# Chapter 6 - Algebra: Equations and Inequalities - 6.2 Linear Equations in One Variable and Proportions - Exercise Set 6.2 - Page 363: 93

The value of ${{x}^{2}}-x$is $2$.

#### Work Step by Step

The equation$4\left( x-2 \right)+2=4x-2\left( 2-x \right)$. Use the distributive property, $a\left( b+c \right)=ab+ac$. \begin{align} & 4x-8+2=4x-4+2x \\ & 4x-6=6x-4 \\ \end{align} Add $4$to both side, \begin{align} & 4x-6+4=6x-4+4 \\ & 4x-2=6x \\ \end{align} Subtract $4x$ from both side, \begin{align} & 4x-4x-2=6x-4x \\ & -2=2x \end{align} Divided by 2 both side, \begin{align} & \frac{-2}{2}=\frac{2x}{2} \\ & -1=x \\ & x=-1 \\ \end{align} The solution set is $\left\{ -1 \right\}$. Check the proposed solution. Substitute $-1$ for x in the original equation $4\left( x-2 \right)+2=4x-2\left( 2-x \right)$. \begin{align} & 4\left( -1-2 \right)+2=4\left( -1 \right)-2\left( 2+1 \right) \\ & 4\left( -3 \right)+2=-4-2\left( 3 \right) \\ & -12+2=-4-6 \\ & -10=-10 \end{align} This true statement $-10=-10$ verifies that the solution set is $\left\{ -1 \right\}$. Now put $x=-1$ in the expression ${{x}^{2}}-x$. \begin{align} & {{x}^{2}}-x={{\left( -1 \right)}^{2}}-\left( -1 \right) \\ & =1+1 \\ & =2 \end{align} Therefore, value of ${{x}^{2}}-x$is $2$.

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