How do you solve $(x-1) /3 - (x-1) /2 = 6$ ?

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Answer 1 · 2016-06-06T17:14:18

$x=-35$

$x-1$ is common to both the terms in the left side of the equation.

Take $x-1$ and write the remaining terms inside brackets:

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

Now, solve the fractions inside the brackets:

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

$(x-1)(-1/6)=6$

$-(x-1)/6=6$

Next, multiply both sides by $-6$ :

$-(x-1)/6color(red)(xx-6)=6color(red)(xx-6)$

$x-1=-36$

Add $1$ to both sides:

$x-1color(red)(+1)=-36color(red)(+1)$

$x=-35$

Let us check our solution

Plug in the value of $x=-35$ in the given equation and solve:

Left side of the equation is $(x-1)/3-(x-1)/2$

$=(-35-1)/3-(-35-1)/2$

$=(-36)/3-(-36/2)$

$=-12-(-18)$

$=-12+18$

$=6=$ right side of the equation.

Hence, our solution is correct.

How do you solve (x-1) /3 - (x-1) /2 = 6(x-1) /3 - (x-1) /2 = 6?