Step-by-Step Guidance for College Algebra Worksheet (Solving Equations, Functions, and Applications)

Q1. Solve the following equation: $x - \frac{7}{8} = 3x + \frac{1}{10}$

Background

Topic: Linear Equations

This question tests your ability to solve linear equations with fractions, a fundamental skill in college algebra.

Key Terms and Formulas:

• Linear equation: An equation involving only the first power of the variable.

• Combine like terms: Move all terms involving $x$ to one side and constants to the other.

• Fractions: Find a common denominator to simplify.

Step-by-Step Guidance

1. Move all terms involving $x$ to one side and constants to the other. For example, subtract $3x$ from both sides.

2. Combine the $x$ terms: $x - 3x$.

3. Move the constants: Add $\frac{7}{8}$ to both sides to isolate constants.

4. Combine the fractions on the right side: $\frac{1}{10} + \frac{7}{8}$.

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Q2. Solve the equation: $-x = 1 - \frac{5(x-5)}{6}$

Background

Topic: Linear Equations with Fractions

This question tests your ability to solve linear equations involving fractions and distribution.

Key Terms and Formulas:

• Distributive property: $a(b + c) = ab + ac$

• Combine like terms and isolate $x$.

Step-by-Step Guidance

1. Expand $5(x-5)$ to $5x - 25$.

2. Rewrite the equation: $-x = 1 - \frac{5x - 25}{6}$.

3. Multiply both sides by 6 to clear the fraction.

4. Combine like terms and solve for $x$.

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Q3. Solve the equation: $x + \frac{4}{5} = \frac{1}{7}x + \frac{1}{3}$

Background

Topic: Linear Equations with Fractions

This question tests your ability to solve linear equations with variable terms and constants as fractions.

Key Terms and Formulas:

• Find the least common denominator (LCD) to clear fractions.

• Combine like terms and solve for $x$.

Step-by-Step Guidance

1. Subtract $\frac{1}{7}x$ from both sides to get all $x$ terms on one side.

2. Subtract $\frac{4}{5}$ from both sides to get constants on the other side.

3. Find the LCD for denominators 5, 7, and 3.

4. Multiply both sides by the LCD to clear fractions.

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Q4. Solve the equation: $4[3(x-13)-2(2x+2)- (x-6)-3(3x-4)] = 5$

Background

Topic: Linear Equations with Parentheses and Distribution

This question tests your ability to use the distributive property and combine like terms in a complex linear equation.

Key Terms and Formulas:

• Distributive property: $a(b + c) = ab + ac$

• Combine like terms.

• Find the LCD if fractions are present.

Step-by-Step Guidance

1. Expand each term inside the brackets using the distributive property.

2. Combine all like terms inside the brackets.

3. Multiply the simplified bracket by 4.

4. Set the equation equal to 5 and solve for $x$.

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Q5. For $f(x) = \frac{8}{3}x - 8$, find (a) the solution to $f(x) = 0$, (b) the x-intercept, and (c) the zero of $f(x)$.

Background

Topic: Functions and Intercepts

This question tests your understanding of zeros of functions and how to find intercepts.

Key Terms and Formulas:

• Zero of a function: The value of $x$ where $f(x) = 0$.

• x-intercept: The point where the graph crosses the x-axis, i.e., $y = 0$.

Step-by-Step Guidance

1. Set $f(x) = 0$: $\frac{8}{3}x - 8 = 0$.

2. Add 8 to both sides: $\frac{8}{3}x = 8$.

3. Solve for $x$ by multiplying both sides by the reciprocal of $\frac{8}{3}$.

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Q6. Given a table: X = 6, Y1 = 0; X = 0, Y1 = -38. Find (a) the x-intercept, (b) the y-intercept, and (c) the solution to $f(x) = 0$.

Background

Topic: Functions and Intercepts from Tables

This question tests your ability to interpret function values from a table and find intercepts.

Key Terms and Formulas:

• x-intercept: Where $y = 0$.

• y-intercept: Where $x = 0$.

• Solution to $f(x) = 0$: The $x$ value where $y = 0$.

Step-by-Step Guidance

1. Look for the row where $Y1 = 0$; the corresponding $X$ is the x-intercept.

2. Look for the row where $X = 0$; the corresponding $Y1$ is the y-intercept.

3. The solution to $f(x) = 0$ is the $X$ value where $Y1 = 0$.

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Q7. Using the graph of $y = f(x)$, and given that the zero of the function is $x = 40$, find (a) the x-intercept and (b) the solution to $f(x) = 0$.

Background

Topic: Graphical Interpretation of Functions

This question tests your ability to interpret zeros and intercepts from a graph.

Key Terms and Formulas:

• Zero of a function: $x$ value where $f(x) = 0$.

• x-intercept: Point where the graph crosses the x-axis.

Step-by-Step Guidance

1. Identify the zero of the function from the graph (given as $x = 40$).

2. The x-intercept is the same as the zero.

3. The solution to $f(x) = 0$ is the $x$ value where the graph crosses the x-axis.

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Q8. For $f(x) = x = 190 + 20$, find (a) the zero of the function, (b) the x-intercept, and (c) solve $f(x) = 0$.

Background

Topic: Linear Functions and Zeros

This question tests your ability to solve for zeros and intercepts in linear functions.

Key Terms and Formulas:

• Zero of a function: Set $f(x) = 0$ and solve for $x$.

• x-intercept: Same as zero for linear functions.

Step-by-Step Guidance

1. Set $f(x) = 0$: $x = 190 + 20$.

2. Solve for $x$ by simplifying the right side.

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Q9. Solve the equation using a graphical method: $16x - 49 = 20 - 7x$

Background

Topic: Solving Linear Equations Graphically

This question tests your ability to solve equations by graphing and finding intersection points.

Key Terms and Formulas:

• Graphical method: Plot both sides as separate functions and find where they intersect.

• Intersection point: The $x$ value where $f(x) = g(x)$.

Step-by-Step Guidance

1. Rewrite the equation as $16x - 49 = 20 - 7x$.

2. Move all terms to one side: $16x + 7x = 20 + 49$.

3. Simplify both sides.

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Q10. Solve the equation using a graphical method: $-4 = \frac{3t + 4}{8t - 1}$

Background

Topic: Solving Rational Equations Graphically

This question tests your ability to solve rational equations using graphical methods.

Key Terms and Formulas:

• Rational equation: An equation involving fractions with variables in the denominator.

• Graphical method: Plot $y_1 = -4$ and $y_2 = \frac{3t + 4}{8t - 1}$ and find intersection.

Step-by-Step Guidance

1. Set $y_1 = -4$ and $y_2 = \frac{3t + 4}{8t - 1}$.

2. Find the value of $t$ where $y_1 = y_2$.

3. Alternatively, multiply both sides by $(8t - 1)$ to clear the denominator.

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Q11. Solve the formula for $t$: $G = P(1 + rt)$

Background

Topic: Solving Formulas for a Variable

This question tests your ability to isolate a variable in a formula, a key algebraic skill.

Key Terms and Formulas:

• Isolate $t$ by using inverse operations.

• Distributive property and solving for a variable.

Step-by-Step Guidance

1. Divide both sides by $P$ to get $\frac{G}{P} = 1 + rt$.

2. Subtract 1 from both sides: $\frac{G}{P} - 1 = rt$.

3. Divide both sides by $r$ to solve for $t$.

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Q12. Solve $B + H = M$ for $x$ (assuming $x$ is one of the variables).

Background

Topic: Solving Formulas for a Variable

This question tests your ability to rearrange equations to solve for a specific variable.

Key Terms and Formulas:

• Isolate $x$ using inverse operations.

Step-by-Step Guidance

1. Subtract $H$ from both sides: $B = M - H$.

2. If $x$ is $B$, then $x = M - H$.

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Q13. Solve for $y$ and graph the equation: $9x - 4y = 9$

Background

Topic: Linear Equations and Graphing

This question tests your ability to solve for $y$ and understand how to graph a linear equation.

Key Terms and Formulas:

• Slope-intercept form: $y = mx + b$

• Isolate $y$ by using inverse operations.

Step-by-Step Guidance

1. Subtract $9x$ from both sides: $-4y = 9 - 9x$.

2. Divide both sides by $-4$ to solve for $y$.

3. Simplify the expression to get $y$ in terms of $x$.

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Q14. The profit from the production and sale of specialty golf hats is given by $P(x) = 20x - 2000$. (a) Producing and selling how many hats will give a profit of $8000$? (b) How many hats must be produced and sold to avoid a loss?

Background

Topic: Applications of Linear Functions

This question tests your ability to apply linear functions to real-world scenarios, such as profit analysis.

Key Terms and Formulas:

• Profit function: $P(x) = 20x - 2000$

• Set $P(x) = 8000$ for part (a).

• Set $P(x) = 0$ for part (b) to avoid a loss.

Step-by-Step Guidance

1. For part (a), set $20x - 2000 = 8000$ and solve for $x$.

2. Add $2000$to both sides: $20x = 10000$.

3. Divide both sides by $20$to solve for$x$.

4. For part (b), set $20x - 2000 = 0$ and solve for $x$.

5. Add $2000$to both sides: $20x = 2000$.

6. Divide both sides by $20$to solve for$x$.

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