# 3(x – 1) < -3(2 – 2x)? A) x > 1 B) x < 1 C) x > -1 D) x < -1

Mar 27, 2016

$\text{B} : x < 1$

#### Explanation:

The given problem is:

$3 \left(x - 1\right) < - 3 \left(2 - 2 x\right)$

The solutions are:

"A")color(white)(i)x>1
"B")color(white)(i)x<1
"C")color(white)(i)x>$- 1$
"D")color(white)(i)x<-1

Solving the Inequality
$1$. Start by factoring $- 2$ from the bracketed terms on the right-hand side of the equation.

$3 \left(x - 1\right) < - 3 \left(2 - 2 x\right)$

$3 \left(x - 1\right) < - 3 \cdot - 2 \left(- 1 + x\right)$

$2$. Multiply $- 3$ and $- 2$ together on the right-hand side of the equation.

$3 \left(x - 1\right) < 6 \left(1 - x\right)$

$3$. Divide both sides by $6$.

$\frac{3 \left(x - 1\right)}{6} < \frac{6 \left(1 - x\right)}{6}$

$\frac{{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}}^{1} \left(x - 1\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{6}}}} ^ 2 < \frac{{\textcolor{red}{\cancel{\textcolor{b l a c k}{6}}}}^{1} \left(1 - x\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{6}}}} ^ 1$

$\frac{x - 1}{2} < 1 - x$

$4$. Multiply the whole inequality by $2$ to get rid of the denominator.

$2 \left(\frac{x - 1}{2}\right) < 2 \left(1 - x\right)$

${\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}}^{1} \left(\frac{x - 1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} ^ 1\right) < 2 \left(1 - x\right)$

$x - 1 < 2 - 2 x$

$5$. Add $2 x$ to both sides of the equation.

$x$ $\textcolor{red}{+ 2 x} - 1 < 2 - 2 x$ $\textcolor{red}{+ 2 x}$

$3 x - 1 < 2$

$6$. Add $1$ to both sides of the equation.

$3 x - 1$ $\textcolor{red}{+ 1} < 2$ $\textcolor{red}{+ 1}$

$3 x < 3$

$7$. Divide both sides by $3$.

$\frac{3 x}{3} < \frac{3}{3}$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} x < 1 \textcolor{w h i t e}{\frac{a}{a}} |}}}$