Keystone Algebra I Practice Questions (2026) — Free Sample Questions with Answers

Answer: D

1. 1. $x^2-9=(x-3)(x+3)$
2. 2. $x^2+5x+6=(x+2)(x+3)$
3. 3. Simplified: $\frac{x-3}{x+2}$, with $x\ne -2$ and $x\ne -3$
4. 4. Check at $x=-3$: original denominator $9-15+6=0$, so the original is undefined while $\frac{-6}{-1}=6$ is defined.

She is correct, because $x=-2$ is the only value that makes the denominator of $\frac{x-3}{x+2}$ equal to 0.: Looks only at the simplified expression. The original denominator $(x+2)(x+3)$ is also 0 at $x=-3$, where the original expression is undefined.

She is not correct, because $x=3$ must also be excluded, since it makes the numerator $x-3$ equal to 0.: Excludes a value that makes the numerator 0; a zero numerator gives a value of 0, which is allowed.

She is not correct, because the expression simplifies to $\frac{-9}{5x+6}$, so $x=-\frac{6}{5}$ is excluded.: Cancels the $x^2$ terms instead of common factors, which changes the expression.

She is not correct, because $x^2+5x+6=(x+2)(x+3)$, so $x=-3$ also makes the original denominator equal to 0.: Correct. $\frac{(x-3)(x+3)}{(x+2)(x+3)}=\frac{x-3}{x+2}$ only when $x\ne -2$ and $x\ne -3$; at $x=-3$ the original expression is undefined, so $-3$ must be excluded too.

Excluded values come from every factor of the original denominator, including factors that divide out during simplifying.

Answer: A

1. 1. 24 = 2³ · 3 and 36 = 2² · 3², so GCF = 2² · 3 = 12
2. 2. Least exponents: $a^2$, $b^2$
3. 3. GCF = $12a^2{b}^2$
4. 4. Check: $24a^3{b}^2÷12a^2{b}^2=2a$ and $36a^2{b}^5÷12a^2{b}^2=3b^3$; 2a and 3b³ share no factor.

$12a^2{b}^2$: Correct. GCF(24, 36) = 12, and the least exponents are $a^2$ and $b^2$, so the GCF is $12a^2{b}^2$.

$12a^3{b}^5$: Uses the greatest exponents, which belong to the LCM; $a^3$ does not divide $36a^2{b}^5$.

$72a^2{b}^2$: Uses the LCM of the coefficients, 72, which is not a factor of 24 or 36.

$72a^3{b}^5$: Gives the LCM of the two monomials instead of the GCF.

The GCF uses the GCF of the coefficients and the least exponent of each common variable.

Answer: B

1. 1. $75x^3{y}^2=(25x^2{y}^2)(3x)$
2. 2. $\sqrt{25x^2{y}^2}=5xy$
3. 3. $\sqrt{75x^3{y}^2}=5xy\sqrt{3x}$
4. 4. Check: $(5xy\sqrt{3x}{)}^2=25x^2{y}^2\cdot 3x=75x^3{y}^2$.

$3xy\sqrt{5x}$: Takes the factor 3 outside and leaves 25 inside, reversing which factor of 75 is the perfect square.

$5xy\sqrt{3x}$: Correct. $75x^3{y}^2=25\cdot x^2\cdot y^2\cdot 3x$, and $\sqrt{25x^2{y}^2}=5xy$, so the expression is $5xy\sqrt{3x}$.

$25xy\sqrt{3x}$: Moves 25 outside the radical without taking its square root; $\sqrt{25}=5$.

$5x^{2}y\sqrt{3x}$: Moves $x^2$ outside without taking its square root; $\sqrt{x^2}=x$.

Take the square root of each perfect-square factor (coefficient and even powers of variables) and leave the remaining factors under the radical.

Answer: C

1. 1. $0.30x+12=0.45x+6$, so $6=0.15x$ and $x=40$, $y=24$
2. 2. Test $x=50$: Alto $27, Brisk $28.50, so Alto is less
3. 3. Test $x=20$: Alto $18, Brisk $15, so Brisk is less
4. 4. Only "longer than 40 miles, Alto charges less" is true.

For a 24-mile trip, both companies charge $40.: Reverses the coordinates; x = 40 miles and y = $24.

For a trip shorter than 40 miles, Alto charges less than Brisk.: Reverses the comparison; for 20 miles Alto charges $18 and Brisk charges $15.

For a trip longer than 40 miles, Alto charges less than Brisk.: Correct. Past the intersection the smaller rate wins: for 50 miles Alto charges $0.30(50)+12=\$27$ and Brisk charges $0.45(50)+6=\$28.50$.

For a trip longer than 40 miles, Brisk charges less than Alto.: Picks the company with the lower starting fee; past 40 miles Brisk's higher rate makes it cost more.

The solution of the system is where the two costs are equal; on either side of it, the plan with the smaller rate costs less for larger inputs.

Answer: C

1. 1. $8x-12-5x=2x+14-3x$
2. 2. $3x-12=-x+14$
3. 3. $4x=26$, so $x=6.5$
4. 4. Check: left $4(13-3)-32.5=7.5$; right $2(13.5)-19.5=7.5$.

4.25: Distributes the 4 to the first term only, writing $8x-3$ instead of $8x-12$.

4.75: Distributes the 2 to the first term only, writing $2x+7$ instead of $2x+14$.

6.5: Correct. The left side is $3x-12$ and the right side is $-x+14$, so $4x=26$ and $x=6.5$.

13: Subtracts $x$ from both sides instead of adding it, getting $2x=26$.

Distribute to every term inside each set of parentheses, combine like terms on each side, then collect the variable on one side.

Answer: D

1. 1. Let x = tacos and y = burritos: $3x+8y\ge 400$ and $x+y\le 120$
2. 2. (90, 15): $270+120=390\ge 400$ false
3. 3. (30, 95): $30+95=125\le 120$ false; (110, 12): $122\le 120$ false
4. 4. (64, 26): $192+208=400\ge 400$ true and $90\le 120$ true

Bruno sells 90 tacos and 15 burritos.: Meets the item limit (105) but earns only $270+120=\$390$, less than $400.

Bruno sells 30 tacos and 95 burritos.: Earns enough ($850) but needs 125 items, more than 120.

Bruno sells 110 tacos and 12 burritos.: Earns enough ($426) but needs 122 items, more than 120.

Bruno sells 64 tacos and 26 burritos.: Correct. $3(64)+8(26)=192+208=\$400\ge \$400$, and $64+26=90\le 120$; earning exactly $400 meets "at least $400".

Test each combination in every inequality of the system; a combination exactly on an "at least" or "at most" boundary still meets it.

Answer: A

1. 1. $y=-\frac{1}{2}x+3$: intercept 3, slope $-\frac{1}{2}$, dashed, shade below
2. 2. $y=x-3$: intercept $-3$, slope 1, solid, shade above
3. 3. Test (0, 0): $0<3$ and $0\ge -3$, both true, so (0, 0) is in the overlap
4. 4. The lines meet at (4, 1): $-2+3=1$ and $4-3=1$.

Graph A: Correct. $y=-\frac{1}{2}x+3$ is dashed (<) with the region below it, and $y=x-3$ is solid (≥) with the region above it; the overlap contains (0, 0).

Graph B: Swaps the line types: $y=-\frac{1}{2}x+3$ must be dashed for <, and $y=x-3$ solid for ≥.

Graph C: Shades above $y=-\frac{1}{2}x+3$; the symbol < calls for the region below the line.

Graph D: Shades below $y=x-3$; the symbol ≥ calls for the region above the line.

Each inequality fixes its boundary line, its line type (dashed for < or >, solid for ≤ or ≥) and its shaded side; the solution is where both shadings overlap.

Answer: D

1. 1. Figure 10: 3(10) + 2 = 32 tiles; twice that is 64.
2. 2. Figure 20: 3(20) + 2 = 62 tiles.
3. 3. 62 ≠ 64, so Paloma is not correct.
4. 4. Check: 62 − 32 = 30 = 10 × 3, ten more steps of 3 tiles.

She is correct, because 20 is twice 10, so the number of tiles also doubles.: Assumes the pattern is proportional. With the constant 2 in t = 3n + 2, doubling n does not double t.

She is not correct, because Figure 20 has 3(20) = 60 tiles, which is less than 64.: Uses 3n alone and drops the constant 2: Figure 20 has 3(20) + 2 = 62 tiles. The conclusion is right, but the reason is wrong.

She is correct, because the pattern adds the same number of tiles each time.: A constant rate makes the pattern linear, but a linear pattern doubles only when its constant term is 0.

She is not correct, because Figure 20 has 62 tiles, and twice Figure 10 is 64.: Correct. t = 3(20) + 2 = 62, while twice Figure 10 is 2 × 32 = 64. Only the 3n part doubles; the constant 2 stays the same.

A linear function y = mx + b is proportional only when b = 0; otherwise doubling the input does not double the output.

Answer: B

1. 1. A relation stays a function if no input gets a second, different output.
2. 2. k = −3, 2 or 6: each input already has an output other than 5, so the relation would fail.
3. 3. k = 0: the pair (0, 5) already exists, so input 0 still has only the output 5.
4. 4. Check: inputs −3, 0, 2, 6 each have one output.

−3: The input −3 is already paired with 1; adding (−3, 5) would give −3 two different outputs.

0: Correct. (0, 5) is already in the relation, so adding it again creates no new pairing; every input still has exactly one output.

2: The input 2 is already paired with 8; adding (2, 5) would give 2 two different outputs.

6: The input 6 is already paired with 1; adding (6, 5) would give 6 two outputs. The repeated output 1 is not the problem.

Repeating an input breaks a function only when the outputs differ; repeating the same ordered pair changes nothing.

Answer: D

1. 1. Rate: (12 − 8) ÷ (5 − 3) = 4 ÷ 2 = 2 seats per table.
2. 2. s = 2n + b; using (2, 6): 6 = 2(2) + b, so b = 2.
3. 3. Equation: s = 2n + 2.
4. 4. Check every row: 2(3) + 2 = 8, 2(5) + 2 = 12, 2(8) + 2 = 18.

$s=n+4$: Fits only the first row: 3 + 4 = 7, not 8. It treats the pattern as adding 1 seat per table.

$s=3n$: Fits only the first row (3 × 2 = 6); for 3 tables it gives 9 seats, not 8.

$s=4n-2$: Uses the change in s between the second and third rows (4) as the rate without dividing by the change in n (2); it fails for n = 3 (10 seats).

$s=2n+2$: Correct. The seats increase by 2 for each added table (from 8 to 12 as n goes from 3 to 5, so 4 ÷ 2 = 2), and 2(2) + 2 = 6 fits the first row.

When the inputs skip values, divide the change in output by the change in input, then test the rule on every row, not just the first.

Answer: B

1. 1. For each pair of neighboring points, compute change in y ÷ change in x.
2. 2. Second set: (11 − 5) ÷ (3 − 1) = 3, (14 − 11) ÷ (4 − 3) = 3, (23 − 14) ÷ (7 − 4) = 3.
3. 3. Fourth set: −3 ÷ 2 = −1.5, −3 ÷ 2 = −1.5, −3 ÷ 3 = −1, not constant.
4. 4. Only the second set has a constant rate, 3.

$(0,2),(1,4),(2,8),(3,16)$: The y-values double each time; the changes, 2, 4 and 8, are not constant.

$(1,5),(3,11),(4,14),(7,23)$: Correct. 6 ÷ 2 = 3, 3 ÷ 1 = 3 and 9 ÷ 3 = 3, so y changes by 3 for every 1 in x even though the x-steps differ.

$(1,2),(2,4),(3,7),(4,11)$: The x-values go up by 1, but y goes up by 2, then 3, then 4, so the rate is not constant.

$(-2,9),(0,6),(2,3),(5,0)$: y drops by 3 each time, but the last x-step is 3, not 2; the rates are −1.5, −1.5 and −1.

A constant rate of change means the ratio of the change in y to the change in x is the same for every pair of points; equal y-steps alone are not enough.

Answer: B

1. 1. Rise: 30 inches ÷ 12 = 2.5 feet.
2. 2. Slope limit: rise ÷ run ≤ 1 ÷ 12, so run ≥ 12 × rise.
3. 3. Run ≥ 12 × 2.5 = 30 feet.
4. 4. Check: 2.5 ÷ 30 = 1 ÷ 12.

2.5: This is the height of the porch in feet (30 ÷ 12); it still must be multiplied by 12 to get the horizontal length.

30: Correct. The rise is 30 inches = 2.5 feet, and the horizontal length must be 12 times the rise: 12 × 2.5 = 30 feet.

36: Converts 30 inches to feet by dividing by 10 instead of 12 (getting 3 feet), then multiplies by 12; 1 foot is 12 inches, so the rise is 2.5 feet.

360: Multiplies 30 by 12 without converting inches to feet; 360 is the length in inches.

A slope is a ratio of rise to run; put both lengths in the same unit before using it.

Answer: C

1. 1. P(blue first) = 5/12.
2. 2. After removing a blue pen: 11 pens, 3 green, so P(green second) = 3/11.
3. 3. P = 5/12 × 3/11 = 15/132 = 5/44.
4. 4. Check: 5/44 ≈ 0.114, a little more than the with-replacement value 5/48 ≈ 0.104, because the total shrank.

$\frac{5}{66}$: Reduces the number of green pens as well as the total, using 2/11; only a blue pen was removed, so 3 green pens remain.

$\frac{5}{48}$: Treats the draws as if the first pen were replaced (5/12 × 3/12); without replacement only 11 pens remain for the second draw.

$\frac{5}{44}$: Correct. P(blue first) = 5/12. With the blue pen removed, 3 of the 11 remaining pens are green: 5/12 × 3/11 = 15/132 = 5/44.

$\frac{2}{3}$: Adds the probabilities (5/12 + 3/12); an "and" probability for two draws is found by multiplying.

For draws without replacement, multiply the first probability by the second probability computed from what remains.

Answer: C

1. 1. n = 40, so the median is the mean of the 20th and 21st values.
2. 2. Cumulative counts: 3, 11, 18, 27, 40.
3. 3. The 20th and 21st values fall in the interval with cumulative count 27: 30–39 minutes.
4. 4. Check: 18 values are below 30 minutes and 13 are 40 minutes or more.

10–19 minutes: This interval holds the 4th through 11th values, so it contains the lower quartile (10th and 11th values), not the median.

20–29 minutes: This is the middle interval label on the axis, but the median depends on the counts: only 18 patients waited less than 30 minutes.

30–39 minutes: Correct. The median of 40 values is between the 20th and 21st values. Counting up, 3 + 8 + 7 = 18 values come before this interval, and it holds values 19 through 27.

40–49 minutes: This is the tallest bar (the mode interval), not the interval that contains the 20th and 21st values.

To find the median interval of a histogram, add bar heights from the left until you pass the middle position; do not use the middle label or the tallest bar.