Lesson 3: Functions and Equations – Answer Key (Page 599)
Functions and equations form the backbone of algebra, enabling us to model relationships and solve real‑world problems. This guide presents a detailed answer key for Lesson 3, complete with step‑by‑step explanations, common pitfalls, and practical tips for mastering the material. Whether you’re a student preparing for exams or a teacher looking for a resource, this article will help you deal with the concepts with confidence.
Introduction
In Lesson 3, we explored functions, equations, and the interplay between them. Understanding these concepts is essential for higher‑level math, science, engineering, and everyday problem‑solving. The exercises on page 599 test a range of skills: identifying function types, solving linear and quadratic equations, interpreting graph data, and applying algebraic manipulation. Below, each problem is tackled in order, with clear reasoning and alternative strategies where helpful.
Problem 1 – Identifying the Function Type
Question:
Given the set of ordered pairs ({(1,3), (2,5), (3,7), (4,9)}), determine whether the relation is a function and, if so, identify its type (linear, quadratic, etc.).
Answer & Explanation:
-
Step 1: Check the function criterion.
A relation is a function if each input (x‑value) maps to exactly one output (y‑value). Here, every x value (1, 2, 3, 4) has a unique y value, so it is a function The details matter here. Turns out it matters.. -
Step 2: Identify the pattern.
Compute differences between successive y‑values:
(5-3 = 2), (7-5 = 2), (9-7 = 2).
Constant first differences indicate a linear function. -
Step 3: Find the equation.
Using two points, say (1, 3) and (2, 5):
Slope (m = \frac{5-3}{2-1} = 2).
Equation: (y = 2x + b).
Plug (1, 3): (3 = 2(1) + b \Rightarrow b = 1).
Final equation: (y = 2x + 1) The details matter here..
Problem 2 – Solving a Linear Equation
Question:
Solve for (x) in the equation:
[3(2x - 5) = 4(x + 1) + 7.]
Answer & Explanation:
- Distribute:
(6x - 15 = 4x + 4 + 7). - Combine constants on the right:
(6x - 15 = 4x + 11). - Move variables to one side:
(6x - 4x = 11 + 15). - Simplify:
(2x = 26). - Solve for (x):
(x = 13).
Check:
Left side: (3(2(13)-5) = 3(26-5) = 3(21) = 63).
Right side: (4(13+1)+7 = 4(14)+7 = 56+7 = 63). ✅
Problem 3 – Quadratic Equation via Factoring
Question:
Factor and solve:
[x^2 - 5x + 6 = 0.]
Answer & Explanation:
- Look for two numbers that multiply to (6) and add to (-5): (-2) and (-3).
- Factor: ((x-2)(x-3) = 0).
- Set each factor to zero:
(x-2 = 0 \Rightarrow x = 2).
(x-3 = 0 \Rightarrow x = 3).
Solutions: (x = 2) or (x = 3) Not complicated — just consistent. Surprisingly effective..
Problem 4 – Graph Interpretation
Question:
A graph shows a line passing through points ((0, -4)) and ((3, 5)). Find the slope and y‑intercept, and write the equation in slope‑intercept form And that's really what it comes down to..
Answer & Explanation:
- Slope (m = \frac{5 - (-4)}{3 - 0} = \frac{9}{3} = 3).
- Y‑intercept is the point where (x = 0), which is ((0, -4)). Thus (b = -4).
- Equation: (y = 3x - 4).
Problem 5 – System of Equations
Question:
Solve the system:
[
\begin{cases}
2x + 3y = 12 \
x - y = 1
\end{cases}
]
Answer & Explanation:
- Express (x) from the second equation:
(x = y + 1). - Substitute into the first:
(2(y+1) + 3y = 12). - Simplify:
(2y + 2 + 3y = 12 \Rightarrow 5y + 2 = 12). - Solve for (y):
(5y = 10 \Rightarrow y = 2). - Find (x):
(x = 2 + 1 = 3).
Solution: ((x, y) = (3, 2)).
Problem 6 – Function Composition
Question:
Let (f(x) = 2x + 1) and (g(x) = x^2). Find ((f \circ g)(x)) and ((g \circ f)(x)).
Answer & Explanation:
-
(f \circ g): Apply (g) first, then (f).
((f \circ g)(x) = f(g(x)) = f(x^2) = 2(x^2) + 1 = 2x^2 + 1). -
(g \circ f): Apply (f) first, then (g).
((g \circ f)(x) = g(f(x)) = g(2x + 1) = (2x + 1)^2 = 4x^2 + 4x + 1).
Problem 7 – Solving a Rational Equation
Question:
Solve for (x):
[\frac{2}{x-1} + 3 = 5.]
Answer & Explanation:
- Isolate the fraction:
(\frac{2}{x-1} = 5 - 3 = 2). - Cross‑multiply:
(2 = 2(x-1)). - Simplify:
(2 = 2x - 2 \Rightarrow 2x = 4 \Rightarrow x = 2).
Check for extraneous solutions:
Denominator (x-1 \neq 0). (x = 2) satisfies this, so it is valid No workaround needed..
Problem 8 – Quadratic Formula
Question:
Solve (2x^2 - 4x - 6 = 0) using the quadratic formula.
Answer & Explanation:
Quadratic formula: (x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}).
Here, (a = 2), (b = -4), (c = -6) Worth keeping that in mind. Less friction, more output..
- Compute discriminant:
(b^2 - 4ac = (-4)^2 - 4(2)(-6) = 16 + 48 = 64). - Apply formula:
(x = \frac{-(-4) \pm \sqrt{64}}{2 \cdot 2} = \frac{4 \pm 8}{4}). - Find two solutions:
- (x = \frac{4 + 8}{4} = \frac{12}{4} = 3).
- (x = \frac{4 - 8}{4} = \frac{-4}{4} = -1).
Solutions: (x = 3) or (x = -1).
Problem 9 – Exponential Function
Question:
Find the value of (x) in the equation (5 \cdot 2^x = 80).
Answer & Explanation:
- Isolate the exponential term:
(2^x = \frac{80}{5} = 16). - Recognize 16 as a power of 2:
(16 = 2^4). - Set exponents equal:
(2^x = 2^4 \Rightarrow x = 4).
Problem 10 – Logarithmic Equation
Question:
Solve (\log_3(x) + \log_3(x-4) = 2).
Answer & Explanation:
- Use log property: (\log_a b + \log_a c = \log_a(bc)).
(\log_3[x(x-4)] = 2). - Exponentiate both sides:
(x(x-4) = 3^2 = 9). - Form quadratic:
(x^2 - 4x - 9 = 0). - Solve using quadratic formula:
(a = 1), (b = -4), (c = -9).
Discriminant (= (-4)^2 - 4(1)(-9) = 16 + 36 = 52).
(x = \frac{4 \pm \sqrt{52}}{2} = \frac{4 \pm 2\sqrt{13}}{2} = 2 \pm \sqrt{13}). - Check domain:
Argument of logarithm must be positive:- (x > 0).
- (x-4 > 0 \Rightarrow x > 4).
Only (x = 2 + \sqrt{13}) (≈ 5.605) satisfies both conditions.
Solution: (x = 2 + \sqrt{13}).
Common Mistakes & How to Avoid Them
| Concept | Typical Error | Fix |
|---|---|---|
| Linear equations | Forgetting to distribute parentheses | Always expand fully before combining like terms |
| Quadratic equations | Mis‑identifying the sign of coefficients | Double‑check each step, especially when factoring |
| Systems of equations | Substituting the wrong variable | Label each equation clearly; write variable expressions separately |
| Rational equations | Ignoring extraneous solutions | After finding a solution, verify that it does not make any denominator zero |
| Logarithms | Applying log rules incorrectly | Remember (\log_a b + \log_a c = \log_a (bc)) and (\log_a b - \log_a c = \log_a (b/c)) |
Tips for Mastery
- Practice Pattern Recognition – Many algebra problems rely on spotting patterns (e.g., constant differences for linear functions, perfect squares for quadratics).
- Check Your Work – Plug solutions back into the original equation to confirm validity.
- Use Graphs When Possible – Visualizing functions can reveal hidden solutions or confirm the nature of the function (linear, quadratic, etc.).
- Keep Units and Domains in Mind – Especially with logs and rational expressions, domain restrictions often eliminate apparent solutions.
- Build Incrementally – Start with simple problems, then layer in complexity (e.g., systems of equations with more variables).
Conclusion
Mastering functions and equations unlocks powerful problem‑solving tools across mathematics and science. Because of that, use the step‑by‑step solutions above as a study guide, and revisit the common pitfalls to strengthen your algebraic intuition. But by understanding the underlying principles—identifying function types, manipulating algebraic expressions, applying the quadratic formula, and respecting domain constraints—you can confidently tackle any challenge presented in Lesson 3. With practice, the patterns will become second nature, and the confidence to solve even the most complex equations will grow.
