OnlyMath / Original learning unit
Rebuild Algebra: From Equations to Functions
Find the step that is holding you back, repair it, and connect equations to graphs and real situations.
Goals and prerequisites
For learners reviewing introductory algebra or returning to math after a break.
This unit covers signed numbers, fractions, linear equations and inequalities, slope, and function notation. It is a foundation for further study, not a complete Algebra 1 course.
Bring these skills
- Multiply and divide whole numbers; recognize a numerator and denominator.
- Read an ordered pair (x, y). Chapter 3 reviews how coordinates connect to a line.
What you will be able to do
- Explain why an algebra step preserves equality, rather than relying on “moving a term.”
- Solve a linear equation or inequality and check the result in its original form.
- Build a linear model from two points, interpret its units, and evaluate function inputs.
Try the five diagnostic questions on paper before opening answers. Use the matching review links for mistakes. Study one chapter at a time: explain the example, attempt both practice questions, then return later for the three-question mastery check. No calculator is needed.
Start with the diagnostic
These are routing questions, not a grade. Count a question as secure only if you can explain the reasoning without looking at its answer.
Question 1
D1 · Signed fractions: calculate and give a simplified fraction.
Answer, reasoning and error check for question 1
Answer: 1/12
Use denominator 12: −9/12 + 10/12 = 1/12. The positive term is slightly larger, so a small positive answer is reasonable.
Common mistake: 2/10 = 1/5 comes from adding the numerators and denominators separately. Denominators name the size of the pieces; make the pieces equal before adding their counts.
Need a review? Go to the matching chapterQuestion 2
D2 · Equations: solve, then substitute your value back.
Answer, reasoning and error check for question 2
Answer: x = 11
Distribute to get 3x − 6 = 2x + 5. Subtract 2x and add 6 on both sides. At x = 11, both original sides equal 27.
Common mistake: x = 7 often comes from writing 3(x − 2) as 3x − 2. The 3 multiplies both terms inside the parentheses.
Need a review? Go to the matching chapterQuestion 3
D3 · Inequalities: describe every solution.
Answer, reasoning and error check for question 3
Answer: x < −3
Subtract 1 to get −2x > 6. Dividing by −2 reverses the ordering. For example, x = −4 works because 9 > 7.
Common mistake: x > −3 forgets that multiplying or dividing an inequality by a negative reverses order. Test x = 0: it would be included, but 1 > 7 is false.
Need a review? Go to the matching chapterQuestion 4
D4 · Rate of change: find the slope of the line through (1, 4) and (3, 10).
Answer, reasoning and error check for question 4
Answer: m = 3
The output rises by 6 while the input rises by 2, so the slope is 6/2 = 3.
Common mistake: 1/3 reverses rise and run. A slope of 3 means 3 output units per input unit, not 3 input units per output unit.
Need a review? Go to the matching chapterQuestion 5
D5 · Function notation: find f(−2).
Answer, reasoning and error check for question 5
Answer: f(−2) = 10
Replace every x with (−2): (−2)² − 3(−2) = 4 + 6 = 10.
Common mistake: 2 comes from treating (−2)² as −4. The parentheses make the whole negative number the base: (−2)(−2) = 4.
Need a review? Go to the matching chapterChoose your starting point
Fractions or signs are unreliable
D1 was wrong or depended on a guess.
Start with Chapter 1. Redo its fraction and distribution checks before solving equations.
Go to this chapterYou lose equality or reverse the wrong sign
D2 or D3 was wrong, but D1 was secure.
Start with Chapter 2. Write the operation applied to both sides on each line.
Go to this chapterSymbols work, but graphs do not
D4 was wrong or you could not explain the units.
Use Chapter 3 to connect a pair of points, a rate, and an equation.
Go to this chapterFunction inputs are confusing
D5 was wrong, or f(x) looks like multiplication.
Start with Chapter 4, then use the mixed mastery questions to check earlier skills.
Go to this chapterChapter 1
1. Keep the quantities and the signs intact
Goal: Add fractions with a common unit and distribute multiplication across every term.
An expression names a quantity; it does not ask for a solution. To simplify an expression, replace it with another expression that has the same value for every allowed input. By contrast, an equation asserts that two quantities are equal and may hold only for some inputs.
To add fractions, express them using equal-sized pieces. With denominators 3 and 4, twelfths work: multiply each numerator and denominator by the same nonzero number. Multiplying by 4/4 or 3/3 changes the name of a number, not its value.
A minus sign before parentheses is multiplication by −1. The distributive law a(b + c) = ab + ac therefore applies to subtraction too. Combine only like terms: 2x and 5x share a variable part; 2x and 5 do not.
Worked example
A fraction outside parentheses
Simplify the expression, then evaluate it at x = −3.
Distribute 2/3 to both terms. Since (2/3) × 6 = 4, the constant term is −4.
Use a common denominator for the x terms. Keep the constant separate.
Substitute x = −3 and combine the remaining numbers.
The simplified expression is 5x/12 − 4. Its value at x = −3 is −21/4. Checking the original gives (2/3)(−9) + 3/4 = −6 + 3/4 = −21/4 too.
Avoid this trap: Cancellation removes common factors of an entire numerator and denominator, not individual terms in a sum. For example, (x + 4)/4 = x/4 + 1; it is not x + 1.
Practice on your own
Question 1
Calculate without a decimal approximation.
Answer, reasoning and error check for question 1
Answer: −7/12
−5/6 = −10/12 and 1/4 = 3/12; their sum is −7/12.
Common mistake: −4/10 adds numerators and denominators separately. The denominators must first agree.
Need a review? Go to the matching chapterQuestion 2
Simplify, then check your expression at x = 2.
Answer, reasoning and error check for question 2
Answer: −5x + 15; at x = 2 the value is 5.
Distribute to get −6x + 15 + x. Combining x terms gives −5x + 15. The original at x = 2 gives −3(−1) + 2 = 5.
Common mistake: −5x − 15 misses that (−3)(−5) is positive. Track the signs of each product separately.
Need a review? Go to the matching chapterBefore moving on: Continue when you can explain the common denominator and distribute a negative factor without looking at the example. A numerical check can catch an error, but checking one input alone does not prove two expressions identical.
Chapter 2
2. Preserve equality; reverse order only when needed
Goal: Solve with reversible operations and distinguish one solution, no solutions, and infinitely many solutions.
Think of an equation as a balance. Adding the same quantity to both sides or multiplying both sides by the same nonzero constant preserves the solution set. “Move and change the sign” is shorthand for an operation on both sides, not a separate rule.
Keep every term when clearing fractions. Multiplying an equation by 6 means multiplying each entire side by 6. Do not divide by an expression that might be zero without treating that case separately.
An inequality describes a range, not usually a single value. Adding the same number keeps its direction. Multiplying by a negative reflects the number line and reverses order: 2 < 5, but −2 > −5. Multiplication by a positive does not reverse it.
Worked example
An equation with fractions on both sides
Solve and verify the result in the original equation.
Multiply both entire sides by 6, the least common multiple of 3 and 2.
Distribute and combine the constants on the left.
Subtract 2x from both sides, then subtract 12 from both sides.
Check both sides of the original, including its denominators.
The unique solution is x = −10. Every transformation used a reversible operation, and substitution confirms it satisfies the starting equation.
Avoid this trap: If variables cancel, read what remains. 2(x + 1) = 2x + 2 becomes 2 = 2, so every real x works. 2(x + 1) = 2x + 5 becomes 2 = 5, so no x works. Neither case means x = 0.
Practice on your own
Question 1
Solve and check.
Answer, reasoning and error check for question 1
Answer: x = 7
4x − 4 = 2x + 10, so 2x = 14. With x = 7, both sides are 24.
Common mistake: x = 3 usually comes from moving −4 without adding 4 to both sides. Write 4x = 2x + 14 first.
Need a review? Go to the matching chapterQuestion 2
Solve and describe its number-line graph.
Answer, reasoning and error check for question 2
Answer: x ≥ −3; a filled point at −3, shaded to the right.
Subtract 5: −3x ≤ 9. Divide by −3 and reverse: x ≥ −3. The endpoint works because 5 − 3(−3) = 14.
Include −3 and every larger number
Common mistake: x ≤ −3 reverses no sign; an open point excludes equality. Both the direction and the included endpoint matter.
Need a review? Go to the matching chapterBefore moving on: Explain which operation produced each line. For an inequality, check the boundary and one point on each side. Review this chapter if either the direction or the equality case is uncertain.
Chapter 3
3. Connect a line to a rate and a starting value
Goal: Use two points to build y = mx + b and explain m and b with units.
A straight-line model has a constant rate of change. Its slope is the change in output divided by the change in input: m = (y₂ − y₁)/(x₂ − x₁), provided the input difference is not zero. Subtract coordinates in the same order in numerator and denominator.
In y = mx + b, b is the output when x = 0. It need not be the first y-value in a data list. A vertical line has no finite slope and cannot be written as y = mx + b; it would give one x multiple y-values.
An algebraic line extends indefinitely; a useful model often does not. State the allowed input range and whether inputs are continuous or whole-number counts. Two measured points alone do not prove a relationship stays linear outside the situation described.
Worked example
A tank draining at a constant rate
A tank contains 26 liters after 2 minutes and 14 liters after 5 minutes. Assume it drains at a constant rate from t = 0 until empty. Find a model and its useful time interval.
Use the points (2, 26) and (5, 14). The units are liters per minute.
Substitute either point into V(t) = −4t + b to recover the initial amount.
Find when the amount reaches zero and restrict the model to the draining period.
The initial volume is 34 liters and the tank loses 4 liters each minute. Plot (0, 34) and (8.5, 0), then join them with a segment. The formula predicts a negative volume after 8.5 minutes, so it is no longer a physical model there.
A falling line, a constant draining rate
Avoid this trap: A negative slope does not mean all outputs are negative. It means outputs decrease as inputs increase. In the tank model, the slope is negative while the volume is positive before the tank is empty.
Practice on your own
Question 1
Find the line through (2, 7) and (6, 15). State its slope and y-intercept.
Answer, reasoning and error check for question 1
Answer: y = 2x + 3; slope 2 and y-intercept 3.
m = (15 − 7)/(6 − 2) = 2. Use 7 = 2(2) + b, so b = 3. Both given points satisfy the result.
Common mistake: y = 2x + 7 uses a given output as the intercept, even though its input is 2 rather than 0.
Need a review? Go to the matching chapterQuestion 2
A bike rental costs $9 plus $4 per hour, with fractional hours charged proportionally. Write C(h), interpret its two constants, and find the cost for 2.5 hours.
Answer, reasoning and error check for question 2
Answer: C(h) = 9 + 4h for h ≥ 0; C(2.5) = $19.
The slope is $4 per hour and the intercept is the $9 starting charge. Substitute h = 2.5: 9 + 4 × 2.5 = 19.
Common mistake: C(h) = 13h charges the starting fee every hour. The fixed fee is added once, so it is not part of the hourly slope.
Need a review? Go to the matching chapterBefore moving on: Given two points, calculate a slope, find the intercept, sketch a line, and explain one real-world restriction. If the algebra is fine but the units are not, repeat the modeling problem in words.
Chapter 4
4. Treat a function as a rule with allowed inputs
Goal: Evaluate, compose, and restrict functions without confusing an input with an output.
A function assigns exactly one output to each allowed input. f(3) means the output of f at input 3; it does not mean f multiplied by 3. Different inputs may share an output: f(x) = x² gives f(−2) = f(2) = 4 and is still a function.
When evaluating, replace every appearance of the input variable and keep parentheses around a negative number or expression. For composition f(g(x)), first find the output g(x), then use that entire output as the input of f.
The domain records allowed inputs. Over the real numbers, division by zero is forbidden and an even root cannot have a negative radicand. A contextual restriction can be stricter than an algebraic one, as with nonnegative rental hours.
Worked example
Composition and a domain restriction
Let f(x) = 2x − 1 and g(x) = x² + 3. Find f(g(−2)). Then state the real domain of r(x) = 1/(x − 5).
Evaluate the inner function at the original input.
Feed 7, not −2, into the outer function.
For r, identify the input that makes the denominator zero and exclude it.
The composition gives 13. The domain of r is every real number except 5. A domain states possible inputs; it does not give the set of outputs, called the range.
Avoid this trap: Order matters: in this example, g(f(−2)) = g(−5) = 28, not 13. Composition is not multiplication and is not generally commutative.
Practice on your own
Question 1
For p(x) = x² − 2x, find p(−3) and simplify p(a + 1).
Answer, reasoning and error check for question 1
Answer: p(−3) = 15; p(a + 1) = a² − 1.
p(−3) = 9 + 6 = 15. For the expression input, (a + 1)² − 2(a + 1) = a² + 2a + 1 − 2a − 2 = a² − 1.
Common mistake: p(a) + 1 is not p(a + 1). Change the input before performing the function operations; do not add 1 to the finished output.
Need a review? Go to the matching chapterQuestion 2
State the real domain of q. Is x = 2 allowed?
Answer, reasoning and error check for question 2
Answer: x ≥ 2 and x ≠ 5. Yes, q(2) = 0.
The square root requires x − 2 ≥ 0. The denominator requires x − 5 ≠ 0. Satisfy both conditions simultaneously.
Common mistake: x > 2 wrongly excludes a zero inside a square root. x ≥ 2 alone wrongly includes division by zero at x = 5.
Need a review? Go to the matching chapterBefore moving on: Explain the difference between f(a + 1) and f(a) + 1, and name every restriction before simplifying a rational expression. You are ready for the mixed check when both explanations are clear.
Mixed mastery check
Close the examples and solve these on paper. Write an explanation, not just an answer. The questions mix skills so that you must choose the method yourself.
Question 1
Solve the equation and explain why each transformation is valid.
Answer, reasoning and error check for question 1
Answer: x = −7
Multiply both sides by 3: 2x − 1 = 3x + 6. Subtract 2x and then 6: x = −7. In the original, both sides equal −5.
Common mistake: 2x − 1 = x + 6 multiplies only part of the right side. Every term on that side must be multiplied by 3.
Need a review? Go to the matching chapterQuestion 2
A linear cost C(h) passes through (1, 11) and (4, 23), with h ≥ 0. Find C(h), then find every h that keeps the cost at most $31. Fractional hours are allowed.
Answer, reasoning and error check for question 2
Answer: C(h) = 4h + 7; 0 ≤ h ≤ 6.
The slope is (23 − 11)/(4 − 1) = 4 and 11 = 4 + b gives b = 7. Solve 4h + 7 ≤ 31 to obtain h ≤ 6, then include h ≥ 0.
Common mistake: h = 6 gives only the boundary. “At most” also includes all allowed smaller times; the contextual lower bound must remain.
Need a review? Go to the matching chapterQuestion 3
For r(x) = (x² − 9)/(x − 3), use the identity x² − 9 = (x − 3)(x + 3) to state the original domain, simplify where allowed, and find r(4). Is r(3) defined?
Answer, reasoning and error check for question 3
Answer: x ≠ 3; r(x) = x + 3 on that domain; r(4) = 7; r(3) is undefined.
Use x² − 9 = (x − 3)(x + 3). Cancel the common factor only when x − 3 is nonzero. Simplifying a formula does not restore an excluded input.
Common mistake: r(3) = 6 uses the simplified expression outside the original domain. In the original rule, x = 3 gives division by zero.
Need a review? Go to the matching chapterUse this as a study checkpoint, not a placement test: aim to solve all three without hints and explain the checks. If one fails, use its review link, correct the reasoning, and try a fresh problem with different numbers later. A correct answer from a memorized pattern is not enough to move on.
Choose your next step
If the foundations are secure
Continue with systems of equations, exponent rules, polynomial operations, and quadratics. Keep translating among words, equations, and graphs before starting a full precalculus course.
Compare algebra resourcesIf calculus is your next goal
First add trigonometry, exponentials, logarithms, and function composition to these skills. Then use the calculus unit’s readiness questions to choose a starting chapter.
Open the calculus learning unitEditorial note and corrections
The explanations and exercises in this unit are original OnlyMath learning material. Linked textbook sections provide optional further reading; their inclusion does not imply endorsement. English and Simplified Chinese cover the same questions and learning goals. Link checks were performed on October 8, 2026; external pages can change.
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