Algebra I - linear equation with one variable

• Mathematics •

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A **linear equation with one variable** is an equation that, after simplifying and collecting like terms on each side, takes the form $ax + b = 0$ or $ax + b = cx + d$. Each side can involve $x$ added to real numbers and multiplied by real numbers, but not multiplied by other $x$'s. * **Linear Example:** $\frac{3}{4}(-\frac{x}{2} - 3) + x = 9 - (x - \frac{5}{8})$ is a linear equation. * **Non-Linear Examples:** $x^2 + 9 = 3$, $x(x + 4) = 2$, and $\sqrt{x} = 5$ are not linear. * **Possible Solutions:** Linear equations in one variable will always have either (a) exactly one real number solution, (b) no solutions, or (c) all real numbers as solutions. --- ### Finding a Unique Solution The PDF illustrates a 6-step procedure using the example $\frac{3}{4}(-\frac{x}{2} - 3) + x = 9 - (x - \frac{5}{8})$: 1. **Get rid of fractions outside parentheses:** Multiply through by the Least Common Multiple (LCM) of the denominators. * *Example:* Multiply by $4$ to get $3(-\frac{x}{2} - 3) + 4x = 36 - 4(x - \frac{5}{8})$. 2. **Simplify using order of operations (PEMDAS):** Use the distributive property and combine like terms on each side, remembering to distribute minus signs. * *Example (Left side):* Distribute the parentheses to get $-\frac{3}{2}x - 9 + 4x = 36 - 4(x - \frac{5}{8})$. Combine like terms to get $\frac{5}{2}x - 9 = 36 - 4(x - \frac{5}{8})$. * *Example (Right side):* Distribute the parentheses to get $\frac{5}{2}x - 9 = 36 - 4x + \frac{5}{2}$. Combine like terms to get $\frac{5}{2}x - 9 = \frac{77}{2} - 4x$. 3. **Repeat as necessary to get the form $ax + b = cx + d$:** Multiply by $2$ to get rid of fractions, resulting in $5x - 18 = 77 - 8x$. 4. **Move variable terms and constant terms to different sides:** Usually, move variables to the side that had the larger variable coefficient to begin with so the equation looks like $ax = b$. * *Example:* Add $8x$ to both sides to get $5x + 8x - 18 = 77$, or $13x - 18 = 77$. Then add $18$ to both sides to get $13x = 77 + 18$, or $13x = 95$. 5. **Divide both sides by the variable's coefficient:** Stop if $a = 0$. * *Example:* Divide by $13$ to get $x = \frac{95}{13}$. 6. **Check the solution by plugging into the original equation:** Plugging $\frac{95}{13}$ back in confirms $\frac{3}{4}(\frac{-95}{2 \cdot 13} - 3) + \frac{95}{13} = 9 - (\frac{95}{13} - \frac{5}{8})$ is true. --- ### Determining if a Unique Solution Exists Any linear equation can be simplified into the form $ax = b$ for some $a$ and $b$: * **Exactly one solution:** If $a \neq 0$, then $x = \frac{b}{a}$. * **No solution:** If $a = 0$ but $b \neq 0$, there is no solution. This happens if, after legal transformations, the new equation is false (e.g., $2 = 3$ or $3x - 7 = 2 + 3x$). * **All real numbers are solutions:** If $a = b = 0$, then all real numbers are solutions. This happens if, after legal transformations, the new equation is an identity (e.g., $2x = 3x - x$ or $1 = 1$). --- ### Inequalities in One Variable To solve linear inequalities, use the exact same procedure as for equalities, **except flip the inequality sign when multiplying or dividing by a negative number** (for example, $-x > 5$ is equivalent to $x < -5$): * **No solution:** The inequality has no solution if it reduces to an impossible statement (e.g., $x + 1 > x + 9$ reduces to $1 > 9$). * **All real numbers as solutions:** The inequality has infinitely many solutions (all real numbers) if it reduces to a statement that is always true (e.g., $5 - x \ge 3 - x$ reduces to $5 \ge 3$). **Solutions Based on the Reduced Form ($ax < b$, $ax \le b$, $ax > b$, $ax \ge b$):** * **When $a > 0$:** Keep the inequality sign in the same direction: * $ax < b$ gives $x < \frac{b}{a}$. * $ax \le b$ gives $x \le \frac{b}{a}$. * $ax > b$ gives $x > \frac{b}{a}$. * $ax \ge b$ gives $x \ge \frac{b}{a}$. * **When $a < 0$:** Flip the direction of the inequality sign: * $ax < b$ gives $x > \frac{b}{a}$. * $ax \le b$ gives $x \ge \frac{b}{a}$. * $ax > b$ gives $x < \frac{b}{a}$. * $ax \ge b$ gives $x \le \frac{b}{a}$. * **When $a = 0$ and $b > 0$:** * $ax < b$ and $ax \le b$ have **all** real numbers as solutions (since $0$ is always less than a positive number $b$). * $ax > b$ and $ax \ge b$ have **none** (no solutions). * **When $a = 0$ and $b < 0$:** * $ax < b$ and $ax \le b$ have **none** (no solutions). * $ax > b$ and $ax \ge b$ have **all** real numbers as solutions (since $0$ is always greater than a negative number $b$). * **When $a = 0$ and $b = 0$:** * Strict inequalities ($ax < b$ and $ax > b$) have **none** (since $0 < 0$ and $0 > 0$ are false). * Inclusive inequalities ($ax \le b$ and $ax \ge b$) have **all** real numbers as solutions (since $0 = 0$ is always true).