Unit 1 Assessment Review: Linear Equations, Literal Formulas, and Modeling

Review Guidelines and Core Objectives

• Assignment Submission Requirements:

  • Complete all review problems directly within a physical notebook.

  • Completed work must be uploaded and turned in as a Google Classroom submission.

• Unit Scope and Content Areas:

  • Solving one-step and two-step linear equations involving integer and rational coefficients.

  • Solving multi-step equations requiring the collection and combining of like terms.

  • Solving equations involving grouping symbols and the distributive property.

  • Solving equations with variables on both sides of the equal sign.

  • Clearing denominators to solve rational linear equations.

  • Rearranging literal equations and geometric formulas for a designated variable.

  • Setting up and solving linear equations and cost functions for real-world modeling scenarios.

Two-Step Linear Equations and Rational Groupings

• Problem 1: Solve $7x - 21 = 14$

  • Given Equation: $7x - 21 = 14$

  • Step 1 (Addition Property of Equality): Add $21$ to both sides of the equation to isolate the variable term:     $7x - 21 + 21 = 14 + 21$     $7x = 35$

  • Step 2 (Division Property of Equality): Divide both sides by the coefficient $7$:     $\frac{7x}{7} = \frac{35}{7}$     $x = 5$

  • Solution: $x = 5$

• Problem 2: Solve $9x + 12 = 48$

  • Given Equation: $9x + 12 = 48$

  • Step 1 (Subtraction Property of Equality): Subtract $12$ from both sides to isolate the variable term:     $9x + 12 - 12 = 48 - 12$     $9x = 36$

  • Step 2 (Division Property of Equality): Divide both sides by the coefficient $9$:     $\frac{9x}{9} = \frac{36}{9}$     $x = 4$

  • Solution: $x = 4$

• Problem 3: Solve $\frac{x}{5} + 3 = 11$

  • Given Equation: $\frac{x}{5} + 3 = 11$

  • Step 1 (Subtraction Property of Equality): Subtract $3$ from both sides:     $\frac{x}{5} + 3 - 3 = 11 - 3$     $\frac{x}{5} = 8$

  • Step 2 (Multiplication Property of Equality): Multiply both sides by the denominator $5$:     $5 \times \left(\frac{x}{5}\right) = 5 \times 8$     $x = 40$

  • Solution: $x = 40$

• Problem 4: Solve $\frac{x}{6} - 4 = 3$

  • Given Equation: $\frac{x}{6} - 4 = 3$

  • Step 1 (Addition Property of Equality): Add $4$ to both sides:     $\frac{x}{6} - 4 + 4 = 3 + 4$     $\frac{x}{6} = 7$

  • Step 2 (Multiplication Property of Equality): Multiply both sides by the denominator $6$:     $6 \times \left(\frac{x}{6}\right) = 6 \times 7$     $x = 42$

  • Solution: $x = 42$

• Problem 5: Solve $\frac{x + 3}{5} = 8$

  • Given Equation: $\frac{x + 3}{5} = 8$

  • Step 1 (Clear the Denominator): Multiply both sides of the equation by $5$:     $5 \times \left(\frac{x + 3}{5}\right) = 5 \times 8$     $x + 3 = 40$

  • Step 2 (Subtraction Property of Equality): Subtract $3$ from both sides:     $x + 3 - 3 = 40 - 3$     $x = 37$

  • Solution: $x = 37$

• Problem 6: Solve $\frac{x - 4}{3} = 7$

  • Given Equation: $\frac{x - 4}{3} = 7$

  • Step 1 (Clear the Denominator): Multiply both sides of the equation by $3$:     $3 \times \left(\frac{x - 4}{3}\right) = 3 \times 7$     $x - 4 = 21$

  • Step 2 (Addition Property of Equality): Add $4$ to both sides:     $x - 4 + 4 = 21 + 4$     $x = 25$

  • Solution: $x = 25$

Multi-Step Equations with Like Terms and Rational Coefficients

• Problem 7: Solve $4x + 9 - 7x = -6$

  • Given Equation: $4x + 9 - 7x = -6$

  • Step 1 (Combine Like Terms): Combine $4x$ and $-7x$ on the left-hand side:     $(4x - 7x) + 9 = -6$     $-3x + 9 = -6$

  • Step 2 (Subtraction Property of Equality): Subtract $9$ from both sides:     $-3x + 9 - 9 = -6 - 9$     $-3x = -15$

  • Step 3 (Division Property of Equality): Divide both sides by $-3$:     $\frac{-3x}{-3} = \frac{-15}{-3}$     $x = 5$

  • Solution: $x = 5$

• Problem 8: Solve $6x - 4 - 2x = 20$

  • Given Equation: $6x - 4 - 2x = 20$

  • Step 1 (Combine Like Terms): Combine $6x$ and $-2x$ on the left-hand side:     $(6x - 2x) - 4 = 20$     $4x - 4 = 20$

  • Step 2 (Addition Property of Equality): Add $4$ to both sides:     $4x - 4 + 4 = 20 + 4$     $4x = 24$

  • Step 3 (Division Property of Equality): Divide both sides by $4$:     $\frac{4x}{4} = \frac{24}{4}$     $x = 6$

  • Solution: $x = 6$

• Problem 9: Solve $-8 = \frac{3}{4}x + 1$

  • Given Equation: $-8 = \frac{3}{4}x + 1$

  • Step 1 (Subtraction Property of Equality): Subtract $1$ from both sides:     $-8 - 1 = \frac{3}{4}x + 1 - 1$     $-9 = \frac{3}{4}x$

  • Step 2 (Multiply by the Reciprocal): Multiply both sides by $\frac{4}{3}$:     $\frac{4}{3} \times (-9) = \frac{4}{3} \times \left(\frac{3}{4}x\right)$     $-\frac{36}{3} = x$     $x = -12$

  • Solution: $x = -12$

• Problem 10: Solve $5 = \frac{2}{3}x - 3$

  • Given Equation: $5 = \frac{2}{3}x - 3$

  • Step 1 (Addition Property of Equality): Add $3$ to both sides:     $5 + 3 = \frac{2}{3}x - 3 + 3$     $8 = \frac{2}{3}x$

  • Step 2 (Multiply by the Reciprocal): Multiply both sides by $\frac{3}{2}$:     $\frac{3}{2} \times 8 = \frac{3}{2} \times \left(\frac{2}{3}x\right)$     $\frac{24}{2} = x$     $x = 12$

  • Solution: $x = 12$

Equations with Variables on Both Sides and the Distributive Property

• Problem 11: Solve $5x - 4 = -3x + 20$

  • Given Equation: $5x - 4 = -3x + 20$

  • Step 1 (Collect Variable Terms on One Side): Add $3x$ to both sides:     $5x + 3x - 4 = -3x + 3x + 20$     $8x - 4 = 20$

  • Step 2 (Addition Property of Equality): Add $4$ to both sides:     $8x - 4 + 4 = 20 + 4$     $8x = 24$

  • Step 3 (Division Property of Equality): Divide both sides by $8$:     $\frac{8x}{8} = \frac{24}{8}$     $x = 3$

  • Solution: $x = 3$

• Problem 12: Solve $7x + 9 = 2x - 16$

  • Given Equation: $7x + 9 = 2x - 16$

  • Step 1 (Collect Variable Terms on One Side): Subtract $2x$ from both sides:     $7x - 2x + 9 = 2x - 2x - 16$     $5x + 9 = -16$

  • Step 2 (Subtraction Property of Equality): Subtract $9$ from both sides:     $5x + 9 - 9 = -16 - 9$     $5x = -25$

  • Step 3 (Division Property of Equality): Divide both sides by $5$:     $\frac{5x}{5} = \frac{-25}{5}$     $x = -5$

  • Solution: $x = -5$

• Problem 13: Solve $-2(x - 5) - 6 = 3(x + 1)$

  • Given Equation: $-2(x - 5) - 6 = 3(x + 1)$

  • Step 1 (Apply the Distributive Property): Expand parentheses across both sides:     $-2 \times x - 2 \times (-5) - 6 = 3 \times x + 3 \times 1$     $-2x + 10 - 6 = 3x + 3$

  • Step 2 (Combine Constant Terms on the Left):     $-2x + 4 = 3x + 3$

  • Step 3 (Collect Variable Terms): Subtract $3x$ from both sides:     $-2x - 3x + 4 = 3x - 3x + 3$     $-5x + 4 = 3$

  • Step 4 (Collect Constants): Subtract $4$ from both sides:     $-5x + 4 - 4 = 3 - 4$     $-5x = -1$

  • Step 5 (Division Property of Equality): Divide both sides by $-5$:     $\frac{-5x}{-5} = \frac{-1}{-5}$     $x = \frac{1}{5}$

  • Solution: $x = \frac{1}{5}$ (or $0.2$)

• Problem 14: Solve $4(2x + 1) - 7 = 3(x - 2)$

  • Given Equation: $4(2x + 1) - 7 = 3(x - 2)$

  • Step 1 (Apply the Distributive Property): Expand parentheses across both sides:     $4 \times 2x + 4 \times 1 - 7 = 3 \times x - 3 \times 2$     $8x + 4 - 7 = 3x - 6$

  • Step 2 (Combine Constant Terms on the Left):     $8x - 3 = 3x - 6$

  • Step 3 (Collect Variable Terms): Subtract $3x$ from both sides:     $8x - 3x - 3 = 3x - 3x - 6$     $5x - 3 = -6$

  • Step 4 (Collect Constants): Add $3$ to both sides:     $5x - 3 + 3 = -6 + 3$     $5x = -3$

  • Step 5 (Division Property of Equality): Divide both sides by $5$:     $\frac{5x}{5} = \frac{-3}{5}$     $x = -\frac{3}{5}$

  • Solution: $x = -\frac{3}{5}$ (or $-0.6$)

• Problem 21: Solve $3(2x + 5) = \frac{2}{3}(27 - 9x)$

  • Given Equation: $3(2x + 5) = \frac{2}{3}(27 - 9x)$

  • Step 1 (Expand Left Side):     $3 \times 2x + 3 \times 5 = 6x + 15$

  • Step 2 (Expand Right Side):     $\frac{2}{3} \times 27 - \frac{2}{3} \times 9x = 2 \times 9 - 2 \times 3x = 18 - 6x$

  • Step 3 (Equate and Collect Variable Terms): Set expressions equal and add $6x$ to both sides:     $6x + 15 = 18 - 6x$     $6x + 6x + 15 = 18 - 6x + 6x$     $12x + 15 = 18$

  • Step 4 (Collect Constant Terms): Subtract $15$ from both sides:     $12x + 15 - 15 = 18 - 15$     $12x = 3$

  • Step 5 (Division Property of Equality): Divide both sides by $12$ and simplify the fraction:     $\frac{12x}{12} = \frac{3}{12}$     $x = \frac{1}{4}$

  • Solution: $x = \frac{1}{4}$ (or $0.25$)

• Problem 22: Solve $4(3x - 2) = \frac{5}{6}(24 + 12x)$

  • Given Equation: $4(3x - 2) = \frac{5}{6}(24 + 12x)$

  • Step 1 (Expand Left Side):     $4 \times 3x - 4 \times 2 = 12x - 8$

  • Step 2 (Expand Right Side):     $\frac{5}{6} \times 24 + \frac{5}{6} \times 12x = 5 \times 4 + 5 \times 2x = 20 + 10x$

  • Step 3 (Equate and Collect Variable Terms): Set expressions equal and subtract $10x$ from both sides:     $12x - 8 = 20 + 10x$     $12x - 10x - 8 = 20 + 10x - 10x$     $2x - 8 = 20$

  • Step 4 (Collect Constant Terms): Add $8$ to both sides:     $2x - 8 + 8 = 20 + 8$     $2x = 28$

  • Step 5 (Division Property of Equality): Divide both sides by $2$:     $\frac{2x}{2} = \frac{28}{2}$     $x = 14$

  • Solution: $x = 14$

Multi-Term Rational Equations and Clearing Denominators

• Problem 15: Solve $\frac{x}{6} + 2 = \frac{x}{3} - \frac{x}{2}$

  • Given Equation: $\frac{x}{6} + 2 = \frac{x}{3} - \frac{x}{2}$

  • Step 1 (Determine Least Common Denominator): The denominators present are $6$, $3$, and $2$. The Least Common Multiple (LCM) is $6$.

  • Step 2 (Multiply Every Term by the LCD): Multiply all terms on both sides by $6$ to eliminate all fractions:     $6 \times \left(\frac{x}{6}\right) + 6 \times (2) = 6 \times \left(\frac{x}{3}\right) - 6 \times \left(\frac{x}{2}\right)$     $x + 12 = 2x - 3x$

  • Step 3 (Combine Like Terms): Simplify the right side of the equation:     $x + 12 = -x$

  • Step 4 (Collect Variable Terms): Subtract $x$ from both sides:     $x - x + 12 = -x - x$     $12 = -2x$

  • Step 5 (Solve for $x$): Divide both sides by $-2$:     $\frac{12}{-2} = \frac{-2x}{-2}$     $-6 = x$

  • Solution: $x = -6$

• Problem 16: Solve $\frac{x}{10} + 4 = \frac{x}{5} + \frac{x}{2}$

  • Given Equation: $\frac{x}{10} + 4 = \frac{x}{5} + \frac{x}{2}$

  • Step 1 (Determine Least Common Denominator): The denominators present are $10$, $5$, and $2$. The LCM is $10$.

  • Step 2 (Multiply Every Term by the LCD): Multiply all terms on both sides by $10$:     $10 \times \left(\frac{x}{10}\right) + 10 \times (4) = 10 \times \left(\frac{x}{5}\right) + 10 \times \left(\frac{x}{2}\right)$     $x + 40 = 2x + 5x$

  • Step 3 (Combine Like Terms): Simplify the right side:     $x + 40 = 7x$

  • Step 4 (Collect Variable Terms): Subtract $x$ from both sides:     $x - x + 40 = 7x - x$     $40 = 6x$

  • Step 5 (Solve for $x$): Divide both sides by $6$ and reduce the fraction to lowest terms:     $\frac{40}{6} = \frac{6x}{6}$     $x = \frac{20}{3}$

  • Solution: $x = \frac{20}{3}$ (or $6\frac{2}{3}$)

Literal Equations and Formula Rearrangement

• Problem 17: Solve for $y$: $6x - y = 15$

  • Given Equation: $6x - y = 15$

  • Step 1 (Isolate the Variable Term): Subtract $6x$ from both sides:     $6x - 6x - y = -6x + 15$     $-y = -6x + 15$

  • Step 2 (Eliminate Negative Coefficient): Multiply or divide the entire equation by $-1$:     $\frac{-y}{-1} = \frac{-6x + 15}{-1}$     $y = 6x - 15$

  • Solution: $y = 6x - 15$

• Problem 18: Solve for $y$: $3x + 2y = 18$

  • Given Equation: $3x + 2y = 18$

  • Step 1 (Isolate the Variable Term): Subtract $3x$ from both sides:     $3x - 3x + 2y = 18 - 3x$     $2y = -3x + 18$

  • Step 2 (Division Property of Equality): Divide every term by the coefficient $2$:     $\frac{2y}{2} = \frac{-3x + 18}{2}$     $y = -\frac{3}{2}x + 9$

  • Alternative Equivalent Form: $y = \frac{18 - 3x}{2}$

  • Solution: $y = -\frac{3}{2}x + 9$

• Problem 19: Solve for $h$: $A = \frac{1}{2}bh$

  • Formula Context: Area of a triangle, where $A$ represents area, $b$ represents base length, and $h$ represents perpendicular height.

  • Given Equation: $A = \frac{1}{2}bh$

  • Step 1 (Clear the Fraction): Multiply both sides by $2$:     $2 \times A = 2 \times \left(\frac{1}{2}bh\right)$     $2A = bh$

  • Step 2 (Isolate Variable $h$): Divide both sides by the base $b$ (assuming $b \neq 0$):     $\frac{2A}{b} = \frac{bh}{b}$     $h = \frac{2A}{b}$

  • Solution: $h = \frac{2A}{b}$

• Problem 20: Solve for $w$: $P = 2l + 2w$

  • Formula Context: Perimeter of a rectangle, where $P$ is the total perimeter, $l$ is the length, and $w$ is the width.

  • Given Equation: $P = 2l + 2w$

  • Step 1 (Isolate the Term Containing $w$): Subtract $2l$ from both sides:     $P - 2l = 2l - 2l + 2w$     $P - 2l = 2w$

  • Step 2 (Isolate Variable $w$): Divide both sides by $2$:     $\frac{P - 2l}{2} = \frac{2w}{2}$     $w = \frac{P - 2l}{2}$

  • Alternative Equivalent Form: $w = \frac{P}{2} - l$

  • Solution: $w = \frac{P - 2l}{2}$

Applied Linear Modeling and Word Problems

• Problem 23: School Club T-Shirt Purchase

  • Scenario Details: A school club buys $36$ T-shirts. A discount coupon of $\$12$ is applied to the total order, resulting in an overall purchase cost of $\$420$. Let $x$ represent the original, non-discounted price of a single T-shirt in dollars.

  • Part A: Formulate the Equation:

  • Total cost before coupon: $36 \times x = 36x$

  • Coupon deduction: $- 12$

  • Final transaction amount: $420$

  • Equation: $36x - 12 = 420$

  • Part B: Solve the Equation:

  • Add $12$ to both sides to reverse the coupon deduction:       $36x - 12 + 12 = 420 + 12$       $36x = 432$

  • Divide by $36$ to determine the single-unit price:       $\frac{36x}{36} = \frac{432}{36}$       $x = 12$

  • Answer: The original price of one T-shirt is $\$12$.

• Problem 24: Camp Water Bottle Purchase

  • Scenario Details: A camp purchases $28$ water bottles. A discount of $\$16$ is applied to the total purchase, and the camp pays an aggregate amount of $\$320$. Let $x$ represent the original price of one water bottle in dollars.

  • Part A: Formulate the Equation:

  • Total price of bottles before discount: $28 \times x = 28x$

  • Discount deducted: $- 16$

  • Net amount paid: $320$

  • Equation: $28x - 16 = 320$

  • Part B: Solve the Equation:

  • Add $16$ to both sides:       $28x - 16 + 16 = 320 + 16$       $28x = 336$

  • Divide both sides by $28$:       $\frac{28x}{28} = \frac{336}{28}$       $x = 12$

  • Answer: The original price of one water bottle is $\$12$.

• Problem 25: Bowling Club Cost Function

  • Scenario Details: The total cost $C$, measured in dollars, for a bowling club is defined by the linear model $C = 24x + 35$, where $x$ represents the total number of bowling trips taken.

  • Part A: Solve the Equation for $x$:

  • Given function: $C = 24x + 35$

  • Subtract the fixed fee ($35$) from both sides:       $C - 35 = 24x$

  • Divide by the rate per trip ($24$):       $x = \frac{C - 35}{24}$

  • Equation for $x$: $x = \frac{C - 35}{24}$

  • Part B: Evaluate the Number of Trips for a Total Cost of $\$179$:

  • Substitute $C = 179$ into the rearranged expression:       $x = \frac{179 - 35}{24}$       $x = \frac{144}{24}$       $x = 6$

  • Answer: The number of bowling trips taken was $6$.

• Problem 26: Hiking Club Cost Function

  • Scenario Details: The total cost $C$, in dollars, for a hiking club is modeled by $C = 45x + 50$, where $x$ represents the number of hiking trips taken.

  • Part A: Solve the Equation for $x$:

  • Given function: $C = 45x + 50$

  • Subtract the fixed membership fee ($50$) from both sides:       $C - 50 = 45x$

  • Divide both sides by the per-trip rate ($45$):       $x = \frac{C - 50}{45}$

  • Equation for $x$: $x = \frac{C - 50}{45}$

  • Part B: Evaluate the Number of Trips for a Total Cost of $\$275$:

  • Substitute $C = 275$ into the isolated expression:       $x = \frac{275 - 50}{45}$       $x = \frac{225}{45}$       $x = 5$

  • Answer: The number of hiking trips taken was $5$.