Unit 4 Lesson 1 Practice Problems Answer Key

Unit 4 Lesson 1 Practice Problems – Answer Key

Introduction

In this article you will find the complete answer key for all of the practice problems introduced in Unit 4, Lesson 1. This set of exercises reinforces the core concepts such as solving linear equations, interpreting inequalities, and applying algebraic reasoning. By working through each example step by step, students build confidence, develop accurate habits, and prepare for more advanced topics that follow in later units. The explanations are written in clear, conversational English so that learners—whether they are high school freshmen or adult educators—can see exactly how to arrive at every correct result Less friction, more output..

What You’ll Find Here

Below is a concise table of contents:

• Overview of the lesson – quick recap of the goals.
• Lesson objectives – three bullet points that define success.
• Practice problems – four representative items with their full solutions.
• Detailed solution walkthrough – where each calculation is broken down.
• Common mistakes – frequent pitfalls and how to sidestep them.
• Study tips – short strategies to improve performance.
• Conclusion – encouragement to keep practicing.

Lesson Objectives

These three targets guide your study of Unit 4, Lesson 1:

1. Solve basic linear equations and verify solutions by substitution.
2. Graph simple linear inequalities on a coordinate plane and identify the shaded region.
3. Translate real‑world word problems into algebraic expressions, then solve for the unknown.

Mastering these skills lays the foundation for quadratic functions, systems of equations, and data analysis in upcoming lessons.

Practice Problems

Problem 1 – Solving a Two‑Term Equation

Find the value of (x) that satisfies

[ 5x - 7 = 3x + 12 ]

Solution:

1. Move all terms containing (x) to the left side: subtract (3x) from both sides.
   [ 5x - 3x - 7 = 12 \quad\Rightarrow\quad 2x - 7 = 12 ]

2. Isolate the variable term by adding 7 to both sides.
   [ 2x = 12 + 7 \quad\Rightarrow\quad 2x = 19 ]

3. Divide both sides by 2.
   [ x = \frac{19}{2} = 9.5 ]

Answer: (\boxed{9.5})

Check: Substitute (x=9.5) back into the original equation:
(5(9.5)-7 = 47.5-7 = 40.5); right side: (3(9.5)+12 = 28.5+12 = 40.5). Both sides match → solution verified.

Problem 2 – Graphing a Linear Inequality

Graph the inequality

[ y > -\frac{2}{3}x + 1 ]

Solution:

1. Identify the slope and intercept. The line has slope (m = -\frac{2}{3}) and y‑intercept ((0,b) = (0,1)).
2. Plot the point ((0,1)) on the coordinate grid.
3. From the origin, move down 2 units and right 3 units (since slope is (-2/3)) to locate another point, say ((3,-1)).
4. Draw a dashed (or solid depending on “>”) line through these two points. Because the inequality uses “>”, the line itself is not included; use a dotted line.
5. Shade the region above the line (the direction that makes test point satisfy). Choose ((0,2)): plug in (x=0), (y=2): (2 > 1) true, so shade upward.

Final graph description: A straight line passing through ((0,1)) and ((3,-1)), shaded above it, with a dotted border.

Answer: The shading region corresponds to all points ((x,y)) satisfying (y > -\frac{2}{3}x + 1) Simple, but easy to overlook..

Problem 3 – Translating a Word Problem into an Equation

A farmer has 48 meters of fencing to enclose a rectangular field along its perimeter. If the length is 10 meters longer than the width, how long is the width?

Solution:

1. Define variables: let (w) = width (meters), then length (l = w + 10).

2. Perimeter formula for a rectangle: (P = 2(l + w)). Set equal to available fencing:

   [ 2\big[(w + 10) + w\big] = 48 ]

3. Simplify inside brackets:

   [ 2(2w + 10) = 48 \quad\Rightarrow\quad 4w + 20 = 48 ]

4. Subtract 20 from

Problem 3 – Translating a Word Problem into an Equation

1. Define the variables: let (w) be the width (in meters).
   Then the length is (l = w + 10) Nothing fancy..

2. Write the perimeter formula for a rectangle:

   [ P = 2(l + w) ]

   Set the total fencing equal to the available length, 48 m:

   [ 2\big[(w + 10) + w\big] = 48 ]

3. Simplify inside the brackets:

   [ 2(2w + 10) = 48 \quad\Rightarrow\quad 4w + 20 = 48 ]

4. Subtract 20 from both sides:

   [ 4w = 48 - 20 \quad\Rightarrow\quad 4w = 28 ]

5. Divide by 4 to solve for (w):

   [ w = \frac{28}{4} = 7 ]

Thus the width of the field is 7 meters, and the corresponding length is

[ l = w + 10 = 7 + 10 = 17\ \text{m}. ]

Check:
Perimeter (= 2(l + w) = 2(17 + 7) = 2 \times 24 = 48) m, which matches the 48 m of fencing provided.

Answer: (\boxed{7}) meters.

Conclusion

The exercises above demonstrate how to move between words and algebra, manipulate one‑term and two‑term equations, interpret the slope and intercept of lines, and translate real‑world constraints (such as fencing a rectangle) into quadratic‑style relationships. Completing Problems 1–3 gives students fluency in:

• Solving linear equations and isolating variables.
• Graphing and shading regions defined by linear inequalities.
• Building and solving word problems by