June 2026 Algebra I Regents: All 35 Worked Solutions

Question 1: Find a term of an arithmetic sequence

1. The sequence increases by 3 each time, so the common difference is d=3 and the first term is a₁=4.
2. Use the arithmetic-sequence formula a_n=a₁+(n−1)d. There are 19 steps from the first term to the twentieth:
   a₂₀=4+(20−1)(3)=4+57=61

Exam detail: Use 19 common differences, not 20.

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Question 2: Solve a linear equation with a decimal

1. Add x to both sides and add 4 to both sides:
   0.5x+x=8+4
2. Combine and divide:
   1.5x=12
   x=8
3. Check: 0.5(8)−4=0 and 8−8=0.

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Question 3: Factor a difference of squares

1. Recognize 4x²=(2x)² and 25=5².
2. Apply a²−b²=(a−b)(a+b) only after identifying the two square roots. For this expression:
   4x²−25=(2x−5)(2x+5)

Exam detail: The middle terms cancel when the conjugate binomials are multiplied.

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Question 4: Keep a relation a function

1. The existing inputs are 3, −4, and 1. Their outputs are already fixed.
2. The new point has output 2. Reusing any of those existing inputs would give that input two different outputs.
3. The unused input 2 avoids that conflict.

Exam detail: The function rule concerns repeated inputs with different outputs. Outputs are allowed to repeat.

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Question 5: Raise every factor in a monomial to a power

1. Apply the cube to the coefficient and both variable factors:
   (4xy²)³=4³x³(y²)³
2. Calculate the coefficient and multiply the nested exponents:
   4³=64
   (y²)³=y⁶

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Question 6: Match all the polynomial conditions

1. A third-degree polynomial has highest exponent 3. A leading coefficient of 4 means its highest-degree term is 4x³.
2. A trinomial has exactly three nonzero terms. The constant term must be +5.
3. The expression 4x³+8x²+5 meets all four requirements.

Exam detail: Degree, number of terms, leading coefficient, and constant are separate features; check each one.

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Question 7: Write a geometric-sequence rule

1. The ratio of consecutive terms is 64128=12, also confirmed by 3264=12.
2. The geometric rule is a_n=a₁rⁿ⁻¹. Substitute the first term and ratio:
   a_n=128(12)ⁿ⁻¹
3. For n=1, the exponent is zero and the rule returns 128.

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Question 8: Distinguish direction from causation

1. Greater brake pressure is associated with a lower speed, so the variables move in opposite directions. This is a negative correlation.
2. The situation also describes a mechanism: applying the brakes slows the bicycle. In this stated setting, brake pressure has a causal effect on speed.

Exam detail: Correlation by itself does not prove causation. Here, the physical action of braking supplies the causal explanation.

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Question 9: Build a cost equation from a total count

1. If d plants are daffodils and the total is 40, the number of tulips is 40−d.
2. Daffodils cost 4d dollars and tulips cost 5(40−d) dollars.
3. Add those costs and set them equal to the $170 budget:
   4d+5(40−d)=170

Exam detail: The 40 belongs in the plant-count expression; 170 is the total cost in dollars.

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Question 10: Evaluate a quadratic at a negative input

1. Substitute −1 everywhere x appears:
   h(−1)=(−1)²−3(−1)+1
2. Square the negative number and evaluate the signed product:
   h(−1)=1+3+1=5

Exam detail: Parentheses ensure that the whole negative input is squared.

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Question 11: Use the zero-product property

1. A product equals zero if at least one factor equals zero. Apply this to all three factors:
   x=0
   3x+2=0
   x−5=0
2. Solve the linear factors:
   x=−23
   x=5

Exam detail: The initial factor x contributes the zero 0.

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Question 12: Compare exponential and linear outputs

1. Use a graphing calculator table for positive integer inputs. Compare the two outputs in each row, rather than comparing the formula coefficients.
2. At 18, the exponential output is still smaller:
   f(18)=1.25¹⁸≈55.51<64=g(18)
3. At 19, it is larger:
   f(19)=1.25¹⁹≈69.39>67=g(19)
4. The integer table has no earlier positive input satisfying the inequality, so the first is 19.

Exam detail: The question asks for an input value. The numbers 67 and approximately 69 are outputs at that input.

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Question 13: Find a point-slope equation

1. Calculate the slope from the two points:
   m=5−(−1)2−(−2)=64=32
2. Use the point (2,5) in y−y₁=m(x−x₁):
   y−5=32(x−2)
3. As a check, substitute (−2,−1): both sides equal −6.

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Question 14: Calculate a percentage within a group

1. The requested group is female students. Its total is 20+40+10=70.
2. Of those 70 students, 40 selected basketball:
   4070×100≈57.14
3. The nearest listed whole-number percentage is 57%.

Exam detail: Use the female-row total as the denominator, not the total number of students or the basketball-column total.

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Question 15: Complete the square correctly

1. Move the constant to the other side:
   x²−10x=24
2. Half of −10 is −5, and its square is 25. Add 25 to both sides:
   x²−10x+25=24+25
3. Factor the perfect square:
   (x−5)²=49

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Question 16: Combine horizontal and vertical translations

1. A shift right by 3 replaces the input x with x−3. This gives f(x−3).
2. A shift up by 4 adds 4 to the output, outside the function:
   f(x−3)+4

Exam detail: Horizontal changes act inside the function; vertical changes act outside it.

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Question 17: Identify the equality property

1. The same number, 16, is added to both sides of the equation.
2. That operation preserves equality, so it uses the addition property of equality. The choice of 16 also prepares a perfect-square trinomial.

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Question 18: Find the upper quartile

1. The ten values are already sorted. Divide them into a lower half of five values and an upper half of five values.
2. The upper half is 32, 34, 34, 40, 42. Its middle value is the third value, 34.
3. Therefore the upper quartile is Q₃=34.

Exam detail: The upper quartile is the median of the upper half, not the maximum.

For more practice, use the quartile calculator with the median-excluded convention.

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Question 19: Subtract every term of a polynomial

1. Write the starting polynomial first and subtract the entire other polynomial:
   (2x²−3x+10)−(−3x²+7x−1)
2. Distribute the subtraction sign:
   2x²−3x+10+3x²−7x+1
3. Combine like terms:
   5x²−10x+11

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Question 20: Read a growth rate from an exponential base

1. For annual growth, the multiplier has the form 1+r, where r is the rate written as a decimal.
2. Here 1+r=1.0448, so r=0.0448.
3. Convert that decimal to a percent: 0.0448×100=4.48.

Exam detail: The base is the whole growth multiplier, not the percentage rate itself.

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Question 21: Convert a distance through several units

1. Convert rods to yards: 2.5×5.5=13.75 yards.
2. Convert yards to feet: 13.75×3=41.25 feet.
3. Convert feet to inches: 41.25×12=495 inches.

Exam detail: Track the unit at each step: rods → yards → feet → inches.

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Question 22: Substitute into the quadratic formula

1. Identify a=2,b=−3,c=−6.
2. Use the quadratic formula with the signed coefficients:
   x=−(−3)±√(−3)²−4(2)(−6)2(2)
3. The discriminant is 9+48=57, so:
   x=3±√574

Exam detail: The negative constant makes the discriminant increase. Keep the sign in the product −4ac explicit.

Check a new example with the quadratic equation calculator.

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Question 23: Find an axis of symmetry from coefficients

1. For y=ax²+bx+c, the symmetry line is x=−b2a.
2. In choice (3), a=1 and b=8. Therefore:
   x=−82(1)=−4

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Question 24: Multiply and simplify radicals

1. Multiply the coefficients and the radicands:
   (2√6)(5√3)=10√18
2. Factor 18 as 9 times 2 and simplify its square root:
   10√18=10√9×2=10(3√2)=30√2

Exam detail: The form 10√18 is equivalent but not in simplest radical form.

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Question 25: Multiply two binomials into standard form

1. Distribute each term in the first binomial across the second:
   (1−2x)(3−5x)=3−5x−6x+10x²
2. Combine the two linear terms and put the powers in descending order:
   10x²−11x+3
3. A quick check at x=1 gives 2 in both the original product and the expanded polynomial.

Exam detail: A negative times a negative gives the positive leading term. Standard form requires descending powers.

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Question 26: Graph a restricted absolute-value function

1. The parent graph y=|x| is a V. Subtracting 3 moves it down, so the vertex is (0,−3).
2. Evaluate the domain endpoints:
   f(−7)=7−3=4
   f(7)=7−3=4
3. Draw straight segments from (−7,4) to (0,−3) to (7,4). The x-intercepts are (−3,0) and (3,0).
4. Use filled endpoint dots because both limits are included. Stop at the stated domain limits.

Exam detail: Do not extend the graph beyond −7 and 7. The official model responses deduct credit for extending beyond the given domain.

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Question 27: Solve an inequality and reverse its direction when needed

1. Distribute and combine terms on the right:
   −4x+1>9+6x+3+x
   −4x+1>7x+12
2. Subtract 7x and subtract 1 from both sides:
   −11x>11
3. Divide by −11. Because the divisor is negative, reverse the inequality sign:
   x<−1
4. Test x=−2: the original left side is 9 and the right side is −2, so the inequality is true. At −1 the sides are equal, so that endpoint is excluded.

Exam detail: An algebraic method is required. A graph or an unsupported final inequality does not show the requested algebraic work.

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Question 28: Make the height the subject of a formula

1. Multiply both sides by 2 to remove the fraction:
   A=12bh
   2A=bh
2. Divide by b to isolate the height:
   h=2Ab

Exam detail: The division requires b≠0. For a triangle, the base length is positive.

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Question 29: Find the average population change per year

1. Use only the two endpoint years specified in the question. The population change is:
   1643734−1428285=215449
2. The elapsed time is 2020−1980=40 years. Divide change by time:
   21544940=5386.225
3. Round to the nearest integer. The result is a positive average change of 5,386 people per year.

Exam detail: Do not report the total change of 215,449 as the annual rate, and do not divide by the number of listed census observations.

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Question 30: Rationalize a square-root denominator

1. Multiply the numerator and denominator by the same nonzero radical. This multiplies the fraction by 1:
   5√3×√3√3
2. The numerator becomes 5√3, and the denominator becomes √3√3=3.
   5√3=5√33

Exam detail: The denominator is now rational. There is no need to replace the exact radical by a decimal.

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Question 31: Graph and interpret the ball’s flight

1. The initial height is D(0)=48 feet. Find the vertex time:
   t=−322(−16)=1
2. Evaluate the height at that time:
   D(1)=−16+32+48=64
3. For the landing time, set the height equal to zero and factor:
   −16t²+32t+48=0
   −16(t−3)(t+1)=0
   The roots are 3 and −1. Only 3 occurs after the toss.
4. Sketch the physical arc through (0,48), (1,64), (2,48), (3,0). Start when the ball is tossed and stop when it reaches the ground.

Exam detail: Include the graph as well as both answers. Extending a height-versus-time curve into negative time or below ground misrepresents the stated flight.

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Question 32: Solve a linear–quadratic system algebraically

1. Both expressions equal y, so set them equal:
   −2x+3=x²−5x+3
2. Move all terms to one side and factor:
   x²−3x=0
   x(x−3)=0
3. The zero-product property gives x=0 or x=3. Substitute each into the line equation:
4. For x=0, y=−2(0)+3=3. For x=3, y=−2(3)+3=−3.
5. Check the second equation: its outputs are 3 at input 0 and −3 at input 3, so both pairs satisfy both equations.

Exam detail: Do not divide x(x−3)=0 by x; doing so would discard the valid zero solution. State both ordered pairs.

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Question 33: Fit and interpret a regression line

1. Enter the six experience values into one calculator list and the six salary values into another, keeping the given pairs aligned. The salary values are in thousands of dollars.
2. Run linear regression. The unrounded slope is 4.6125 and the intercept is approximately 11.933333. Round both to the nearest hundredth:
   y=4.61x+11.93
3. The correlation coefficient is approximately 0.975544, so r≈0.98 to the nearest hundredth.
4. This is a strong positive linear relationship. In these data, more experience is associated with a higher salary, and the points lie fairly close to an upward-sloping line.

Exam detail: Keep salary in the units specified by the table. State the strength of the fit, not merely its direction. Correlation alone does not prove causation.

Explore the data with the regression analysis calculator.

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Question 34: Graph a system of inequalities and justify a point

1. Rewrite each inequality with y isolated:
   y<12x−4
   y≥−3x+6
2. Draw y=12x−4 as a dashed line and shade below it. Draw y=−3x+6 as a solid line and shade above it. The overlap is the solution region.
3. Label both boundaries and mark a point in the overlap, for example (4,−4).
4. Check that point in both original inequalities:
   2(−4)<4−8
   −8<−4
   3(4)+(−4)≥6
   8≥6
   Both statements are true.

Exam detail: A correct point needs a justification, and the graph must include a labeled boundary. Other points are valid if they satisfy both inequalities.

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Question 35: Model and graph clothing prices

1. Let x be the price of one tank top and y the price of one sweatshirt. The two purchase options give:
   3x+2y=52
   2x+y=30
2. Rearrange for graphing:
   y=26−32x
   y=30−2x
3. For the first line, plot (0,26), (4,20), (8,14). For the second, plot (0,30), (5,20), (8,14). Draw and label the two lines in the nonnegative-price region.
4. The lines intersect at (8,14). Check the point in both original models:
   3(8)+2(14)=52
   2(8)+14=30
5. The first coordinate means that one tank top costs $8. The second coordinate means that one sweatshirt costs $14.

Exam detail: This question continues on exam page 19. Full work includes both model equations, the requested graph with line labels, the intersection coordinates, and their meaning in context. Algebra alone does not replace the required graph.

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