College Algebra Test Review Guidance

Q1. Simplify and write the answer using only positive exponents. Leave the answer in exponential notation: $\frac{m^8 n^7}{m^5 n^4}$

Background

Topic: Exponent Rules and Simplification

This question tests your understanding of how to simplify expressions with exponents, specifically using the quotient rule for exponents and ensuring all exponents are positive.

Key Terms and Formulas:

• Quotient Rule: $\frac{a^m}{a^n} = a^{m-n}$

• Positive Exponents: If you get a negative exponent, rewrite it as a reciprocal to make it positive.

Step-by-Step Guidance

1. Identify the bases in the numerator and denominator: $m$ and $n$.

2. Apply the quotient rule to each base: $\frac{m^8}{m^5} = m^{8-5}$ and $\frac{n^7}{n^4} = n^{7-4}$.

3. Simplify the exponents: $m^{8-5}$ and $n^{7-4}$.

4. Check if any exponents are negative. If so, rewrite them as positive exponents using reciprocals.

Try solving on your own before revealing the answer!

Q2. Solve the system using the elimination (addition) method: $x - 2y = 6$ $-4x - 2y = -26$

Background

Topic: Systems of Linear Equations (Elimination Method)

This question tests your ability to solve a system of two linear equations using the elimination (addition) method, which involves adding or subtracting equations to eliminate one variable.

Key Terms and Formulas:

• Elimination Method: Add or subtract equations to eliminate one variable.

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

Step-by-Step Guidance

1. Write both equations in standard form:

   • $x - 2y = 6$

   • $-4x - 2y = -26$

2. Decide which variable to eliminate. Here, $y$ has the same coefficient in both equations.

3. Subtract or add the equations to eliminate $y$. (Consider multiplying one equation if needed.)

4. Solve for the remaining variable.

Try solving on your own before revealing the answer!

Final Answer for Q1:

$m^3 n^3$

We subtracted the exponents for each base and ensured all exponents are positive.

Final Answer for Q2:

$x = 4$, $y = -1$

After eliminating $y$, we solved for $x$ and substituted back to find $y$.