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

Jun 10, 2016

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

Explanation:

We observe that modulii in the given equation (eqn.) vanish at $x = 1 \mathmr{and} x = 5.$

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

Case (i) $x < 1$.
$\therefore$(x-1)$< 0$, $\left(x - 5\right)$<0$.$:. $| x - 1 | = 1 - x , | x - 5 | = 5 - x$
$\therefore$ given eqn. becomes, $5 - x + 2 \left(1 - x\right) = 7 , \mathmr{and} , 7 - 3 \cdot x = 7 , s o , x = 0.$ We verify that this also satisfies the given eqn.

Case (ii) $1 <$x$< 5$.
In this case, $\left(x - 1\right)$>$0$ and $\left(x - 5\right)$<$0$, so that $| x - 1 | = x - 1$, $| x - 5 |$=$5 - x$.
$\therefore$ Eqn. becomes $5 - x + 2 \cdot x - 2 = 7$. $\therefore$ $x = 4$.
We verify that $1$<$4$<$5$, and this satisfy the eqn.

Case (iii) $x > 5$
Here, $\left(x - 5\right)$>$0$, $\left(x - 1\right)$>4>0, so, $| x - 5 | = x - 5 , | x - 1 | = x - 1$
$\therefore$ eqn. becomes, $x - 5 + 2 \cdot x - 2 = 7$, giving $x = \frac{14}{3}$which is non-permissible as $x > 5$. Hence no root in this case!

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