Linear equations in one variable on the SAT

Example 1: distribute, then collect

Problem Solve $3(x - 2) + 5 = 2x + 9$.

1. Distribute the 3 to both terms in the parentheses.
   $$3x - 6 + 5 = 2x + 9$$
2. Combine the numbers on the left.
   $$3x - 1 = 2x + 9$$
3. Subtract $2x$ from both sides so x is on one side only.
   $$x - 1 = 9$$
4. Add 1 to both sides.
   $$x = 10$$
5. Check: put 10 back into both sides of the original equation.
   $$3(10 - 2) + 5 = 29 = 2(10) + 9$$

Answer $x = 10$

Example 2: a word problem

Problem A repair shop charges a flat fee of $65 plus $40 for each hour of labor. A customer's bill was $225. How many hours of labor were on the bill?

1. Let $h$ be the number of hours. Fee plus hourly cost equals the bill.
   $$65 + 40h = 225$$
2. Subtract 65 from both sides.
   $$40h = 160$$
3. Divide both sides by 40.
   $$h = 4$$
4. Check: $65 + 40(4) = 65 + 160 = 225$. It matches the bill.

Answer 4 hours of labor.

Example 3: when is there no solution?

Problem In the equation $ax + 3 = 5x - 2(x - 4)$, $a$ is a constant. For what value of $a$ does the equation have no solution?

1. Simplify the right side first. Distribute the $-2$ to both terms.
   $$5x - 2x + 8 = 3x + 8$$
2. Now the equation is $ax + 3 = 3x + 8$. No solution means the x terms must cancel and leave a false statement.
3. The x terms cancel only when the coefficients match, so $a = 3$. Then the equation becomes $3 = 8$, which is false. So there is no solution.
   $$3x + 3 = 3x + 8$$

Answer $a = 3$

Example 4 (SAT-hard): solve for an expression, not for x

Problem If $\frac{2}{5}(10x - 15) = 18$, what is the value of $2x - 3$?

1. Look before you solve. $10x - 15$ is 5 times $2x - 3$.
   $$\frac{2}{5} \cdot 5(2x - 3) = 18$$
2. The 5s cancel.
   $$2(2x - 3) = 18$$
3. Divide both sides by 2. You are done, and you never found x.
   $$2x - 3 = 9$$
4. Check (optional): $2x - 3 = 9$ gives $x = 6$, and $\frac{2}{5}(60 - 15) = \frac{2}{5}(45) = 18$.

Answer $2x - 3 = 9$