Intermediate Algebra Practice Questions & Step-by-Step Solutions
Sample Set 1

Problem: Write \( 230{,}000{,}000{,}000 \) in scientific notation.

Solution:

Write the number in the form \( a \times 10^n \), where \( 1 \leq |a| < 10 \) and \( n \in \mathbb{Z} \):
\[ 230{,}000{,}000{,}000 = 2.3 \times 10^{11} \]

Problem: Evaluate: \[ 30 - \dfrac{12}{3} \times 2 \]

Solution:

First, perform division and multiplication from left to right: \[ \dfrac{12}{3} \times 2 = 4 \times 2 = 8 \] Then subtract: \[ 30 - 8 = 22 \]

Problem: Evaluate: \[ |4 - 8(3 - 12)| - |5 - 11| \]

Solution:

Start with the inner parentheses: \[ 3 - 12 = -9 \] Then multiply: \[ 8 \times (-9) = -72 \] Evaluate absolute values: \[ |4 - (-72)| = |76| = 76 \] \[ |5 - 11| = |-6| = 6 \] Subtract the results: \[ 76 - 6 = 70 \]

Problem: Evaluate: \[ -18 + 4\left(\dfrac{6}{2}\right)^2 \]

Solution:

Evaluate inside parentheses: \[ \dfrac{6}{2} = 3 \] Square the result: \[ 3^2 = 9 \] Multiply by 4: \[ 4 \times 9 = 36 \] Add to -18: \[ -18 + 36 = 18 \]

Problem: Evaluate: \[ 11 + \sqrt{-4 + \dfrac{6 \times 4}{3}} \]

Solution:

Simplify the numerator under the radical: \[ 6 \times 4 = 24 \implies \dfrac{24}{3} = 8 \] Combine with -4: \[ -4 + 8 = 4 \implies \sqrt{4} = 2 \] Add to 11: \[ 11 + 2 = 13 \]

Problem: Simplify: \[ 12x^3 - 3(2x^3 + 4x - 1) - 5x + 7 \]

Solution:

Distribute the -3: \[ 12x^3 - 6x^3 - 12x + 3 - 5x + 7 \] Combine like terms: \[ (12x^3 - 6x^3) + (-12x - 5x) + (3 + 7) = 6x^3 - 17x + 10 \]

Problem: Simplify: \[ \left(\dfrac{x^4}{x^3}\right)^3 \]

Solution:

Simplify inside the parentheses using the quotient rule: \[ \dfrac{x^4}{x^3} = x^{4-3} = x \] Raise to the power of 3: \[ x^3 \]

Problem: Simplify: \[ \dfrac{(3x^2y^{-2})^3}{(9xy^3)^3} \]

Solution:

Expand powers in numerator and denominator: \[ (3x^2y^{-2})^3 = 27x^6y^{-6} \] \[ (9xy^3)^3 = 729x^3y^9 \] Divide and simplify: \[ \dfrac{27x^6y^{-6}}{729x^3y^9} = \dfrac{1}{27}x^{6-3}y^{-6-9} = \dfrac{x^3}{27y^{15}} \]

Problem: Simplify: \[ \dfrac{(2x^{-3}y^4)^3(x^3 + y)^0}{(4xy^{-2})^3} \]

Solution:

Note that any non-zero expression to the power of 0 equals 1: \( (x^3 + y)^0 = 1 \).
Expand numerator: \[ (2x^{-3}y^4)^3 = 8x^{-9}y^{12} \] Expand denominator: \[ (4xy^{-2})^3 = 64x^3y^{-6} \] Combine and simplify: \[ \dfrac{8x^{-9}y^{12}}{64x^3y^{-6}} = \dfrac{1}{8}x^{-9-3}y^{12 - (-6)} = \dfrac{1}{8}x^{-12}y^{18} = \dfrac{y^{18}}{8x^{12}} \]

Problem: Write as a mathematical inequality: "9 is less than the product of M and N".

Solution:

\[ 9 < M \cdot N \]

Problem: Find the slope of the line perpendicular to the line with equation: \[ y = \dfrac{1}{3}x - 7 \]

Solution:

The slope of the given line is \( m = \dfrac{1}{3} \).
The perpendicular slope is the negative reciprocal: \[ m_{\perp} = -\dfrac{1}{\dfrac{1}{3}} = -3 \]

Problem: Write an equation of the line with slope \( -3 \) and y-intercept at \( (0, -5) \).

Solution:

Using slope-intercept form \( y = mx + b \): \[ y = -3x - 5 \]

Problem: Solve the equation: \[ -5x + 20 = 25 \]

Solution:

Subtract 20 from both sides: \[ -5x = 5 \] Divide by -5: \[ x = -1 \]

Problem: Solve the inequality: \[ -3x + 4 < -8 \]

Solution:

Subtract 4 from both sides: \[ -3x < -12 \] Divide by -3 and reverse the inequality sign: \[ x > 4 \]

Problem: Solve the equation: \[ 2x^2 - 32 = 0 \]

Solution:

Add 32 to both sides: \[ 2x^2 = 32 \] Divide by 2: \[ x^2 = 16 \] Take the square root: \[ x = \pm 4 \quad (\text{i.e., } x = -4 \text{ or } x = 4) \]

Problem: Solve the equation: \[ -0.25x + 1.3 = -0.55x - 0.2 \]

Solution:

Add \( 0.55x \) to both sides: \[ 0.3x + 1.3 = -0.2 \] Subtract 1.3 from both sides: \[ 0.3x = -1.5 \] Divide by 0.3: \[ x = -5 \]

Problem: Solve the equation: \[ -0.25x^2 + 1.5 = -10.75 \]

Solution:

Subtract 1.5 from both sides: \[ -0.25x^2 = -12.25 \] Divide by -0.25: \[ x^2 = 49 \] Take the square root: \[ x = \pm 7 \quad (\text{i.e., } x = -7 \text{ or } x = 7) \]

Problem: What is the slope of a line perpendicular to the line with equation \( x = -3 \)?

Solution:

The equation \( x = -3 \) represents a vertical line. A line perpendicular to a vertical line is a horizontal line.
The slope of a horizontal line is 0.

Problem: What is the slope of a line parallel to the line with equation \( x = 5 \)?

Solution:

The equation \( x = 5 \) represents a vertical line. Any line parallel to a vertical line is also vertical.
The slope of a vertical line is undefined.

Problem: What is the slope of a line perpendicular to the line with equation \( y = 6 \)?

Solution:

The equation \( y = 6 \) represents a horizontal line. A line perpendicular to a horizontal line is a vertical line.
The slope of a vertical line is undefined.