Algebra I: Linear Equations & Formulas

How to do it:

1. Name the unknown with a letter and a unit.

2. Translate each phrase: "more than" → $+\text{,}$ "of" a percent → multiply.

3. Write $=$ between two expressions that describe the same quantity.

Worked example: A gym plan costs $\$20$ plus $\$8$ per class. You have $\$68\text{.}$ $20 + 8c = 68\text{.}$

Check: $c$ is classes, not dollars. Undo in the Equation solver: subtract 20, then divide by 8 to get $c = 6\text{.}$

How to do it:

1. Start from "if the original equation is true..."

2. Name the property (add the same number, multiply by a nonzero number).

3. Stop when the variable is isolated.

Worked example: $3x - 5 = 16\text{.}$ Add 5: $3x = 21\text{.}$ Divide by 3: $x = 7\text{.}$ Each step keeps equality because both sides change the same way.

Check: plug $x = 7$ back: $3(7)-5 = 16\text{.}$ If a step was multiplying by 0, you would have destroyed information.

How to do it:

1. Use the same inverse steps as an equation.

2. If you multiply or divide by a negative, reverse the inequality mark.

3. Graph the solution on a number line: open circle for $<$ or $>\text{,}$ closed for $\leq$ or $\geq\text{.}$

Worked example: $-2x + 3 > 11\text{.}$ Subtract 3: $-2x > 8\text{.}$ Divide by $-2$ and flip: $x < -4\text{.}$ Graph an open ray left of $-4\text{.}$

Check: test $x = -5\text{:}$ $-2(-5)+3 = 13 > 11\text{,}$ true. Test $x = 0\text{:}$ $3 > 11\text{,}$ false. Forgetting the flip is the usual error.

How to do it:

1. Identify the letter you want.

2. Undo operations in reverse order, treating other letters as known.

3. Keep units: if $d$ is miles and $t$ is hours, $r = d/t$ is miles per hour.

Worked example: $V = IR$ for $R\text{:}$ divide both sides by $I$ (if $I \neq 0\text{)}$ to get $R = V/I\text{.}$ Same move as Ohm's law in the Formula Library.

Check: do not "move" a letter by subtracting it from the wrong side. Always apply the same operation to the entire side.

How to do it:

1. Collect the variable terms on one side and the constants on the other, using inverse steps.

2. Divide by the coefficient that is left.

3. If the variables cancel entirely, read what remains. A true statement like $6=6$ means every value works (an identity); a false one like $5=1$ means no value works.

Worked example: $5x-8=2x+7\text{.}$ Subtract $2x\text{:}$ $3x-8=7\text{.}$ Add $8\text{:}$ $3x=15\text{.}$ Divide by $3\text{:}$ $x=5\text{.}$ Now the surprises. $3x+6=3(x+2)$ expands to $3x+6=3x+6\text{,}$ and subtracting $3x$ leaves $6=6\text{:}$ true for every $x\text{,}$ infinitely many solutions. $2x+5=2x+1$ leaves $5=1\text{:}$ no solution.

Check: substitute $x=5\text{:}$ $5(5)-8=17$ and $2(5)+7=17\text{.}$ When the variables vanish, do not write $x=0\text{;}$ the equation is telling you about all numbers or none.

How to do it:

1. Find the least common denominator of every fraction in the equation.

2. Multiply every term on both sides by it, so the fractions disappear.

3. For decimals, multiply every term by $10\text{,}$ $100\text{,}$ or whatever moves each decimal point past the last digit.

Worked example: $\dfrac{x}{3}+\dfrac{x}{4}=7\text{.}$ The LCD is $12\text{:}$ $4x+3x=84\text{,}$ so $7x=84$ and $x=12\text{.}$ With decimals, $0.25x+0.5=3\text{:}$ multiply by $100$ to get $25x+50=300\text{,}$ so $x=10\text{.}$ (Multiplying by $4$ works too: $x+2=12\text{.)}$

Check: $\dfrac{12}{3}+\dfrac{12}{4}=4+3=7\text{.}$ The usual slip is multiplying only the fraction terms and forgetting the $7\text{;}$ every term gets the LCD.

How to do it:

1. And (a sandwich like $-3<2x+1\leq 7\text{):}$ do each inverse step to all three parts at once.

2. Or (two separate conditions): solve each inequality on its own, and keep both rays.

3. Graph it: an "and" is one segment between the two ends; an "or" is two rays pointing away from each other. Open circle for $<$ or $>\text{,}$ closed for $\leq$ or $\geq\text{.}$

Worked example: $-3<2x+1\leq 7\text{.}$ Subtract $1$ everywhere: $-4<2x\leq 6\text{.}$ Divide by $2\text{:}$ $-2<x\leq 3\text{.}$ Open circle at $-2\text{,}$ closed circle at $3\text{,}$ shade between. For $x+1<0$ or $2x\geq 8\text{:}$ $x<-1$ or $x\geq 4\text{,}$ two rays.

Check: test one number inside and one outside. $x=0$ makes $-3<1\leq 7$ true; $x=4$ makes $-3<9\leq 7$ false. If you divide a sandwich by a negative, flip both inequality signs.

How to do it:

1. Treat every letter except the one you want as a known number that happens to be spelled with a letter.

2. Undo operations in reverse order, exactly as with numbers.

3. State any restriction you needed, such as dividing by a letter that cannot be $0\text{.}$

Worked example: solve $ax+b=c$ for $x\text{:}$ subtract $b\text{,}$ then divide by $a\text{:}$ $x=\dfrac{c-b}{a}\text{,}$ with $a\neq 0\text{.}$ Solve $3x+2y=12$ for $y\text{:}$ subtract $3x\text{,}$ divide by $2\text{:}$ $y=\dfrac{12-3x}{2}=6-1.5x\text{.}$ Solve $P=2\ell+2w$ for $w\text{:}$ subtract $2\ell\text{,}$ divide by $2\text{:}$ $w=\dfrac{P-2\ell}{2}\text{.}$

Check: put the answer back. If $a=3\text{,}$ $b=-5\text{,}$ $c=16\text{,}$ then $x=\dfrac{16-(-5)}{3}=7\text{,}$ matching $3x-5=16$ from earlier. A letter coefficient follows the same rules as $3$ did; only the bookkeeping changes.