June 2024 Algebra I Regents — Answer Key & Worked Solutions

Every question from the June 2024 Algebra I Regents exam, with the correct answer and a full step-by-step solution. Check your work, learn the method, then convert your raw score.

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These are real questions from the June 2024 Algebra I New York State Regents examination, published by NYSED. Worked solutions are written by RegentsLab.

Question 1 · 2 pts

A ball was launched into the air, and its height above the ground was recorded each second, as shown in the table below. | Time (sec) | 0 | 1 | 2 | 3 | 4 | |---|---|---|---|---|---| | Height (ft) | 11 | 59 | 75 | 59 | 11 | Based on these data, which statement is a valid conclusion?

(A) The ball lands on the ground at 4 seconds.
(B) The ball reaches a maximum height of 11 feet.
(C) The ball was launched from a height of 0 feet.
(D) The ball reaches its maximum height at 2 seconds. ✓

Worked solution

$\text{Heights: } 11, 59, 75, 59, 11 \text{ for } t=0,1,2,3,4$

$\text{The data are symmetric about } t=2$ Height increases up to t=2 then decreases the same way, so the maximum occurs at t=2 seconds.

The height values rise to a peak of 75 ft at t = 2 seconds and then mirror themselves going back down, showing the maximum occurs at t = 2.

Question 2 · 2 pts

A tour bus can seat, at most, 48 passengers. An adult ticket costs $18 and a child ticket costs $12. The bus company must collect at least $650 to make a profit. If $a$ represents the number of adult tickets sold and $c$ represents the number of child tickets sold, which system of inequalities models this situation if they make a profit?

(A) $a + c < 48$ $18a + 12c > 650$
(B) $a + c \le 48$ $18a + 12c \ge 650$ ✓
(C) $a + c < 48$ $18a + 12c < 650$
(D) $a + c \le 48$ $18a + 12c \le 650$

Worked solution

$a + c \le 48$ "At most, 48 passengers" means the total is less than or equal to 48.

$18a + 12c \ge 650$ "Must collect at least $650" means the revenue is greater than or equal to 650.

"At most" translates to $\le$, and "at least" translates to $\ge$, giving the system $a+c \le 48$ and $18a+12c \ge 650$.

Question 3 · 2 pts

Which equation is always true?

(A) $x^2 \cdot x^3 = x^5$ ✓
(B) $3^x \cdot 3^2 = 9^{2x}$
(C) $-z^2 = z^2$
(D) $7^a \cdot 7^b = 7^{ab}$

Worked solution

$x^2 \cdot x^3 = x^{2+3} = x^5$ Product of powers rule: multiplying like bases adds exponents.

By the product of powers rule, $x^2 \cdot x^3 = x^{2+3} = x^5$ for all values of $x$, so it is always true; the others fail for at least some values.

Question 4 · 2 pts

The expression $-2(x^2 - 2x + 1) + (3x^2 + 3x - 5)$ is equivalent to

(A) $x^2 + x - 4$
(B) $x^2 - x - 7$
(C) $x^2 + 7x - 4$
(D) $x^2 + 7x - 7$ ✓

Worked solution

$-2(x^2-2x+1) = -2x^2+4x-2$

$-2x^2+4x-2+3x^2+3x-5 = x^2+7x-7$

Distributing the $-2$ and combining like terms with the second polynomial gives $x^2 + 7x - 7$.

Question 5 · 2 pts

Which sum is irrational?

(A) $-2\sqrt{12} + \sqrt{100}$ ✓
(B) $-\sqrt{4} + \frac{1}{3}\sqrt{900}$
(C) $\frac{1}{2}\sqrt{25} + \sqrt{64}$
(D) $\sqrt{49} + 3\sqrt{121}$

Worked solution

$-2\sqrt{12}+\sqrt{100} = -4\sqrt{3}+10$ $\sqrt{12}=2\sqrt3$, and $\sqrt3$ is irrational, so the whole sum is irrational.

$-\sqrt4+\tfrac13\sqrt{900} = -2+10 = 8 \ (\text{rational})$

$\tfrac12\sqrt{25}+\sqrt{64} = 2.5+8 = 10.5 \ (\text{rational})$

$\sqrt{49}+3\sqrt{121} = 7+33 = 40 \ (\text{rational})$

Only the first expression simplifies to a rational number plus an irrational multiple of $\sqrt3$, so its sum is irrational; the rest simplify fully to rational numbers.

Question 6 · 2 pts

The solution to $\frac{4(x-5)}{3} + 2 = 14$ is

(A) 15
(B) 14 ✓
(C) 6
(D) 4

Worked solution

$\frac{4(x-5)}{3} = 12$

$4(x-5) = 36$

$x-5 = 9 \Rightarrow x = 14$

Subtracting 2 then multiplying by 3 and dividing by 4 isolates $x$, giving $x=14$.

Question 7 · 2 pts

On an island, a rare breed of rabbit doubled its population each month for two years. Which type of function best models the increase in population at the end of two years?

(A) linear growth
(B) linear decay
(C) exponential growth ✓
(D) exponential decay

Worked solution

$P(t) = P_0 \cdot 2^t$ Doubling repeatedly (multiplying by a constant factor each period) is the signature of exponential growth.

A quantity that doubles every fixed time period grows by a constant multiplicative factor, which is exponential growth, not linear change.

Question 8 · 2 pts

What is the degree of the polynomial $2x - x^2 + 4x^3$?

(A) 1
(B) 2
(C) 3 ✓
(D) 4

Worked solution

$\text{Terms: } 2x^1,\ -x^2,\ 4x^3$ The degree of a polynomial is the highest exponent present.

The highest power of $x$ in the polynomial is $x^3$, so the degree is 3.

Question 9 · 2 pts

The zeros of the function $f(x) = x(x-5)(3x+6)$ are

(A) 0, $-5$, and 2
(B) 0, 5, and $-2$ ✓
(C) $-5$ and 2, only
(D) 5 and $-2$, only

Worked solution

$x=0$

$x-5=0 \Rightarrow x=5$

$3x+6=0 \Rightarrow x=-2$

Setting each factor equal to zero gives the three zeros $x=0$, $x=5$, and $x=-2$.

Question 10 · 2 pts

What is the $y$-intercept of the line that passes through the points $(-1,5)$ and $(2,-1)$?

(A) $-1$
(B) $-2$
(C) 3 ✓
(D) 5

Worked solution

$m = \frac{-1-5}{2-(-1)} = \frac{-6}{3} = -2$

$y-5 = -2(x-(-1)) \Rightarrow y = -2x+3$

$\text{y-intercept} = 3$

Finding the slope between the two points and using point-slope form gives the line $y=-2x+3$, whose $y$-intercept is 3.

Question 11 · 2 pts

Nancy has just been hired for her first job. Her company gives her four choices for how she can collect her annual salary over the first eight years of employment. Each function below represents the four choices she has for her annual salary in thousands of dollars, where $t$ represents the number of years after she is hired. $a(t) = 2^t + 25$ $b(t) = 10t + 75$ $c(t) = \sqrt{400t} + 80$ $d(t) = 2(t+1)^2 - 10t + 50$ Which pay plan should Nancy choose in order to have the highest salary in her eighth year?

(A) $a(t)$ ✓
(B) $b(t)$
(C) $c(t)$
(D) $d(t)$

Worked solution

$a(8) = 2^8+25 = 256+25 = 281$

$b(8) = 10(8)+75 = 155$

$c(8) = \sqrt{400(8)}+80 = \sqrt{3200}+80 \approx 136.6$

$d(8) = 2(9)^2-10(8)+50 = 162-80+50 = 132$

Evaluating all four functions at $t=8$ shows $a(t)$ gives by far the highest value (281), because exponential growth eventually outpaces linear, square-root, and quadratic-linear combinations.

Question 12 · 2 pts

The third term in a sequence is 25 and the fifth term is 625. Which number could be the common ratio of the sequence?

(A) $\frac{1}{5}$
(B) 5 ✓
(C) $\frac{1}{25}$
(D) 25

Worked solution

$a_5 = a_3 \cdot r^2$

$625 = 25r^2 \Rightarrow r^2 = 25$

$r = 5 \ (\text{taking the positive root among the choices})$

Since two terms apart in a geometric sequence differ by a factor of $r^2$, solving $625=25r^2$ gives $r^2=25$, so $r=5$ matches one of the given choices.

Question 13 · 2 pts

The box plot below summarizes the data for the amount of snowfall, in inches, during the winter of 2021 for 12 locations in western New York. What is the interquartile range?

(A) 30
(B) 50 ✓
(C) 80
(D) 110

Worked solution

$IQR = Q_3 - Q_1$

$Q_1 \approx 60, \quad Q_3 \approx 110$

$IQR \approx 110-60 = 50$

Reading the left and right edges of the box (the first and third quartiles) off the box plot and subtracting gives an interquartile range of 50 inches.

Question 14 · 2 pts

Four quadratic functions are represented below. I: $a(x) = (x-3)^2 - 7$ II: the graph of $b(x)$ III: $c(x) = x^2 + 6x + 3$ IV: a table of values for $d(x)$: | x | $d(x)$ | |---|---| | $-4$ | $-1$ | | $-3$ | $-4$ | | $-2$ | $-5$ | | $-1$ | $-4$ | | 0 | $-1$ | Which function has the smallest minimum value?

(A) I ✓
(B) II
(C) III
(D) IV

Worked solution

$a(x): \text{minimum} = -7 \ (\text{vertex form, } k=-7)$

$c(x) = x^2+6x+3: \text{vertex at } x=-3,\ c(-3) = 9-18+3 = -6$

$d(x): \text{minimum from table at } x=-2 \text{ is } -5$

$b(x): \text{graph shows a minimum greater than } -7$

Function I has a minimum value of $-7$, which is lower than the minima of $-6$ (III), $-5$ (IV), and the minimum visible on the graph of $b(x)$ (II), so I has the smallest minimum.

Question 15 · 2 pts

The equation that represents the sequence $-2, -5, -8, -11, -14, \ldots$ is

(A) $a_n = -3 + (-2)(n-1)$
(B) $a_n = -2 + (-3)(n-1)$ ✓
(C) $a_n = 3 + (-2)(n-1)$
(D) $a_n = -2 + (3)(n-1)$

Worked solution

$\text{Common difference } d = -5-(-2) = -3$

$a_1 = -2$

$a_n = -2 + (-3)(n-1)$

The sequence decreases by 3 each term with a first term of $-2$, so the explicit formula is $a_n = -2 + (-3)(n-1)$.

Question 16 · 2 pts

The dot plot below shows the number of goals Jessica scored in each lacrosse game last season. Which statement about the dot plot is correct?

(A) mean > mode
(B) mean = median ✓
(C) mode = median
(D) median > mean

Worked solution

$\text{Read the frequency of each value (0 through 6 goals) from the dot plot}$

$\text{Compute mean } = \frac{\sum(\text{goals} \times \text{frequency})}{\text{total games}}$

$\text{Compute the median (middle value of the ordered data)}$

$\text{The computed mean and median come out equal}$

Using the frequencies shown in the dot plot, the mean number of goals per game turns out equal to the median number of goals per game.

Question 17 · 2 pts

The students in Mrs. Smith's algebra class were asked to describe the graph of $g(x) = 2(x-3)^2$ compared to the graph of $f(x) = x^2$. Which student response is correct?

(A) Ashley said that the graph of $g(x)$ is wider and shifted left 3 units.
(B) Beth said that the graph of $g(x)$ is narrower and shifted left 3 units.
(C) Carl said that the graph of $g(x)$ is wider and shifted right 3 units.
(D) Don said that the graph of $g(x)$ is narrower and shifted right 3 units. ✓

Worked solution

$g(x) = 2(x-3)^2$ The coefficient 2 (with $|2|>1$) makes the parabola narrower than $f(x)=x^2$.

$(x-3) \text{ shifts the graph right 3 units}$

A leading coefficient greater than 1 vertically stretches (narrows) the parabola, and $(x-3)$ shifts the graph 3 units to the right, matching Don's description.

Question 18 · 2 pts

One Saturday, Dave took a long bike ride. The graph below models his trip, plotting Miles Traveled versus Hours. What was Dave's average rate of change, in miles per hour, on this trip?

(A) 10 ✓
(B) 11
(C) 11.6
(D) 14.5

Worked solution

$\text{Starting point: } (0, 0), \ \text{ending point: } (5.5, 55)$

$\text{average rate of change} = \frac{55-0}{5.5-0} = 10$

The overall average rate of change is total distance divided by total time, from the start point (0,0) to the end point (5.5, 55), giving 10 miles per hour.

Question 19 · 2 pts

Which expression is equivalent to $(x-5)(2x+7) - (x+5)$?

(A) $2x^2 - 2x - 30$
(B) $2x^2 - 2x - 40$
(C) $2x^2 - 4x - 30$
(D) $2x^2 - 4x - 40$ ✓

Worked solution

$(x-5)(2x+7) = 2x^2+7x-10x-35 = 2x^2-3x-35$

$2x^2-3x-35-(x+5) = 2x^2-3x-35-x-5$

$= 2x^2-4x-40$

Expanding the product and then subtracting $(x+5)$, combining like terms, gives $2x^2-4x-40$.

Question 20 · 2 pts

The functions $f(x)$ and $g(x)$ are graphed on the set of axes below. What is the solution to the equation $f(x) = g(x)$?

(A) 1 and 5 ✓
(B) $-5$ and 0
(C) $-3$ and 5
(D) 0 and 4

Worked solution

$\text{The solution to } f(x)=g(x) \text{ is the set of } x\text{-values where the two graphs intersect}$

$\text{The two curves cross at } x=1 \text{ and } x=5$

The solutions to $f(x)=g(x)$ occur exactly where the two graphs intersect, which the graph shows happening at $x=1$ and $x=5$.

Question 21 · 2 pts

When babysitting, Nicole charges an hourly rate and an additional charge for gas. She uses the function $C(h) = 6h + 5$ to determine how much to charge for babysitting. The constant term of this function represents

(A) the additional charge for gas ✓
(B) the hourly rate Nicole charges
(C) the number of hours Nicole babysits
(D) the total Nicole earns from babysitting

Worked solution

$C(h) = 6h + 5$ 6 is the rate per hour (coefficient of h), 5 is the fixed constant term

$\text{constant term} = 5 \rightarrow \text{flat gas charge, independent of } h$

Answer: the additional charge for gas

In a linear function of the form (rate)(h) + constant, the constant term is the value added regardless of the input, here representing the fixed gas charge.

Question 22 · 2 pts

When solved for $x$ in terms of $a$, the solution to the equation $3x - 7 = ax + 5$ is

(A) $\frac{12}{3a}$
(B) $\frac{12}{3-a}$ ✓
(C) $\frac{3a}{12}$
(D) $\frac{3-a}{12}$

Worked solution

$3x - 7 = ax + 5$

$3x - ax = 5 + 7$

$x(3-a) = 12$

$x = \frac{12}{3-a}$

Answer: $\frac{12}{3-a}$

Collect the x-terms on one side and constants on the other, then factor out x and divide.

Question 23 · 2 pts

Wayde van Niekerk, a runner from South Africa, ran 400 meters in 43.03 seconds to set a world record. Which calculation would determine his average speed, in miles per hour?

(A) $\frac{400\text{ m}}{43.03\text{ sec}} \cdot \frac{1000\text{ m}}{0.62\text{ mi}} \cdot \frac{1\text{ hr}}{3600\text{ sec}}$
(B) $\frac{400\text{ m}}{43.03\text{ sec}} \cdot \frac{0.62\text{ mi}}{1000\text{ m}} \cdot \frac{1\text{ hr}}{3600\text{ sec}}$
(C) $\frac{400\text{ m}}{43.03\text{ sec}} \cdot \frac{0.62\text{ mi}}{1000\text{ m}} \cdot \frac{3600\text{ sec}}{1\text{ hr}}$ ✓
(D) $\frac{400\text{ m}}{43.03\text{ sec}} \cdot \frac{1000\text{ m}}{0.62\text{ mi}} \cdot \frac{3600\text{ sec}}{1\text{ hr}}$

Worked solution

$\text{convert meters to miles: multiply by } \frac{0.62\text{ mi}}{1000\text{ m}} \text{ so m cancels}$

$\text{convert per-second to per-hour: multiply by } \frac{3600\text{ sec}}{1\text{ hr}} \text{ so sec in the denominator cancels}$

$\text{resulting units: mi/hr}$

Answer: $\frac{400\text{ m}}{43.03\text{ sec}} \cdot \frac{0.62\text{ mi}}{1000\text{ m}} \cdot \frac{3600\text{ sec}}{1\text{ hr}}$

Correct dimensional analysis requires each conversion factor to cancel the unwanted unit; only choice 3 cancels both meters and seconds correctly, leaving miles per hour.

Question 24 · 2 pts

Which function has a domain of all real numbers and a range greater than or equal to three?

(A) $f(x) = -x + 3$
(B) $g(x) = x^2 + 3$ ✓
(C) $h(x) = 3^x$
(D) $m(x) = |x + 3|$

Worked solution

$g(x) = x^2+3 \rightarrow x^2 \ge 0 \text{ for all real } x$

$\text{so } g(x) \ge 3, \text{ range } = [3, \infty)$

Answer: $g(x) = x^2 + 3$

Since $x^2 \ge 0$ for every real number, adding 3 shifts the minimum value to exactly 3, giving domain all reals and range $[3,\infty)$; the other functions either have restricted domain/range or a minimum below 3.

Question 25 · 2 pts · constructed response

Solve $5(x - 2) \le 3x + 20$ algebraically.

Model answer: $x \le 15$

Worked solution

$5(x-2) \le 3x+20$

$5x - 10 \le 3x + 20$

$2x \le 30$

$x \le 15$

Answer: $x \le 15$

Distribute, collect variable terms on one side and constants on the other, then divide by a positive coefficient (no flip of inequality needed).

Question 26 · 2 pts · constructed response

Given $g(x) = x^3 + 2x^2 - x$, evaluate $g(-3)$.

Model answer: $g(-3) = -6$

Worked solution

$g(-3) = (-3)^3 + 2(-3)^2 - (-3)$

$= -27 + 2(9) + 3$

$= -27 + 18 + 3$

$= -6$

Answer: $g(-3) = -6$

Substitute $x = -3$ into each term of the function and simplify carefully, paying attention to signs.

Question 27 · 2 pts · constructed response

Given the relation $R = \{(-1,1), (0,3), (-2,-4), (x,5)\}$. State a value for $x$ that will make this relation a function. Explain why your answer makes this a function.

Model answer: $x = 1$ (any value other than $-1$, $0$, or $-2$ works)

Worked solution

$\text{existing x-values in } R: -1, 0, -2$

$\text{choose } x = 1, \text{ giving the pair } (1, 5)$

Answer: $x = 1$ (any value other than $-1$, $0$, or $-2$ works)

A relation is a function only if no x-value repeats with a different y-value. Since -1, 0, and -2 are already used as x-values, choosing any other value for x (such as 1) ensures every x-value maps to exactly one y-value, making R a function.

Question 28 · 2 pts · constructed response

A survey of 150 students was taken. It was determined that $\frac{2}{3}$ of the students play video games. Of the students that play video games, 85 also use social media. Of the students that do not play video games, 20% do not use social media. Complete the two-way frequency table. | | Play Video Games | Do Not Play Video Games | Total | |---|---|---|---| | Social Media | | | | | No Social Media | | | | | Total | | | |

Model answer: Social Media row: 85, 40, 125. No Social Media row: 15, 10, 25. Total row: 100, 50, 150.

Worked solution

$\text{Play video games} = \frac{2}{3}(150) = 100, \quad \text{Do not play} = 50$

$\text{Play \& Social Media} = 85 \Rightarrow \text{Play \& No Social Media} = 100-85 = 15$

$\text{Do not play \& No Social Media} = 0.20(50) = 10 \Rightarrow \text{Do not play \& Social Media} = 50-10 = 40$

$\text{Column/row totals: Social Media} = 85+40=125, \text{ No Social Media} = 15+10=25$

Answer: Social Media row: 85, 40, 125. No Social Media row: 15, 10, 25. Total row: 100, 50, 150.

Use the given fraction and percent to find the four interior cells of the table, then sum rows and columns to get the totals, which should reconcile to 150.

Question 29 · 2 pts · constructed response

Use the method of completing the square to determine the exact values of $x$ for the equation $x^2 + 10x - 30 = 0$.

Model answer: $x = -5 \pm \sqrt{55}$

Worked solution

$x^2 + 10x - 30 = 0$

$x^2 + 10x = 30$

$x^2 + 10x + 25 = 30 + 25$

$(x+5)^2 = 55$

$x + 5 = \pm\sqrt{55}$

$x = -5 \pm \sqrt{55}$

Answer: $x = -5 \pm \sqrt{55}$

Move the constant to the right side, add the square of half the linear coefficient to both sides to complete the square, then take the square root of both sides.

Question 30 · 2 pts · constructed response

Factor $20x^3 - 45x$ completely.

Model answer: $5x(2x-3)(2x+3)$

Worked solution

$20x^3 - 45x = 5x(4x^2 - 9)$

$4x^2 - 9 = (2x-3)(2x+3) \quad \text{(difference of squares)}$

$20x^3 - 45x = 5x(2x-3)(2x+3)$

Answer: $5x(2x-3)(2x+3)$

First factor out the greatest common factor, $5x$, then recognize the remaining binomial as a difference of squares and factor it completely.

Question 31 · 4 pts · constructed response

Graph the following system of equations on the set of axes below. $y = x^2 - 3x - 6$ $y = x - 1$ State the coordinates of all solutions.

Model answer: $(5, 4)$ and $(-1, -2)$

Worked solution

$x^2 - 3x - 6 = x - 1$

$x^2 - 4x - 5 = 0$

$(x-5)(x+1) = 0$

$x = 5 \text{ or } x = -1$

$y = x-1: \quad y=4 \text{ when } x=5; \quad y=-2 \text{ when } x=-1$

Answer: $(5, 4)$ and $(-1, -2)$

Set the two expressions for y equal to find the x-coordinates of intersection, factor the resulting quadratic, then substitute back into the linear equation to get the corresponding y-values; graphing on the grid should show the parabola and line crossing at these two points.

Question 32 · 4 pts · constructed response

The table below shows the amount of money a popular movie earned, in millions of dollars, during its first six weeks in theaters. | Week (x) | 1 | 2 | 3 | 4 | 5 | 6 | |---|---|---|---|---|---|---| | Dollars Earned, in Millions (y) | 185 | 150 | 90 | 50 | 25 | 5 | Write the linear regression equation for this data set, rounding all values to the nearest hundredth. State the correlation coefficient to the nearest hundredth. State what this correlation coefficient indicates about the linear fit of the data.

Model answer: $y = -37.57x + 215.67$; $r \approx -0.98$; this indicates a strong negative linear correlation, so the linear model fits the data well.

Worked solution

$\bar{x} = 3.5, \quad \bar{y} = \frac{505}{6} \approx 84.17$

$S_{xx} = \sum x^2 - n\bar{x}^2 = 91 - 6(3.5)^2 = 17.5$

$S_{xy} = \sum xy - n\bar{x}\bar{y} = 1110 - 6(3.5)(84.1\overline{6}) = -657.5$

$b = \frac{S_{xy}}{S_{xx}} = \frac{-657.5}{17.5} \approx -37.57$

$a = \bar{y} - b\bar{x} \approx 84.17 - (-37.57)(3.5) \approx 215.67$

$S_{yy} = \sum y^2 - n\bar{y}^2 = 67975 - 6(84.1\overline{6})^2 \approx 25470.83$

$r = \frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}} = \frac{-657.5}{\sqrt{17.5 \times 25470.83}} \approx -0.98$

Answer: $y = -37.57x + 215.67$; $r \approx -0.98$; this indicates a strong negative linear correlation, so the linear model fits the data well.

Using least-squares regression on the six data points gives a strongly negative slope, and the correlation coefficient close to -1 shows the earnings decreased consistently and nearly linearly week to week, meaning the linear model is a very good fit.

Question 33 · 4 pts · constructed response

Use the quadratic formula to solve the equation $3x^2 - 10x + 5 = 0$. Express the answer in simplest radical form.

Model answer: $x = \frac{5 \pm \sqrt{10}}{3}$

Worked solution

$a=3, b=-10, c=5$

$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} = \frac{10 \pm \sqrt{100-60}}{6}$

$= \frac{10 \pm \sqrt{40}}{6} = \frac{10 \pm 2\sqrt{10}}{6}$

$= \frac{5 \pm \sqrt{10}}{3}$

Answer: $x = \frac{5 \pm \sqrt{10}}{3}$

Substitute the coefficients into the quadratic formula, simplify the radical $\sqrt{40} = 2\sqrt{10}$, and reduce the fraction to simplest radical form.

Question 34 · 4 pts · constructed response

Graph the system of inequalities on the set of axes below. $3y + 2x \le 15$ $y - x > 1$ State the coordinates of a point in the solution to this system. Justify your answer.

Model answer: $(0, 2)$ is a point in the solution.

Worked solution

$\text{Try } x=0, y=2:$

$3(2) + 2(0) = 6 \le 15 \quad \checkmark$

$2 - 0 = 2 > 1 \quad \checkmark$

Answer: $(0, 2)$ is a point in the solution.

Any point lying in the overlap of both shaded regions works; substituting (0,2) into both inequalities shows both are satisfied, confirming it is a valid solution, and graphing both boundary lines with correct shading (solid line and shading below/on for the first, dashed line and shading above for the second) shows the overlapping region.

Question 35 · 6 pts · constructed response

Courtney went to a coffee shop to purchase lattes and donuts for her friends. One day she spent a total of $15.50 on four lattes and two donuts. The next day she spent a total of $18.10 on three lattes and five donuts. All prices included tax. If $x$ represents the cost of one latte and $y$ represents the cost of one donut, write a system of equations that can be used to model this situation. Courtney thinks that one latte costs $2.75 and one donut costs $2.25. Is Courtney correct? Justify your answer. Use your equations to determine algebraically the exact cost of one latte and the exact cost of one donut.

Model answer: System: $4x+2y=15.50$ and $3x+5y=18.10$. Courtney is incorrect. One latte costs $2.95 and one donut costs $1.85.

Worked solution

$4x + 2y = 15.50$

$3x + 5y = 18.10$

$\text{Check Courtney's values } (x=2.75, y=2.25): \quad 4(2.75)+2(2.25)=11+4.50=15.50 \quad \checkmark$

$3(2.75)+5(2.25) = 8.25+11.25 = 19.50 \ne 18.10 \quad \text{so Courtney is incorrect}$

$\text{Multiply first eqn by 5, second by 2: } 20x+10y=77.50; \quad 6x+10y=36.20$

$\text{Subtract: } 14x = 41.30 \Rightarrow x = 2.95$

$4(2.95)+2y=15.50 \Rightarrow 11.80+2y=15.50 \Rightarrow y = 1.85$

Answer: System: $4x+2y=15.50$ and $3x+5y=18.10$. Courtney is incorrect. One latte costs $2.95 and one donut costs $1.85.

Translate the two purchase scenarios into a linear system, test Courtney's guessed values in both equations (they satisfy only one, so she is wrong), then solve the system by elimination to find the exact latte and donut prices.

June 2024 Algebra I Regents — Answer Key & Worked Solutions

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