Algebra curriculum

Put the value in parentheses wherever the variable appears, then follow the order of operations. For 3x² − 2x with x = −4: 3(−4)² − 2(−4) = 3(16) + 8 = 56. Without parentheses, −4² becomes −16 and the answer becomes −40. Parentheses keep a negative sign attached to the number it belongs to.

Evaluate parentheses first, then exponents, then multiplication and division from left to right, then addition and subtraction from left to right. So 2 + 3 × 4² = 2 + 3 × 16 = 50, not (2 + 3) × 16 = 80 or 2 + 12² = 146. Multiplication and division share a level and run left to right: 24 ÷ 4 × 2 = 12.

Like terms share the same variable raised to the same power, and only they combine: 5x + 3x − 2 is 8x − 2, and 4x² + x − x² is 3x² + x. Terms such as x and x², or x and y, stay separate. Adding the coefficients of unlike terms, as in 2x + 3x² = 5x³, is the common error.

A factor outside parentheses multiplies every term inside: 4(2x + 3) = 8x + 12. A negative factor flips each sign: −3(x − 4) = −3x + 12. The usual errors are multiplying only the first term, as in 8x + 3, and losing the sign of the second term, as in −3x − 12.

Factoring reverses distribution: find the largest factor every term shares and divide it out. In 6x² + 9x the terms share 3x, so 6x² + 9x = 3x(2x + 3). Check by distributing back. Taking out only part of the common factor, as in 3(2x² + 3x), is correct but not fully factored.

Solving an equation means undoing what was done to the variable, in reverse order, and doing the same thing to both sides. In 2x + 13 = 45, the x was doubled and then 13 was added, so subtract 13 (2x = 32) and then divide by 2 (x = 16). Dividing only the x term by 2, and not the 13, gives 9.5.

A one-step equation needs one inverse operation. In x − 23 = 41, add 23 to both sides: x = 64. In 13x = 156, divide both sides by 13: x = 12. Applying the operation to one side only, or repeating the operation instead of inverting it, is the usual slip. Check by substituting the answer back.

With two operations on the variable, clear the added or subtracted number first, then the coefficient. For 4x − 9 = 27, add 9 to get 4x = 36, then divide by 4 to get x = 9. Getting the variable term alone first means the last step is always a single division or multiplication.

Tidy each side first. In 6x − 2x + 5 = 33, combine 6x − 2x into 4x, giving 4x + 5 = 33, so 4x = 28 and x = 7. Combining the coefficients wrongly, or dividing 33 by one coefficient before combining, gives an answer that fails when you substitute it back.

When x appears on both sides, subtract the smaller x term from both sides. In 11x − 3 = 4x + 25, subtract 4x to get 7x − 3 = 25, add 3 to get 7x = 28, and divide to get x = 4. Moving a term across the equals sign without changing its sign is the classic error.

For 5(x − 3) = 40, either distribute (5x − 15 = 40, so 5x = 55 and x = 11) or divide both sides by 5 first (x − 3 = 8, so x = 11). Dividing first is faster when the right side divides evenly. Distributing to only the first term, writing 5x − 3, is the common error.

Fractions are easier to remove than to work with. Multiply every term by the least common denominator: in x/3 + x/4 = 14, multiply by 12 to get 4x + 3x = 168, so 7x = 168 and x = 24. Forgetting to multiply the right side, or a term with no fraction, unbalances the equation.

When two ratios are equal, their cross products are equal. For 6/15 = x/40, cross-multiply to get 15x = 240, so x = 16. Simplifying first also works: 6/15 is 2/5, and 2/5 of 40 is 16. Setting up the ratios with matching quantities in mismatched positions is the error to watch for.

Substitute your answer into the original equation, not into a later step. If x = 4 solves 11x − 3 = 4x + 25, both sides give 41. Checking a later line only confirms the arithmetic after any mistake was made. A failed check points to a dropped sign or a term not carried to both sides.

Inequalities are solved like equations: add, subtract, multiply, or divide both sides by the same number. For 4x − 5 ≤ 23, add 5 (4x ≤ 28) and divide by 4 (x ≤ 7). The answer is a range of values, so 7 and every smaller number work. Writing x = 7 alone misses that the solution is a whole set.

Multiplying or dividing both sides by a negative number reverses the inequality. From −2x > 10, dividing by −2 gives x < −5, not x > −5. Check with a value: x = −6 gives 12, which is greater than 10, so it is a solution, as x < −5 says. Adding or subtracting never flips the sign.

To confirm an answer such as x < −5, test one value inside the range and one outside in the original inequality. With −2x > 10, x = −6 gives 12 > 10, which is true, and x = 0 gives 0 > 10, which is false. If the tests come out the other way round, the sign was flipped when it should not have been, or not flipped when it should.

When one variable has opposite coefficients in two equations, adding the equations cancels it. From 3x + 2y = 16 and 5x − 2y = 8, adding gives 8x = 24, so x = 3; substituting back gives 2y = 7 and y = 3.5. If no coefficients match, multiply one equation first so that they do.

If you know the sum and the difference of two numbers, add the equations. From x + y = 30 and x − y = 8, adding gives 2x = 38, so x = 19 and y = 11. In general the larger number is half of the sum plus the difference, and the smaller is half of the sum minus the difference.

When one equation already gives a variable on its own, substitute it into the other. With y = 2x + 1 and 3x + y = 21, replace y to get 3x + 2x + 1 = 21, so 5x = 20, x = 4, and y = 9. Put the substituted expression in parentheses when it is being multiplied, so every term is included.

f(4) means substitute 4 for x in the rule. For f(x) = 2x² − 3, f(4) = 2(16) − 3 = 29. The notation f(4) is not f times 4; it is the output when the input is 4. Use parentheses for a negative input: f(−1) = 2(1) − 3 = −1.

In f(g(x)), find g(x) first and feed the result into f. With f(x) = 3x − 1 and g(x) = x + 5, g(2) = 7 and f(7) = 20, so f(g(2)) = 20. Order matters: g(f(2)) = g(5) = 10. Working from the outside in, or multiplying the two functions, gives a wrong answer.

f(f(x)) applies the same rule twice. With f(x) = 2x − 3, f(5) = 7, and then f(7) = 11, so f(f(5)) = 11. Squaring the first output, or doubling it, is not the same as applying the rule again. Write the intermediate value down so the second step uses it.

An inverse undoes a function: if f takes x to y, the inverse takes y back to x. For f(x) = 4x − 7, f multiplies by 4 and then subtracts 7, so the inverse adds 7 and then divides by 4. The inverse at 21 is (21 + 7) ÷ 4 = 7, and indeed f(7) = 21. An inverse is not the reciprocal 1/f(x).

An arithmetic sequence adds the same difference each step, so the nth term is the first term plus (n − 1) differences. Starting at 11 with a difference of 4, the 15th term is 11 + 14 × 4 = 67. Using n differences instead, to get 71, is the classic off-by-one error: the first term has had no difference added yet.

The sum of an arithmetic sequence is the number of terms times the average of the first and last terms. For the first 30 terms of 4, 15, 26, and so on, the last term is 4 + 29 × 11 = 323, so the sum is 30 × (4 + 323) ÷ 2 = 4,905. Pairing the first term with the last, the second with the second-to-last, gives the same result.

A geometric sequence multiplies by the same ratio each step, so the nth term is the first term times the ratio raised to the power n − 1. For 7, 14, 28, and so on, the 8th term is 7 × 2⁷ = 896. Adding the ratio instead of multiplying by it, or raising it to the nth power, are the common errors.

Check the differences between consecutive terms: if they are constant, the sequence is arithmetic. If the ratios are constant instead, it is geometric. In 5, 15, 45, 135 the differences are 10, 30, and 90, which change, but every ratio is 3, so the sequence is geometric. Using the arithmetic formula on it badly underestimates later terms.

Multiplying powers of the same base adds the exponents: 7³ × 7⁵ = 7⁸, because there are 3 + 5 factors of 7. Multiplying the exponents, to get 7^15, or multiplying the bases, to get 49⁸, are the standard errors. The rule needs a shared base; 2³ × 3² cannot combine this way.

A power of a power multiplies the exponents: (5²)³ = 5⁶. Dividing powers of the same base subtracts them: 2^10 ÷ 2⁴ = 2⁶ = 64. Any nonzero base to the power 0 is 1, which follows from the quotient rule, since 2⁴ ÷ 2⁴ = 2⁰ = 1.

If both sides can be written as powers of the same base, set the exponents equal. In 4^x = 8^(x − 1), write 4 as 2² and 8 as 2³, so 2^(2x) = 2^(3x − 3). Then 2x = 3x − 3 and x = 3. Check: 4³ = 64 and 8² = 64. Equating exponents before the bases match gives a wrong answer.

When a quantity multiplies by the same factor each period, the amount after t periods is the starting amount times the factor to the power t. A colony of 150 that triples every day reaches 150 × 3⁴ = 12,150 after four days. Adding the same amount each day instead treats the growth as linear and badly underestimates it.

To factor x² + bx + c, find two numbers that multiply to c and add to b. For x² − 5x − 14, the numbers are −7 and 2, since −7 × 2 = −14 and −7 + 2 = −5, so the expression is (x − 7)(x + 2). Watch the signs: a negative c means the two numbers have opposite signs.

If a product equals zero, one of its factors must be zero. From (x − 7)(x + 2) = 0, either x − 7 = 0 or x + 2 = 0, so x = 7 or x = −2. This works only when one side is zero; (x − 7)(x + 2) = 8 cannot be split this way. Report both roots unless the question restricts them.

The quadratic formula solves ax² + bx + c = 0 even when factoring is hard: x = (−b ± √(b² − 4ac)) ÷ 2a. For 2x² + 3x − 2 = 0, b² − 4ac = 9 + 16 = 25, so x = (−3 ± 5) ÷ 4, which gives 1/2 or −2. Dividing only the square root by 2a, rather than the whole numerator, is a common error.

The discriminant b² − 4ac, the part under the square root, tells you how many real roots a quadratic has: two if it is positive, one if it is zero, none if it is negative. For x² + 4x + 7 it is 16 − 28 = −12, so there are no real roots. Checking it first can save a calculation that cannot succeed.

For x² + bx + c = 0, the roots add to −b and multiply to c. For x² + 2x − 35 = 0 the sum is −2 and the product is −35, which the roots 5 and −7 confirm. The usual trap is reporting the sum as b. For ax² + bx + c, divide by a first: the sum is −b/a and the product c/a.

A difference of two squares factors as a² − b² = (a + b)(a − b). It turns hard arithmetic into easy arithmetic: 53² − 47² = (53 + 47)(53 − 47) = 100 × 6 = 600. Given x + y and x − y, the value of x² − y² is simply their product. A sum of squares, a² + b², does not factor this way.

Since (x + y)² = x² + 2xy + y², it follows that x² + y² = (x + y)² − 2xy. If x + y = 15 and xy = 50, then x² + y² = 225 − 100 = 125, with no need to find x and y, which are 5 and 10. Forgetting the 2 in 2xy gives 175, and stopping at the squared sum gives 225.

The remainder when a polynomial p(x) is divided by (x − a) equals p(a). Dividing x³ − 2x + 5 by (x − 2) leaves 8 − 4 + 5 = 9, found without long division. Watch the sign: dividing by (x + 3) means evaluating at x = −3, not at 3.

Start by writing what each letter stands for, with units: let a be the number of adult tickets and c the number of child tickets. Then translate one sentence at a time. "Is" becomes =, "more than" adds, "times" multiplies, and "less than" subtracts in reverse order, so 5 less than n is n − 5, not 5 − n.

Two people's ages rise by the same amount each year, so their difference is constant. When will a parent of 40 be exactly twice the age of a child of 14? Write 40 + t = 2(14 + t), so t = 12: the parent will be 52 and the child 26. The difference, 26, is the same then as it is now.

Turn each worker's time into a rate in jobs per hour, then add the rates. Someone who paints a fence in 9 hours does 1/9 of it per hour; someone who takes 18 hours does 1/18. Together they do 1/9 + 1/18 = 1/6 per hour, so the job takes 6 hours. Averaging the times gives 13.5 hours, longer than the faster worker needs alone, which cannot be right.

Problems mixing two kinds of item give two equations, one for the count and one for the value. For 20 coins, all quarters and dimes, worth $3.65, q + d = 20 and 25q + 10d = 365, in cents. Substituting d = 20 − q gives 15q + 200 = 365, so there are 11 quarters and 9 dimes. Mixing cents and dollars in one equation breaks it.

Most algebra errors are sign errors. Watch three places: distributing a negative, as in −2(x − 5) = −2x + 10; subtracting an expression in parentheses, as in 7 − (x − 3) = 10 − x; and moving a term across the equals sign, which changes its sign. Recheck any step where a negative touches parentheses.

Check that the answer fits its context: a number of people must be a whole, nonnegative number; a time cannot be negative; a cyclist moving at 600 mph signals a units error. When a quadratic gives two roots, keep only the ones that fit the situation, such as the positive length of a side.

Match the form of your answer to the question: both roots of a quadratic, an inequality rather than a single value, a simplified expression rather than a number, or the value of x² − y² rather than x itself. Many wrong choices are correct intermediate results, such as x when the question asks for 2x + 1.