Praxis Algebra I common mistakes
Praxis Algebra I · Understand the traps and how to avoid them.
When solving 3(x − 2) = 5x + 4, the candidate distributes the 3 to the first term only, writing 3x − 2 = 5x + 4, then carries that wrong constant to the final answer.
The distributive property is applied so often by pattern-matching that the second factor gets skipped under time pressure; the candidate's eye locks onto the term adjacent to the parentheses and moves on.
What to do instead
Adopt a written checkpoint: after distributing, count terms inside the parentheses and count the terms you produced - they must match. Practice ten equations of the form a(bx + c) = dx + e, writing the distribution step on its own line before solving, until the two-line habit is automatic.
When solving −2x > 6, the candidate divides by −2 and writes x > −3, keeping the inequality direction unchanged.
The same motor routine used for equations (divide both sides by the coefficient) is reused, and the direction-reversal rule is memorized as a sentence rather than tied to a number-line picture, so it does not fire in the moment.
What to do instead
Test every inequality answer with one substituted value. For x > −3, substitute x = 0 into −2x > 6: you get 0 > 6, which is false, so the direction must be wrong. Make this one-substitution check a fixed final step on every inequality item.
When asked for the equation of the line through (2, 5) and (6, 13), the candidate computes slope as (13 − 5)/(6 − 2) correctly, then substitutes the wrong point's coordinates or reverses the subtraction order in only one of the numerator and denominator, producing slope 2/4 or −2.
Two subtractions must be done in the same order, and the candidate tracks the order mentally instead of on paper; mixing the order in one place flips the sign or scrambles the ratio.
What to do instead
Write the slope formula with the points labeled above it - (x₁, y₁) = (2, 5) and (x₂, y₂) = (6, 13) - and fill the formula vertically, numerator from y's, denominator from x's, in the same left-to-right order. Then verify by counting rise over run on a quick sketch.
When converting y − 5 = 3(x − 2) to slope-intercept form, the candidate distributes to get y − 5 = 3x − 6 but then moves the 5 by subtracting it from the right side, giving y = 3x − 11.
The candidate knows 'do the same thing to both sides' but applies the inverse operation to the wrong side or the wrong sign, because the goal (isolate y) is not held in mind while manipulating.
What to do instead
State the goal aloud before each step: 'I need y alone, so I must undo −5 by adding 5 to both sides.' Write the operation in the margin next to the line it changes. Then check by substituting the original point (2, 5) into the final equation - it must satisfy it.
Given f(x) = 2x² − 3x + 1, the candidate evaluates f(−2) as 2(−2)² − 3(−2) + 1 = −8 + 6 + 1 = −1, squaring only the 2 and keeping the negative sign outside the square.
The notation −2² and (−2)² look nearly identical in handwriting, and the candidate applies exponent-before-sign from memory without tracking which symbol the exponent attaches to.
What to do instead
When substituting a negative value, always write it in parentheses at the moment of substitution: replace every x with (−2). Then evaluate the exponent first on the parenthesized quantity. Do five substitution drills with negative inputs until parenthesizing is reflex.
When simplifying (x³y²)² · x⁻¹, the candidate multiplies the exponents inside by the outside 2 for x but forgets the y, and adds the −1 to 6 instead of applying it to the correct base, ending with x⁵y² instead of x⁵y⁴.
Multiple exponent rules (power of a power, product rule, negative exponent) are triggered in one expression, and the candidate applies them sequentially from memory without a written rule list, dropping one rule per pass.
What to do instead
Before simplifying, write the three rules you will need at the top of the workspace: (aᵐ)ⁿ = aᵐⁿ, aᵐ·aⁿ = aᵐ⁺ⁿ, a⁻ⁿ = 1/aⁿ. Then handle one base at a time - do all x work, then all y work - and box each result before combining.
When factoring x² − 7x + 12, the candidate writes (x − 3)(x + 4) and stops, without checking that the middle term matches.
The candidate finds two numbers that multiply to 12 and assumes the job is done, forgetting that the sum must equal −7; the sign pattern is chosen by guess rather than by solving the pair of conditions.
What to do instead
Write both conditions explicitly before choosing factors: product = +12, sum = −7. List factor pairs of 12 with signs and cross off any pair whose sum is not −7. Then expand the factored form as a final check - the middle term must come back as −7x.
When solving x² = 5x, the candidate divides both sides by x and reports only x = 5.
Dividing by the variable feels like normal simplifying, and the candidate does not register that division by x is invalid when x = 0, so the zero solution silently disappears.
What to do instead
Never divide by a variable expression. Move everything to one side: x² − 5x = 0, factor to x(x − 5) = 0, and apply the zero-product property to get both solutions. State the rule as 'factor, don't divide' and apply it to every quadratic, even ones that look trivially divisible.
Given a table of x and y values, the candidate reads the y-intercept from the first row of the table rather than from the row where x = 0, or extrapolates the intercept by continuing the pattern incorrectly.
Tables are often sorted so the first row looks like the starting value, and the candidate conflates 'first entry' with 'value at x = 0' without checking the x-column.
What to do instead
Before reading anything off a table, scan the x-column for a 0. If none exists, compute slope from two rows, then substitute one point into y = mx + b and solve for b algebraically instead of guessing.
When a word problem asks for the cost after a 20% discount followed by 8% tax, the candidate computes 0.20 + 0.08 and applies a single 28% adjustment, or applies the discount and tax in the wrong order without noticing the results differ.
Percent operations look commutative, and the candidate treats successive percent changes as additive rather than multiplicative; the compounding structure of the problem is not modeled.
What to do instead
Translate each percent change into a multiplier: a 20% discount is ×0.80, and 8% tax is ×1.08. Multiply the multipliers in sequence and keep the original price as a variable, e.g. P × 0.80 × 1.08. Test with P = 100 to confirm the result matches an intuitive dollar figure.
On a system of equations solved by substitution, the candidate solves 2x + y = 11 for y = 11 − 2x, then substitutes back into the same equation instead of into the other one, producing a true-but-useless identity like 11 = 11 and concluding there are infinitely many solutions.
Substituting into the equation just rearranged always yields an identity, and the candidate does not recognize that the identity is a sign of a procedural error rather than a property of the system.
What to do instead
After isolating a variable, circle the equation you used and strike it out; the substitution must go into the remaining equation. If you get an identity, stop and re-check which equation you substituted into before interpreting the result.
When a quadratic has no real roots (discriminant negative), the candidate still writes two real solutions by forcing the square root of a negative number into a decimal, or by miscomputing the discriminant as b² + 4ac.
The quadratic formula is recalled as a string of symbols, and the sign inside the discriminant is the easiest part to flip; the candidate also lacks a habit of interpreting the discriminant before computing roots.
What to do instead
Compute the discriminant b² − 4ac as a separate labeled step and interpret it first: positive means two real roots, zero means one, negative means none. Only then proceed. On multiple-choice items, this often eliminates answers before any root computation.
When graphing y = −2(x + 3)² + 1, the candidate places the vertex at (−3, −1) or at (3, 1), misreading the signs inside the squared term and the constant outside.
Vertex form y = a(x − h)² + k has a subtraction inside, so a plus sign in the problem corresponds to a negative h; the candidate applies the sign literally rather than inverting it.
What to do instead
Rewrite the given form to match the template before reading values: −2(x + 3)² + 1 becomes −2(x − (−3))² + 1, so h = −3 and k = +1. Say 'the sign inside flips, the sign outside stays' and verify by plugging x = h into the function to confirm you get k.