# Solve algebraically for x: (x + 2)/6 = 3/(x - 1)

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We have to solve (x + 2)/6 = 3/(x - 1) for x.

(x + 2)/6 = 3/(x - 1)

=> (x + 2)(x - 1) = 3*6

=> x^2 + 2x - x - 2 = 18

=> x^2 + x - 20 = 0

=> x^2 + 5x - 4x - 20 = 0

=> x(x + 5) - 4(x + 5) = 0

=> (x - 4)(x + 5) = 0

=> x = 4 and x = -5

**The solution of the equation is x = 4 and x = -5**

Given: (x + 2)/6 = 3/(x - 1).

To find x.

(x + 2)/6 = 3/(x - 1)

=> (x + 2)(x - 1) = 3*6

=> x^2 + 2x - x - 2 = 18

=> x^2 + 2x - x - 2 -18 = 0

=> x^2 + x - 20 = 0

=> x^2 + 5x - 4x - 20 = 0 (writing x=5x-4x)

=> x(x + 5) - 4(x + 5) = 0

=> (x - 4)(x + 5) = 0

Therefore, **x = 4 and x = -5**

`(x + 2)/6 = 3/(x - 1) `

`3 xx 6=(x - 1)(x + 2) `

multiply the numbers and unfoil the parentheses

`3 xx 6 = 18 `

`x xx x = x^2 `

`x xx 2 = 2x `

`-1 xx x= -x `

`-1 xx 2 = -2x `

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

move the 18 to the other side by minusing it

`x^2 + 2x - x - 2-18 `

`x^2 + 2x - x - 20 `

combine like terms

`x^2 + x - 20 `

find facts of -20 that add up to 1

5 and -4

plug them in as b

`x^2 + 5x - 4x - 20 `

group the numbers

`(x^2 + 5x) (- 4x - 20 ) `

find greatest common factors

`x (x + 5) -4 (x + 5) `

`(x - 4) ( x + 5) `

set them equal to zero

x + 5 = 0

-5 -5

**x = -5**

x - 4 = 0

+4 +4

**x=4**

**so x is 4 and -5**

Given:-

(x + 2)/6 = 3/(x - 1)

or, (x+2)*(x-1) = 3*6

or, (x^2) + x - 2 = 18

or, (x^2) + x - 20 = 0

or, (x^2) + 5x - 4x - 20 = 0

or, x(x+5) - 4(x+5) = 0

or, (x-4)*(x+5) = 0

or, x = 4

or, x = -5

For this question, I would suggest that you first cross multiply.

(x+2)(x-1)= 3X6

x^2+x-2= 18

+2 on both sides

x^2+ x=20

x^2+x-20= 0

factor: (x+5)(x-4), x= -5 or x=4

To solve x + 2/ 6 = 3/ x - 1

We assume both sides are monomials like (x+2)/6 = 3/(x-1) and not binomials**.**

Now we multiply both sides by 6(x-1) and get:

(x-1)(x+2) = 3*6

x^2+x-2 = 18

x^2+x-2-18 = 18-18

x^2+x-20 = (x+5)(x-4) = 0

Therefore either of the factors x+5 = 0 or x-4 = 0.

**This gives x = -5 or x= 4.**