1. Simplifying expressions: Examples:
  • Simplify 3x + 2x – 5y + y
  • Simplify 4(x + 3) – 5(x – 2)

Exercises: a) Simplify 5x – 2y + 3x + y b) Simplify 2(3x – 4) – 3(x – 2) c) Simplify 2a + 3b – 4a + 5c – b

Answers: a) 8x – y b) 3x – 2 c) -2a + 2b + 5c

Explanation: Simplifying expressions involves combining like terms. Like terms are those that have the same variables raised to the same powers. In the examples above, we combine like terms to simplify the expressions.

  1. Solving linear equations: Examples:
  • Solve for x: 2x + 3 = 7x – 1
  • Solve for y: 5(y + 2) = 3(y – 4)

Exercises: a) Solve for x: 4x – 5 = 3x + 10 b) Solve for y: 2(y – 3) = y + 6 c) Solve for z: 6z + 2 = 5z – 3

Answers: a) x = 15 b) y = 12 c) z = -5

Explanation: To solve a linear equation, isolate the variable on one side of the equation. Use inverse operations (opposite operations) to do so. In the examples above, we used inverse operations to isolate the variable and solve for it.

  1. Graphing linear equations: Examples: (need to find a better way to demonstrate this topic on my blog…)
  • Graph the equation y = 2x + 3
  • Graph the equation 3y – 6x = 9

Exercises: a) Graph the equation y = -3x + 2 b) Graph the equation 2y + 4x = 8 c) Graph the equation y – 5 = -2(x – 3)

  1. Factoring quadratic expressions: Examples:
  • Factor x^2 + 5x + 6
  • Factor 2x^2 – 7x – 15

Exercises: a) Factor x^2 – 4x – 12 b) Factor 3x^2 + 10x + 7 c) Factor 2x^2 + 3x

Answers: a) (x – 6)(x + 2) b) (3x + 7)(x + 1) c) x(2x + 3)

Explanation: To factor a quadratic expression, look for two numbers that multiply to the constant term and add to the coefficient of the linear term. Then use those two numbers to write the quadratic expression as a product of two linear expressions. In the examples above, we factored quadratic expressions by finding the two numbers that multiply to the constant term and add to the coefficient of the linear term.

  1. Solving systems of linear equations: Examples:
  • Solve the system of equations: 2x – 3y = 7 4x + 5y = 1
  • Solve the system of equations: 3x + y = 5 2x – y = 3

Exercises: a) Solve the system of equations: 3x – y = 4 x + 2y = 7 b) Solve the system of equations: 2x + y = 5 3x – 2y = 8

Answers: a) x = 2, y = 2 b) x = 2, y = -1

Explanation: To solve a system of linear equations, use one of the following methods:

  • Elimination method: Multiply one or both equations by a constant to get opposite coefficients for one of the variables, then add or subtract the equations to eliminate that variable.
  • Substitution method: Solve one of the equations for one variable in terms of the other variable, then substitute that expression into the other equation and solve for the remaining variable. In the examples above, we solved systems of linear equations using one of these methods.
  1. Working with inequalities: Examples:
  • Solve the inequality 2x + 5 < 11
  • Solve the inequality -3x > 12

Exercises: a) Solve the inequality 4x – 7 > 5x – 3 b) Solve the inequality 2(x – 3) ≤ 4x + 1

Answers: a) x < 4 b) x ≥ 2

Explanation: To solve an inequality, treat it like an equation and use inverse operations to isolate the variable. However, if you multiply or divide both sides of an inequality by a negative number, you need to reverse the direction of the inequality. In the examples above, we solved inequalities by isolating the variable and applying the rules for reversing the direction of the inequality when necessary.

  1. Working with functions: Examples:
  • Evaluate the function f(x) = 3x – 2 for x = 4
  • Find the domain of the function f(x) = 1/(x – 3)

Exercises: a) Evaluate the function g(x) = 2x^2 + 5x – 1 for x = -2 b) Find the domain of the function h(x) = √(4 – x)

Answers: a) g(-2) = 5 b) Domain: x ≤ 4 and x ≥ -4

Explanation: A function is a rule that assigns each input (x-value) to exactly one output (y-value). To evaluate a function, substitute the given input value into the function and simplify. The domain of a function is the set of all possible input values (x-values) that the function can take. In the examples above, we evaluated functions by substituting values into the function and found the domain of a function by considering the values that make the function undefined.

I hope this study guide helps you prepare for the algebra skills that may be on the ACT!