Elementary algebra: solving linear equations in one variable  (Page 3/3)

Solve for a: $-4a+2-a=3+5a-2$ $$\begin{array}{cccc}-4a+2-a& =& 3+5a-2& \mathit{\text{Add same side like terms first.}}\\ -5a+2& =& 5a+1& \hfill \\ -5a+2-5a& =& 5a+1-5a& \mathit{\text{Subtract 5a from both sides}}\text{.}\hfill \\ -10a+2& =& 1& \\ -10a+2-2& =& 1-2& \mathit{\text{Subtract 2 from both sides}}\text{.}\hfill \\ -10a& =& -1& \\ \frac{-10a}{-10}& =& \frac{-1}{-10}& \mathit{\text{Divide both sides by -10}}\text{.}\hfill \\ a& =& \frac{1}{10}& \end{array}$$ The solution set is $\left\{\frac{1}{10}\right\}$

Solve for x: $5(3x+2)-2=-2(1-7x)$ $$\begin{array}{cccc}5(3x+2)-2& =& -2(1-7x)& \mathit{\text{Distribute}}\text{.}\hfill \\ 15x+10-2& =& -2+14x& \mathit{\text{Add same side like terms.}}\hfill \\ 15x+8& =& -2+14x& \\ 15x+8-14x& =& -2+14x-14x& \mathit{\text{Subtract 14x on both sides.}}\hfill \\ x+8& =& -2& \\ x+8-8& =& -2-8& \mathit{\text{Subtract 8 on both sides.}}\hfill \\ x& =& -10& \end{array}$$ The solution set is $\left\{-10\right\}$ .

Solve for x: $4(x+5)+6=2(2x+3)$ $$\begin{array}{cccc}4(x+5)+6& =& 2(2x+3)& \mathit{\text{Distribute}}\hfill \\ 4x+20+6& =& 4x+6& \mathit{\text{Add same side like terms}}\text{.}\hfill \\ 4x+26& =& 4x+6& \\ 4x+26-4x& =& 4x+6-4x& \mathit{\text{Subtract 4x on both sides.}}\hfill \\ 26& =& 6& \mathit{\text{False}}\hfill \end{array}$$ There is no solution, $\varnothing$ .

Solve for y: $3(3y+5)+5=10(y+2)-y$ $$\begin{array}{cccc}3(3y+5)+5& =& 10(y+2)-y& \mathit{\text{Distribute}}\hfill \\ 9y+15+5& =& 10y+20-y& \mathit{\text{Add same side like terms}}\text{.}\hfill \\ 9y+20& =& 9y+20& \\ 9y+20-20& =& 9y+20-20& \mathit{\text{Subtract 20 on both sides}}\text{.}\hfill \\ 9y& =& 9y& \\ 9y-9y& =& 9y-9y& \mathit{\text{Subtract 9y on both sides}}\text{.}\hfill \\ 0& =& 0& \mathit{\text{True}}\hfill \end{array}$$ The equation is an identity, the solution set consists of all real numbers, $\Re$ .

Solve for a: $P=2a+b$ $$\begin{array}{cccc}P& =& 2a+b& \\ P-b& =& 2a+b-b& \mathit{\text{Subtract b on both sides.}}\hfill \\ P-b& =& 2a& \\ \frac{P-b}{2}& =& \frac{2a}{2}& \mathit{\text{Divide both sides by 2.}}\hfill \\ \frac{P-b}{2}& =& a& \\ & & & \end{array}$$ Solution: $a=\frac{P-b}{2}$

Solve for x: $z=\frac{x+y}{2}$ $$\begin{array}{cccc}z& =& \frac{x+y}{2}& \\ 2\cdot z& =& 2\cdot \frac{x+y}{2}& \mathit{\text{Multiply both sides by 2}}\text{.}\hfill \\ 2z& =& x+y& \\ 2z-y& =& x+y-y& \mathit{\text{Subtract y on both sides}}\text{.}\hfill \\ 2z-y& =& x& \end{array}$$ Solution $x=2z-y$