College Algebra Exam 1 Study Guide – Step-by-Step Guidance

Q1. Solve the linear equation for the variable x: 3x - 7 = 17

Background

Topic: Solving Linear Equations

This question tests your ability to isolate the variable x in a simple linear equation.

Key Terms and Formulas

• Linear Equation: An equation of the form $ax + b = c$.

• To solve, use inverse operations to isolate x.

Step-by-Step Guidance

1. Start by adding 7 to both sides of the equation to move the constant term.

2. Next, divide both sides by 3 to solve for x.

Try solving on your own before revealing the answer!

Final Answer: x = 8

$3x - 7 = 17$

Add 7 to both sides: $3x = 24$

Divide both sides by 3: $x = 8$

The solution is x = 8.

Q2. Solve the linear equation for the variable x: 5x - 10 = 40

Background

Topic: Solving Linear Equations

This question also tests your ability to solve for x in a linear equation.

Key Terms and Formulas

• Linear Equation: $ax + b = c$

Step-by-Step Guidance

1. Add 10 to both sides to move the constant term.

2. Divide both sides by 5 to isolate x.

Try solving on your own before revealing the answer!

Final Answer: x = 10

$5x - 10 = 40$

Add 10: $5x = 50$

Divide by 5: $x = 10$

Q3. Determine the restriction of the following rational expression: $\frac{6}{x-9}$

Background

Topic: Rational Expressions – Domain Restrictions

This question tests your understanding of when a rational expression is undefined.

Key Terms and Formulas

• Rational Expression: An expression of the form $\frac{a}{b}$, where b cannot be zero.

• Restriction: Values of x that make the denominator zero.

Step-by-Step Guidance

1. Set the denominator equal to zero: $x - 9 = 0$.

2. Solve for x to find the restriction.

Try solving on your own before revealing the answer!

Final Answer: x ≠ 9

The expression is undefined when $x - 9 = 0$, so $x = 9$ is not allowed.

Q4. Determine the restriction, then solve the rational equation: $\frac{9}{x-7} + 1 = \frac{8}{x-7}$

Background

Topic: Rational Equations and Restrictions

This question tests your ability to find restrictions and solve rational equations.

Key Terms and Formulas

• Restriction: Set denominator not equal to zero.

• Rational Equation: An equation involving rational expressions.

Step-by-Step Guidance

1. First, set $x - 7 \neq 0$ to find the restriction.

2. Combine like terms or move all terms to one side to simplify the equation.

3. Solve for x by isolating the variable.

Try solving on your own before revealing the answer!

Final Answer: Restriction: x ≠ 7; Solution: x = 8

Restriction: $x - 7 \neq 0 \implies x \neq 7$

Combine terms: $\frac{9}{x-7} + 1 = \frac{8}{x-7}$

Subtract $\frac{8}{x-7}$ from both sides: $\frac{1}{x-7} + 1 = 0$

Solve: $x = 8$

Q5. Determine the restriction, then solve the rational equation: $\frac{5}{x-8} + 5 = \frac{10}{x-8}$

Background

Topic: Rational Equations and Restrictions

This question is similar to the previous one, focusing on restrictions and solving rational equations.

Key Terms and Formulas

• Restriction: Denominator cannot be zero.

Step-by-Step Guidance

1. Set $x - 8 \neq 0$ to find the restriction.

2. Combine like terms or move all terms to one side.

3. Solve for x.

Try solving on your own before revealing the answer!

Final Answer: Restriction: x ≠ 8; Solution: x = 10

Restriction: $x - 8 \neq 0 \implies x \neq 8$

Combine terms: $\frac{5}{x-8} + 5 = \frac{10}{x-8}$

Solve: $x = 10$

Q6. Solve the equation, then determine if it is conditional, inconsistent, or an identity equation: $2x + 10 = 3(x + 3) - x + 1$

Background

Topic: Types of Equations (Conditional, Inconsistent, Identity)

This question tests your ability to solve linear equations and classify the type of solution.

Key Terms and Formulas

• Conditional Equation: True for some values of x.

• Inconsistent Equation: No solution.

• Identity Equation: True for all values of x.

Step-by-Step Guidance

1. Distribute and combine like terms on both sides.

2. Move all terms involving x to one side and constants to the other.

3. Solve for x and analyze the result to classify the equation.

Try solving on your own before revealing the answer!

Final Answer: x = 2 (Conditional Equation)

After simplifying, you get $x = 2$, so the equation is conditional.

Q7. Solve the equation, then determine if it is conditional, inconsistent, or an identity equation: $3(x + 6) = 3x - 16$

Background

Topic: Types of Equations

This question is similar to the previous one, focusing on solving and classifying the equation.

Key Terms and Formulas

• Distribute and combine like terms.

Step-by-Step Guidance

1. Distribute the 3 on the left side.

2. Move all x terms to one side and constants to the other.

3. Solve for x and classify the equation.

Try solving on your own before revealing the answer!

Final Answer: x = -6 (Conditional Equation)

After simplifying, you get $x = -6$, so the equation is conditional.

Q8. Solve the equation, then determine if it is conditional, inconsistent, or an identity equation: $3(x - 1) + 4 = -2(x + 3)$

Background

Topic: Types of Equations

This question tests your ability to solve and classify a linear equation.

Key Terms and Formulas

• Distribute and combine like terms.

Step-by-Step Guidance

1. Distribute on both sides.

2. Combine like terms and isolate x.

3. Solve for x and determine the type of equation.

Try solving on your own before revealing the answer!

Final Answer: x = -1 (Conditional Equation)

After simplifying, you get $x = -1$, so the equation is conditional.

Q9. Solve the linear equation: $14 = 3(x - 3) - (x - 5)$

Background

Topic: Solving Linear Equations

This question tests your ability to solve a linear equation with parentheses and distribution.

Key Terms and Formulas

• Distribute and combine like terms.

Step-by-Step Guidance

1. Distribute the 3 and the negative sign.

2. Combine like terms.

3. Isolate x and solve.

Try solving on your own before revealing the answer!

Final Answer: x = 6

After distributing and combining like terms, you solve for x and get 6.

Q10. Solve the linear equation using the zero product principle: $x^2 - 5x = -6$

Background

Topic: Quadratic Equations – Zero Product Principle

This question tests your ability to solve a quadratic equation by factoring.

Key Terms and Formulas

• Zero Product Principle: If $ab = 0$, then $a = 0$ or $b = 0$.

• Quadratic Equation: $ax^2 + bx + c = 0$

Step-by-Step Guidance

1. Move all terms to one side to set the equation to zero.

2. Factor the quadratic expression.

3. Set each factor equal to zero and solve for x.

Try solving on your own before revealing the answer!

Final Answer: x = 6 or x = -1

After factoring, you get two solutions: x = 6 and x = -1.