0.1 Solving linear equations and inequalities: solving equations  (Page 2/2)

Solving $x+a=b$ For $x$

$\begin{array}{llll}\hfill x+a& =\hfill & b\hfill & \text{The}a\text{is}\text{associated}\text{with}x\text{by}\text{addition}\text{.}\text{Undo}\text{the}\text{association}\hfill \\ \hfill x+a-a& =\hfill & b-a\hfill & \text{by}\text{subtracting}a\text{from}both\text{sides}\text{.}\hfill \\ \hfill x+0& =\hfill & b-a\hfill & a-a=0\text{and}0\text{is}\text{the}\text{additive}\text{identity}\text{.}x+0=x.\hfill \\ \hfill x& =\hfill & b-a\hfill & \text{This}\text{equation}\text{is}\text{equivalent}\text{to}\text{the}\text{first}\text{equation,}\text{and}\text{it}\text{is}\hfill \\ \hfill & \hfill & \hfill & \text{solved}\text{for}x.\hfill \end{array}$

Solving $x-a=b$ For $x$

$\begin{array}{llll}\hfill x-a& =\hfill & b\hfill & \text{The}a\text{is}\text{associated}\text{with}x\text{by}\text{subtraction}\text{.}\text{Undo}\text{the}\text{association}\hfill \\ \hfill x-a+a& =\hfill & b+a\hfill & \text{by}\text{adding}a\text{to}both\text{sides}\text{.}\hfill \\ \hfill x+0& =\hfill & b+a\hfill & -a+a=0\text{and}0\text{is}\text{the}\text{additive}\text{identity}\text{.}x+0=x.\hfill \\ \hfill x& =\hfill & b+a\hfill & \text{This}\text{equation}\text{is}\text{equivalent}\text{to}\text{the}\text{first}\text{equation,}\text{and}\text{it}\text{is}\hfill \\ \hfill & \hfill & \hfill & \text{solved}\text{for}x.\hfill \end{array}$

Solve $x+7=10$ for $x$ .

$\begin{array}{llll}\hfill x+7& =\hfill & 10\hfill & 7\text{is}\text{associated}\text{with}x\text{by}\text{addition}\text{.}\text{Undo}\text{the}\text{association}\hfill \\ \hfill x+7-7& =\hfill & 10-7\hfill & \text{by}\text{subtracting}7\text{from}both\text{sides}\text{.}\hfill \\ \hfill x+0& =\hfill & 3\hfill & 7-7=0\text{and}0\text{is}\text{the}\text{additive}\text{identity}.x+0=x.\hfill \\ \hfill x& =\hfill & 3\hfill & x\text{is}\text{isolated,}\text{and}\text{the}\text{equation}x=3\text{is}\text{equivalent}\text{to}\text{the}\hfill \\ \hfill & \hfill & \hfill & \text{original}\text{equation}x+7=10.\text{Therefore,}\text{these}\text{two}\hfill \\ \hfill & \hfill & \hfill & \text{equation}\text{have}\text{the}\text{same}\text{solution}\text{.}\text{The}\text{solution}\text{to}x=3\hfill \\ \hfill & \hfill & \hfill & \text{is}\text{clearly}3.\text{Thus,}\text{the}\text{solution}\text{to}x+7=10\text{is}\text{also}3.\hfill \end{array}$

Check : Substitute 3 for $x$ in the original equation. $$\begin{array}{llll}\hfill x+7& \text{=}\hfill & 10\hfill & \hfill \\ \hfill 3+7& \text{=}\hfill & 10\hfill & \text{Is}\text{this}\text{correct?}\hfill \\ \hfill 10& \text{=}\hfill & 10\hfill & \text{Yes,}\text{this}\text{is}\text{correct}\text{.}\hfill \end{array}$$

Solve $m-2=-9$ for $m$ .

$\begin{array}{llll}\hfill m-2& =\hfill & -9\hfill & 2\text{is}\text{associated}\text{with}m\text{by}\text{subtraction}\text{.}\text{Undo}\text{the}\text{association}\hfill \\ \hfill m-2+2& =\hfill & -9+2\hfill & \text{by}\text{adding}2\text{from}both\text{sides}\text{.}\hfill \\ \hfill m+0& =\hfill & -7\hfill & -2+2=0\text{and}0\text{is}\text{the}\text{additive}\text{identity}.m+0=m.\hfill \\ \hfill m& =\hfill & -7\hfill & \hfill \end{array}$

Check : Substitute $-7$ for $m$ in the original equation. $$\begin{array}{llll}\hfill m-2& \text{=}\hfill & -9\hfill & \hfill \\ \hfill -7-2& \text{=}\hfill & -9\hfill & \text{Is}\text{this}\text{correct?}\hfill \\ \hfill -9& \text{=}\hfill & -9\hfill & \text{Yes,}\text{this}\text{is}\text{correct}\text{.}\hfill \end{array}$$

Solve $y-2.181=-16.915$ for $y$ .

$\begin{array}{lll}\hfill y-2.181& =\hfill & -16.915\hfill \\ \hfill y-2.181+2.181& =\hfill & -16.915+2.181\hfill \\ \hfill y& =\hfill & -14.734\hfill \end{array}$

On the Calculator
$\begin{array}{ll}\hfill & \hfill \\ \text{Type}\hfill & 16.915\hfill \\ \text{Press}\hfill & \begin{array}{||}\hline +/-\\ \hline\end{array}\hfill \\ \text{Press}\hfill & \begin{array}{||}\hline +\\ \hline\end{array}\hfill \\ \text{Type}\hfill & 2.181\hfill \\ \text{Press}\hfill & \begin{array}{||}\hline =\\ \hline\end{array}\hfill \\ \text{Display}\text{reads:}\hfill & -14.734\hfill \end{array}$

Solve $y+m=s$ for $y$ .

$\begin{array}{llll}\hfill y+m& =\hfill & s\hfill & m\text{is}\text{associated}\text{with}y\text{by}\text{addition}\text{.}\text{Undo}\text{the}\text{association}\hfill \\ \hfill y+m-m& =\hfill & s-m\hfill & \text{by}\text{subtracting}m\text{from}both\text{sides}\text{.}\hfill \\ \hfill y+0& =\hfill & s-m\hfill & m-m=0\text{and}0\text{is}\text{the}\text{additive}\text{identity}.y+0=y.\hfill \\ \hfill y& =\hfill & s-m\hfill & \hfill \end{array}$

Check : Substitute $s-m$ for $y$ in the original equation. $$\begin{array}{lllll}\hfill y+m& \text{=}\hfill & s\hfill & \hfill & \hfill \\ \hfill s-m+m& \text{=}\hfill & s\hfill & \hfill & \text{Is}\text{this}\text{correct?}\hfill \\ \hfill s& \text{=}\hfill & s\hfill & \text{True}\hfill & \text{Yes,}\text{this}\text{is}\text{correct}\text{.}\hfill \end{array}$$

Solve $k-3h=-8h+5$ for $k$ .

$\begin{array}{llll}\hfill k-3h& =\hfill & -8h+5\hfill & 3h\text{is}\text{associated}\text{with}k\text{by}\text{subtraction}\text{.}\text{Undo}\text{the}\text{association}\hfill \\ \hfill k-3h+3h& =\hfill & -8h+5+3h\hfill & \text{by}\text{adding}3h\text{to}both\text{sides}\text{.}\hfill \\ \hfill k+0& =\hfill & -5h+5\hfill & -3h+3h=0\text{and}0\text{is}\text{the}\text{additive}\text{identity}.k+0=k.\hfill \\ \hfill k& =\hfill & -5h+5\hfill & \hfill \end{array}$