Algebra Mastery

Master the art of solving and graphing equations to conquer the ACT Math section!

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Algebra Fundamentals

Linear Equations

Linear equations are the foundation of algebra and appear frequently on the ACT. They take the form:

\[ y = mx + b \]

Where:

  • \(m\) is the slope (rate of change)
  • \(b\) is the y-intercept (where the line crosses the y-axis)

Example:

Find the slope and y-intercept of the line \(2x + 3y = 6\)

Step 1: Rearrange to slope-intercept form (y = mx + b)

\[ 2x + 3y = 6 \] \[ 3y = 6 - 2x \] \[ y = \frac{6 - 2x}{3} \] \[ y = \frac{6}{3} - \frac{2x}{3} \] \[ y = 2 - \frac{2}{3}x \] \[ y = -\frac{2}{3}x + 2 \]

Step 2: Identify the slope and y-intercept

Slope (m) = \(-\frac{2}{3}\)

y-intercept (b) = 2

Key ACT Concept: Parallel and Perpendicular Lines

  • Parallel lines have the same slope
  • Perpendicular lines have slopes that are negative reciprocals of each other (product = -1)

Quadratic Equations

Quadratic equations are second-degree equations that take the form:

\[ ax^2 + bx + c = 0 \]

The solutions to a quadratic equation can be found using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Example:

Solve the quadratic equation \(2x^2 - 5x - 3 = 0\)

Step 1: Identify the values of a, b, and c

a = 2, b = -5, c = -3

Step 2: Substitute into the quadratic formula

\[ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(-3)}}{2(2)} \] \[ x = \frac{5 \pm \sqrt{25 + 24}}{4} \] \[ x = \frac{5 \pm \sqrt{49}}{4} \] \[ x = \frac{5 \pm 7}{4} \]

Step 3: Calculate the two solutions

\[ x = \frac{5 + 7}{4} = \frac{12}{4} = 3 \] \[ x = \frac{5 - 7}{4} = \frac{-2}{4} = -\frac{1}{2} \]

The solutions are x = 3 and x = -\(\frac{1}{2}\)

Key ACT Concept: Discriminant

The discriminant (\(b^2 - 4ac\)) tells you about the nature of the solutions:

  • If \(b^2 - 4ac > 0\), there are two real solutions
  • If \(b^2 - 4ac = 0\), there is one real solution (repeated root)
  • If \(b^2 - 4ac < 0\), there are no real solutions (complex solutions)

Systems of Equations

Systems of equations involve solving multiple equations simultaneously. On the ACT, you'll often see systems of two linear equations:

\[ \begin{cases} a_1x + b_1y = c_1 \\ a_2x + b_2y = c_2 \end{cases} \]

Example:

Solve the system of equations:

\[ \begin{cases} 2x + y = 7 \\ 3x - 2y = 4 \end{cases} \]

Step 1: Solve for y in the first equation

\[ y = 7 - 2x \]

Step 2: Substitute this expression for y into the second equation

\[ 3x - 2(7 - 2x) = 4 \] \[ 3x - 14 + 4x = 4 \] \[ 7x - 14 = 4 \] \[ 7x = 18 \] \[ x = \frac{18}{7} = 2\frac{4}{7} \]

Step 3: Find y by substituting the value of x back into the first equation

\[ y = 7 - 2(2\frac{4}{7}) \] \[ y = 7 - 2(\frac{18}{7}) \] \[ y = 7 - \frac{36}{7} \] \[ y = \frac{49 - 36}{7} = \frac{13}{7} = 1\frac{6}{7} \]

The solution is \(x = 2\frac{4}{7}\) and \(y = 1\frac{6}{7}\)

Key ACT Concept: Types of Solutions for Systems

  • One unique solution: Lines intersect at exactly one point
  • No solution: Lines are parallel (inconsistent system)
  • Infinitely many solutions: Lines are the same (dependent system)

Inequalities

Inequalities involve relationships using <, >, ≤, or ≥ instead of equals signs. Key rules to remember:

  • When multiplying or dividing both sides by a negative number, reverse the inequality sign
  • You can add or subtract the same value from both sides without changing the inequality

Example:

Solve the inequality \(3x - 4 < 5x + 8\)

Step 1: Rearrange to isolate x terms on one side

\[ 3x - 4 < 5x + 8 \] \[ 3x - 5x < 8 + 4 \] \[ -2x < 12 \]

Step 2: Divide both sides by -2 (remember to flip the inequality sign)

\[ \frac{-2x}{-2} > \frac{12}{-2} \] \[ x > -6 \]

The solution is x > -6

Key ACT Concept: Compound Inequalities

Compound inequalities combine multiple conditions:

  • "And" inequalities: Both conditions must be true (intersection)
  • "Or" inequalities: At least one condition must be true (union)

Absolute Value Equations and Inequalities

Absolute value represents the distance from zero on a number line. For any real number x:

\[ |x| = \begin{cases} x, & \text{if } x \geq 0 \\ -x, & \text{if } x < 0 \end{cases} \]

Example:

Solve the absolute value equation \(|2x - 3| = 7\)

Step 1: Set up two equations

\[ 2x - 3 = 7 \text{ or } 2x - 3 = -7 \]

Step 2: Solve each equation

\[ 2x - 3 = 7 \] \[ 2x = 10 \] \[ x = 5 \]
\[ 2x - 3 = -7 \] \[ 2x = -4 \] \[ x = -2 \]

The solutions are x = 5 and x = -2

Key ACT Concept: Absolute Value Inequalities

  • |x| < a means -a < x < a (values between -a and a)
  • |x| > a means x < -a or x > a (values less than -a or greater than a)

Ready to Practice?

Now that you've learned the fundamentals of algebra, test your knowledge with practice problems!

Practice Problems

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Algebra Challenge

Put your algebra skills to the test with these timed challenges. Complete them to earn badges and level up your character!

Linear Equations Challenge

100 XP

Solve 5 linear equation problems in 5 minutes.

5 minutes
5 problems

Quadratic Equations Challenge

150 XP

Solve 5 quadratic equation problems in 7 minutes.

7 minutes
5 problems

Systems of Equations Challenge

200 XP

Solve 5 systems of equations problems in 10 minutes.

10 minutes
5 problems

Algebra ACT Test Simulation

Take a timed test that simulates the algebra questions you'll see on the ACT. This will help you identify your strengths and areas for improvement.

Test Details

  • Time Limit: 15 minutes
  • Questions: 10 algebra questions in ACT format
  • Topics Covered: Linear equations, quadratic equations, systems of equations, inequalities, and absolute value
  • Scoring: Instant results with detailed explanations

Ready to Test Your Skills?

This simulation will help you prepare for the actual ACT math section.