📐 Algebra Assessment

Mathematics Training Certification — Five 9s Communications

1
2
3
4
5
6
0 of 6 Sections Complete

Welcome to Algebra Assessment Training

Course Duration: 35–45 minutes

Algebra is the everyday math of low-voltage work. Solving Ohm's Law for an unknown current, estimating cable footage and cost, sizing a power budget, and reading a voltage-drop chart all rely on the same handful of skills: isolating a variable, evaluating a formula, working with exponents, and reading a straight-line relationship. This course reviews those skills in six short sections, each followed by a knowledge check, and ends with a 10-question final assessment.

What You'll Learn:

  • Solve one- and two-step linear equations using inverse operations
  • Solve quadratic equations by factoring and by taking square roots
  • Use function notation and evaluate a function for a given input
  • Apply the product, power, and quotient rules for exponents
  • Solve a system of two equations by substitution or elimination
  • Identify slope and y-intercept and calculate slope from two points

Course Structure

  1. Linear Equations
  2. Quadratic Equations
  3. Functions
  4. Exponents and Powers
  5. Systems of Equations
  6. Slope-Intercept Form
  7. Final Assessment — 10 questions, 80% to pass

Each knowledge check gives you two attempts. You will enter your name and email just before you submit the final assessment; your results are then emailed automatically to the training administrator, with a copy to you.

Section 1: Linear Equations

A linear equation is an equation in which the variable appears only to the first power — no x², no √x. Examples are 2x + 5 = 13 and V = I × R when you are solving for one of the letters. Every linear equation in one variable has exactly one solution (unless it simplifies to something always true or never true).

The balance principle

An equation is a balance: whatever you do to one side, you must do to the other. To solve, you undo the operations around the variable using inverse operations, working in the reverse order of operations:

  1. Simplify each side (distribute, combine like terms).
  2. Move variable terms to one side and constants to the other by adding or subtracting.
  3. Divide (or multiply) to leave the variable by itself.
  4. Check by substituting your answer back into the original equation.
Example: 4x − 3 = 17 Add 3 to both sides: 4x = 20 Divide both sides by 4: x = 5 Check: 4(5) − 3 = 20 − 3 = 17 ✓

When the variable appears on both sides, collect it on one side first:

Example: 5x + 2 = 2x + 14 Subtract 2x from both sides: 3x + 2 = 14 Subtract 2: 3x = 12 Divide by 3: x = 4
On the job: Ohm's Law, V = I × R, is a linear equation in each variable. If a 24 V circuit has 120 Ω of resistance, the current is found by dividing both sides by R: I = V ÷ R = 24 ÷ 120 = 0.2 A (200 mA).
Knowledge Check 1
Solve for x: 2x + 5 = 13
Attempts: 0/2

Section 2: Quadratic Equations

A quadratic equation has the variable squared as its highest power. Its standard form is ax² + bx + c = 0, where a ≠ 0. Unlike a linear equation, a quadratic can have two solutions, one repeated solution, or no real solution.

Method 1 — Factoring and the zero-product property

If two numbers multiply to zero, at least one of them must be zero. So if you can write the quadratic as a product of two factors, set each factor equal to zero. For x² + bx + c, look for two numbers that multiply to c and add to b.

Example: x² + 2x − 15 = 0 Two numbers that multiply to −15 and add to +2: 5 and −3 Factor: (x + 5)(x − 3) = 0 Solve: x + 5 = 0 → x = −5 x − 3 = 0 → x = 3

Method 2 — Taking the square root

When there is no x term, isolate x² and take the square root of both sides. Remember that both a positive and a negative number square to the same value, so the answer carries a ±.

Example: x² = 36 → x = ±6 (both 6² and (−6)² equal 36)

Method 3 — The quadratic formula

Any quadratic can be solved with x = (−b ± √(b² − 4ac)) ÷ 2a. Factoring is faster when it works; the formula always works.

On the job: Power in a resistor is P = I²R. Solving for current gives I² = P ÷ R, then I = √(P ÷ R). In a circuit only the positive root has physical meaning, but on a test the ± matters.
Knowledge Check 2
What are the solutions to x² − 5x + 6 = 0?
Attempts: 0/2

Section 3: Functions

A function is a rule that assigns exactly one output to each input. We write it with function notation: f(x) = 2x + 3 reads "f of x equals two x plus three." The letter in parentheses is the input; f(x) is the output. Note that f(x) does not mean f times x.

Evaluating a function

To evaluate, replace every x in the rule with the input value — in parentheses — and simplify using the order of operations.

Example: g(x) = 3x − 1. Find g(5). g(5) = 3(5) − 1 = 15 − 1 = 14

Watch the negatives

Always wrap a negative input in parentheses before squaring. (−3)² = 9, but −3² = −9 because the exponent is applied before the negative sign.

Example: h(x) = x² + 1. Find h(−3). h(−3) = (−3)² + 1 = 9 + 1 = 10

Domain and range

The domain is the set of allowed inputs; the range is the set of outputs the function can produce. A function cannot divide by zero, and (with real numbers) cannot take the square root of a negative.

On the job: A spreadsheet formula is a function. If cable cost is C(f) = 0.18f + 25 (18¢ per foot plus a $25 termination kit), a 300-ft run costs C(300) = 0.18(300) + 25 = $79.
Knowledge Check 3
If f(x) = 2x + 3, what is f(4)?
Attempts: 0/2

Section 4: Exponents and Powers

An exponent tells you how many times to multiply the base by itself: 2⁴ = 2 × 2 × 2 × 2 = 16. A small set of rules lets you simplify expressions with exponents without multiplying everything out.

RulePatternExample
Product ruleaᵐ × aⁿ = aᵐ⁺ⁿ3² × 3⁴ = 3⁶
Quotient ruleaᵐ ÷ aⁿ = aᵐ⁻ⁿ7⁵ ÷ 7² = 7³
Power rule(aᵐ)ⁿ = aᵐˣⁿ(x²)⁴ = x⁸
Zero exponenta⁰ = 1 (a ≠ 0)9⁰ = 1
Negative exponenta⁻ⁿ = 1 ÷ aⁿ10⁻³ = 0.001

The product and quotient rules only apply when the bases are the same. When terms have coefficients, multiply the coefficients and add the exponents of like bases separately:

Example: (4x³)(5x²) Coefficients: 4 × 5 = 20 Exponents: x³ × x² = x⁵ Result: 20x⁵
On the job: Engineering prefixes are powers of ten. 1 mA = 10⁻³ A and 1 kΩ = 10³ Ω, so 5 mA × 2 kΩ = 10 × 10⁻³⁺³ = 10 × 10⁰ = 10 V.
Knowledge Check 4
Simplify: (2³)²
Attempts: 0/2

Section 5: Systems of Equations

A system of equations is two (or more) equations that share the same unknowns. The solution is the pair of values that makes both equations true at the same time — graphically, the point where the two lines cross.

Method 1 — Substitution

Solve one equation for a variable, then substitute that expression into the other equation. This works best when one variable is already alone or has a coefficient of 1.

Example: y = x + 1 and 2x + y = 10 Substitute: 2x + (x + 1) = 10 → 3x + 1 = 10 → x = 3 Back-solve: y = 3 + 1 = 4 Solution: (3, 4)

Method 2 — Elimination

Add or subtract the equations so that one variable cancels. If no variable cancels directly, multiply one equation by a constant first.

Example: 3x + y = 11 and x − y = 1 Add the equations: 4x = 12 → x = 3 Back-solve: 3 − y = 1 → y = 2 Solution: (3, 2)

Always check the pair in both original equations.

On the job: A mixed order of 40 jacks costs $260. Cat 6 jacks are $5 and Cat 6A jacks are $8. With c + a = 40 and 5c + 8a = 260, elimination gives a = 20 and c = 20.
Knowledge Check 5
Solve the system: x + y = 10 and x − y = 2. What is x?
Attempts: 0/2

Section 6: Slope-Intercept Form

Every non-vertical straight line can be written in slope-intercept form: y = mx + b. The two numbers tell you everything about the line:

  • m — the slope: how much y changes for each 1-unit increase in x ("rise over run"). Positive slopes rise left to right; negative slopes fall.
  • b — the y-intercept: the value of y where the line crosses the y-axis, i.e. the value of y when x = 0.
Example: y = −2x + 9 Slope m = −2 (y drops 2 for every 1 step right) y-intercept b = 9 (the line crosses at (0, 9))

Slope from two points

Given points (x₁, y₁) and (x₂, y₂), the slope is the change in y divided by the change in x: m = (y₂ − y₁) ÷ (x₂ − x₁). Subtract in the same order top and bottom.

Example: points (1, 4) and (5, 12) m = (12 − 4) ÷ (5 − 1) = 8 ÷ 4 = 2

If the equation is not already in y = mx + b form, solve it for y first. For example, 2x + y = 7 becomes y = −2x + 7, so the slope is −2 and the intercept is 7.

On the job: The cable-cost function C(f) = 0.18f + 25 from Section 3 is a line. The slope (0.18) is the cost per foot; the intercept (25) is the fixed cost before any cable is pulled.
Knowledge Check 6
A line passes through the points (0, 2) and (3, 8). What is its slope?
Attempts: 0/2

Final Assessment

Answer all 10 questions. You need 80% to pass. Enter your name and email below the questions, then submit.

1. Solve: 3x − 7 = 14
2. Factor: x² + 7x + 12
3. If f(x) = x² − 4, what is f(−2)?
4. Simplify: 5² × 5³
5. Solve: 2x + 3y = 12 and x = 3. What is y?
6. What is the slope of y = 4x − 3?
7. Solve: x² = 49
8. Simplify: (3x²)(2x³)
9. If 2x − 5 = 15, what is x + 3?
10. What is the y-intercept of y = 2x + 7?