Algebra 2 by Richard Wright
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Objectives:
SDA NAD Content Standards (2018): AII.4.1, AII.5.1
Do unto others what you would have them do unto you.
Do unto one side what you have done to the other side.
Solve equations getting all of the desired variables on same side of the equation. Then get everything else away from the variable you are solving for. Usually you work from the most outside piece away from the variable. Whatever you do to one side of the equation, do unto the other side.
Solve x + 3 = 7.
Solution
$$ \begin{array}{rll} x+3 & =7 & \\ x+3\color{red}{-3} & =7\color{red}{-3} & ←\text{Subtract 3 from both sides so } x \text{ is by itself} \\ x & =4 & \end{array} $$
Solve 2x − 4 = 10.
Solution
$$ \begin{array}{rll} 2x-4 & =10 & \\ 2x-4\color{red}{+4} & =10\color{red}{+4} & ←\text{Add 4 to both sides} \\ 2x & =14 & \\ \frac{2x}{2} & =\frac{14}{2} & ←\text{Divide both sides by 2} \\ x & =7 & \end{array} $$
Solve 3x − 1 = x + 5.
Solution
$$ \begin{array}{rll} 3x-1 & =x+5 & \\ 3x\color{red}{-x}-1 & =x\color{red}{-x}+5 & ←\text{Subtract } x \text{ to get all the } x \text{'s on the same side} \\ 2x-1 & =5 & ←\text{Combine like terms} \\ 2x-1\color{red}{+1} & =5\color{red}{+1} & ←\text{Add 1 to both sides} \\ 2x & =6 & \\ \frac{2x}{2} & =\frac{6}{2} & ←\text{Divide both sides by 2} \\ x & =3 & \end{array} $$
Inequalities are like equations, but one side has a larger value than the other side. Solve inequalities like you do with equations. The one difference is if you multiply or divide both sides by a negative, then flip the inequality sign.
Solve −2(x − 4) > 12.
Solution
$$ \begin{array}{rll}-2\left(x-4\right)&>12&\\\frac{-2\left(x-4\right)}{-2}&<\frac{12}{-2}&\ \ \ \gets\mathrm{Divide\ both\ sides\ by\ 2\ because\ it\ is\ the\ most\ outside\ piece\ from\ the\ }x\\&&\ \ \ \ \mathrm{Flip\ the\ inequality\ because\ divided\ by\ negative}\\x-4&<-6&\\x-4+4&<-6+4&\ \ \ \ \gets\mathrm{Add\ 4\ to\ each\ side}\\x&<-2&\\\end{array} $$
Solve 3(x + 5) < x − 8.
Solution
$$ \begin{array}{rll} 3(x+5) & < x-8 & \\ 3x+15 & < x-8 & \gets\mathrm{Distribute\ the\ 3}\\3x-x+15& < x-x-8&\gets\mathrm{Subtract\ }x\mathrm{\ from\ both\ sides}\\2x+15& < -8&\\2x+15-15&<-8-15&\gets\mathrm{Subtract\ }x\mathrm{\ from\ both\ sides}\\2x& < -23&\\\frac{2x}{2}& < -\frac{23}{2}&\gets\mathrm{Divide\ both\ sides\ by\ 2}\\x& < -\frac{23}{2}&\\\end{array} $$
Sometimes there is more than one variable. Treat the unwanted variable as a number and solve just like you solved the other equations and inequalities.
Solve for h. \(A_1 < \frac{1}{2}\left(b_1+b_2\right)h < A_2\)
Solution
$$ \begin{array}{rcll} A_1 < & \frac{1}{2}\left(b_1+b_2\right)h & < A_2& \\ \color{red}{2\cdot} A_1< & \color{red}{2\cdot}\frac{1}{2}\left(b_1+b_2\right)h & < \color{red}{2\cdot} A_2 & \gets\mathrm{Multiply\ all\ parts\ by\ 2\ to\ remove\ }\frac{1}{2} \\ 2A_1 < & \left(b_1+b_2\right)h & < 2A_2 & \\ \frac{2A_1}{\color{red}{b_1+b_2}} < & \frac{\left(b_1+b_2\right)h}{\color{red}{b_1+b_2}} & < \frac{2A_2}{\color{red}{b_1+b_2}} & \gets\mathrm{Divide\ by\ }\left(b_1+b_2\right) \\ \frac{2A_1}{b_1+b_2} < & h & < \frac{2A_2}{b_1+b_2} & \end{array} $$
Solve for ℓ. \(P=2\ell+2w\).
Solution
$$ \begin{array}{rll} P & =2\ell+2w & \\ P\color{red}{-2w} & =2\ell+2w\color{red}{-2w} & \gets\mathrm{Subtract\ }2w\mathrm{\ from\ both\ sides\ to\ get\ l\ by\ itself} \\ P-2w & =2\ell & \\ \frac{P-2w}{\color{red}{2}} & =\frac{2\ell}{\color{red}{2}} & \gets\mathrm{Divide\ both\ sides\ by\ 2} \\ \frac{P-2w}{2} & =\ell & \\ \ell & =\frac{P-2w}{2} & \gets\mathrm{Write\ the\ equation\ with\ the\ chosen\ variable\ on\ the\ left} \end{array} $$
Solve for z. 2xz + 5z < 15.
Solution
$$ \begin{array}{rll} 2xz+5z & < 15 & \\ z\left(2x+5\right) & < 15 & \gets\mathrm{Distributive\ property} \\ \frac{z\left(2x+5\right)}{\color{red}{2x+5}} & < \frac{15}{\color{red}{2x+5}} & \gets\mathrm{Divide\ both\ sides\ by\ }2x+5\mathrm{\ to\ get\ the\ }z\mathrm{\ by\ itself} \\ z & < \frac{15}{2x+5} & \end{array} $$
Solve the equation. Check your solution.
Solve the inequality.
Solve the equation or formula for the given variable.
Mixed Review: Write an expression to answer the question.