 1) Associate Educator

Posted July 21, 2009 at 12:08 AM (Answer #1)

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(x+1)/(x+4)=(2x-1)/(x+6)

(x + 1)(x+6) = (2x - 1)(x+4)     x both sides by (x + 4)(x + 6)

x^2 + 7x + 6 = 2x^2 + 7x - 4   Expand the parenthesis

x^2 + 6 = 2x^2 - 4                 Subtract 7x from both sides

10 = x^2                               Add (4 - x^2) to both sides

+- sqrt(10) = x                       Take square root of both sides

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neela | High School Teacher | Valedictorian

Posted July 21, 2009 at 12:21 AM (Answer #2)

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Multiply both sides of the given expression by the LCM (x+6) (x+4) to get rid of the denominator. Then we get:

(x+1)(x+6) =(2x-1)(x+4)

Expand both sides :

x^2+6x+x+6 =2x^2+8x-x-4

Simplify both sides:

x^2+7x+6=2x^2+7x-4.

Subtract x^2+7x+6 from both sides:

0=2x^2+7x-4-(x^2+7x+6)

0=x^2-10

x^2 = 10  or

x= + sqrt(10) or  x=-sqrt(10)

 

 

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giorgiana1976 | College Teacher | Valedictorian

Posted July 21, 2009 at 1:04 AM (Answer #3)

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We'll use the property of equal ratios, so we'll subtract the denominator from numerator, at both sides of the equality:

(x+1-x-4)/(x+4)=(2x-1-x-6)/(x+6)

-3/(x+4)=(x-7)/(x+6)

We'll cross multiplying and the result will be:

-3*(x+6)=x^2-3*x-28

We'll reduce the similar terms, both sides of the equality,

"-3x"

-18=x^2-28

x^2-10=0

x^2=10

x1=+ sqrt 10

x2=- sqrt 10

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Wiggin42 | TA , Grade 11 | Valedictorian

Posted March 23, 2014 at 4:11 PM (Answer #4)

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`(x+1)/(x+4)=(2x-1)/(x+6)`

Cross multiply: 

(x+1)(x+6) = (x+4)(2x-1)

FOIL both sides

x^2 +7x + 6 = 2x^2 + 7x - 4

Bring all terms to one side

x^2 - 10 = 0

x^2 = 10

x = plus or minus sqrt(10)

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atyourservice | TA , Grade 10 | Valedictorian

Posted April 16, 2014 at 4:08 PM (Answer #5)

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(x+1)/(x+4)=(2x-1)/(x+6)

cross multiply the numbers

(x+1)(x+6) = (x+4)(2x-1)

unfoil the problem by multiplying every number from the first parenthesis by the numbers inside the second parenthesis

you should end up with 

`x^2 +7x + 6 = 2x^2 + 7x - 4  `

Add the opposite of each number to combine like terms

x^2 +7x + 6 = 2x^2 + 7x - 4

               +4                     +4            

x^2 +7x + 10 = 2x^2 + 7x 

       -7x                       -7x

x^2 + 10 = 2x^2

-x^2           -x^2

`10=x^2 `

find the square root 

`sqrt(10) =sqrt(x^2)`

`x= +-sqrt(10)`

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