Question

How do you solve `|x-5|+|2-2x|=7`?

Answer 1

The roots are `x=0, x=4`.

Explanation:

We observe that modulii in the given equation (eqn.) vanish at `x=1 and x=5.`

So, very naturally, we consider these 3 cases :
(i) `x<1`, (ii) `1<`x`<5` (iii) `x>5`

Case (i) `x<1`.
`:.`(x-1)`<0`, `(x-5)`<0`.`:. `|x-1|=1-x, |x-5|=5-x`
`:.` given eqn. becomes, `5-x+2(1-x)=7, or, 7-3*x=7, so, x=0.` We verify that this also satisfies the given eqn.

Case (ii) `1<`x`<5`.
In this case, `(x-1)`>`0` and `(x-5)`<`0`, so that `|x-1|=x-1`, `|x-5|`=`5-x`.
`:.` Eqn. becomes `5-x+2*x-2=7`. `:.` `x=4`.
We verify that `1`<`4`<`5`, and this satisfy the eqn.

Case (iii) `x>5`
Here, `(x-5)`>`0`, `(x-1)`>4>0, so, `|x-5|=x-5, |x-1|=x-1`
`:.` eqn. becomes, `x-5+2*x-2=7`, giving `x=14/3`which is non-permissible as `x>5`. Hence no root in this case!

Altogether, the roots are `x=0, x=4`.