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June 2023 Algebra I Regents — Answer Key & Worked Solutions

Every question from the June 2023 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 2023 Algebra I New York State Regents examination, published by NYSED. Worked solutions are written by RegentsLab.

Question 1 · 2 pts

The expression $9m^2 - 100$ is equivalent to

Worked solution
$9m^2-100=(3m)^2-(10)^2$
$=(3m-10)(3m+10)$ difference of squares
Answer: $(3m-10)(3m+10)$

This is a difference of squares, $a^2-b^2=(a-b)(a+b)$, with $a=3m$ and $b=10$.

Question 2 · 2 pts

Which expression represents an irrational number?

Worked solution
$\sqrt{36}=6$
$6+\sqrt{7}$ since 7 is not a perfect square, \sqrt{7} is irrational, and a rational number plus an irrational number is irrational
Answer: $\sqrt{36}+\sqrt{7}$

All choices except this one simplify to sums of two perfect square roots, which are rational. Here $\sqrt{7}$ is irrational, so the sum is irrational.

Question 3 · 2 pts

Which linear equation represents a line that passes through the point $(-3,-8)$?

Worked solution
$y=2(-3)-2=-6-2=-8$ substitute x=-3 and check that y=-8
Answer: $y=2x-2$

Substitute $x=-3$ into each equation; only $y=2x-2$ gives $y=-8$, matching the given point.

Question 4 · 2 pts

The expression $(5x^2-x+4)-3(x^2-x-2)$ is equivalent to

Worked solution
$(5x^2-x+4)-3x^2+3x+6$
$=(5x^2-3x^2)+(-x+3x)+(4+6)$
$=2x^2+2x+10$
Answer: $2x^2+2x+10$

Distribute the $-3$ across the second parenthesis, then combine like terms.

Question 5 · 2 pts

The $24^{\text{th}}$ term of the sequence $-5,-11,-17,-23,\ldots$ is

Worked solution
$d=-11-(-5)=-6$ common difference
$a_{24}=-5+(24-1)(-6)$
$a_{24}=-5+23(-6)=-5-138=-143$
Answer: $-143$

This is an arithmetic sequence with first term $-5$ and common difference $-6$; use $a_n=a_1+(n-1)d$.

Question 6 · 2 pts

When completing the square for $x^2-18x+77=0$, which equation is a correct step in this process?

Worked solution
$x^2-18x=-77$
$x^2-18x+81=-77+81$ add \left(\frac{-18}{2}\right)^2=81 to both sides
$(x-9)^2=4$
Answer: $(x-9)^2=4$

Move the constant, then add the square of half the linear coefficient, $81$, to both sides to complete the square.

Question 7 · 2 pts

Which function will have the greatest value when $x > 1$?

Worked solution
$g(2)=2(5)^2=50,\ h(2)=2(2)^2+5=13,\ k(2)=2(2)^3+5=21,\ f(2)=2(2)+5=9$ compare values just past x=1
Answer: $g(x)=2(5)^x$

An exponential function with base greater than 1 eventually grows faster than any linear, quadratic, or cubic function, and it already exceeds them for $x>1$.

Question 8 · 2 pts

Mike uses the equation $b=1300(2.65)^x$ to determine the growth of bacteria in a laboratory setting. The exponent represents

Worked solution
$b=1300(2.65)^x$ 1300 is the initial amount, 2.65 is the growth factor per period, and x (the exponent) is the number of periods
Answer: the number of time periods

In an exponential growth model $b=a(r)^x$, the exponent $x$ counts the number of time periods that have elapsed.

Question 9 · 2 pts

A company ships an average of 30,000 items each week. The approximate number of items shipped each minute is calculated using the conversion

Worked solution
$\frac{30{,}000\text{ items}}{1\text{ week}}\cdot\frac{1\text{ week}}{7\text{ days}}\cdot\frac{1\text{ day}}{24\text{ hrs}}\cdot\frac{1\text{ hr}}{60\text{ min}}$ Each factor cancels the previous unit: week, then day, then hr, leaving items per minute.
$= \frac{30{,}000}{7\cdot24\cdot60}\text{ items/min} \approx 2.98\text{ items/min}$
Answer: Choice (2)

Only choice (2) is set up so the units cancel correctly (week, then day, then hour), leaving items per minute; it gives 30,000 / (7 x 24 x 60) which is about 2.98 items per minute.

Question 10 · 2 pts

A function is graphed below. A possible equation for this function is

alg1 question 10 figure
Worked solution
$\text{The graph shows a curve touching the x-axis (a double root) and crossing it once elsewhere, consistent with a cubic with a repeated root at }x=2\text{ and a single root at }x=-3$
Answer: $f(x)=(x-2)^2(x+3)$

The graph shows end behavior and root pattern of a cubic function with a repeated (tangent) zero at $x=2$ and a simple zero at $x=-3$, matching $(x-2)^2(x+3)$.

Question 11 · 2 pts

If $g(x)=-x^2-x+5$, then $g(-4)$ is equal to

Worked solution
$g(-4)=-(-4)^2-(-4)+5$
$=-16+4+5=-7$
Answer: $-7$

Substitute $-4$ for $x$ and simplify using order of operations.

Question 12 · 2 pts

A movie theater's popcorn box is a rectangular prism with a base that measures 6 inches by 4 inches and has a height of 8 inches. To create a larger box, both the length and the width will be increased by $x$ inches. The height will remain the same. Which function represents the volume, $V(x)$, of the larger box?

Worked solution
$V=\text{length}\times\text{width}\times\text{height}$
$V(x)=(6+x)(4+x)(8)$ height stays 8 since only length and width increase
Answer: $V(x)=(6+x)(4+x)(8)$

Volume of a rectangular prism is length times width times height; only the length and width get the $+x$, and the height stays 8.

Question 13 · 2 pts

The expression $300(4)^{x+3}$ is equivalent to

Worked solution
$4^{x+3}=4^x\cdot 4^3$ product rule of exponents
$300(4)^{x+3}=300(4)^x(4)^3$
Answer: $300(4)^x(4)^3$

By the product rule for exponents, $a^{m+n}=a^m\cdot a^n$, so $4^{x+3}=4^x\cdot4^3$.

Question 14 · 2 pts

Ashley only has 7 quarters and some dimes in her purse. She needs at least \$3.00 to pay for lunch. Which inequality could be used to determine the number of dimes, $d$, she needs in her purse to be able to pay for lunch?

Worked solution
$7\text{ quarters}=7(0.25)=1.75$ value of quarters
$1.75+0.10d\ge 3.00$ each dime is worth 0.10, and she needs at least (\ge) 3.00
Answer: $1.75+0.10d\ge 3.00$

Seven quarters equal \$1.75, each dime adds \$0.10, and "at least" means the total must be greater than or equal to \$3.00.

Question 15 · 2 pts

The formula for the area of a trapezoid is $A=\frac{1}{2}(b_1+b_2)h$. The height, $h$, of the trapezoid may be expressed as

Worked solution
$A=\frac{1}{2}(b_1+b_2)h$
$2A=(b_1+b_2)h$
$h=\frac{2A}{b_1+b_2}$
Answer: $\dfrac{2A}{b_1+b_2}$

Multiply both sides by 2 to clear the fraction, then divide by $(b_1+b_2)$ to isolate $h$.

Question 16 · 2 pts

The function $f(x)=|x|$ is multiplied by $k$ to create the new function $g(x)=k|x|$. Which statement is true about the graphs of $f(x)$ and $g(x)$ if $k=\frac{1}{2}$?

Worked solution
$g(x)=\frac{1}{2}|x|$ multiplying by a factor between 0 and 1 vertically compresses the graph, making it wider/flatter
Answer: $g(x)$ is wider than $f(x)$.

Since $0<k<1$, the graph is vertically compressed compared to $f(x)=|x|$, which makes it appear wider.

Question 17 · 2 pts

Some adults were surveyed to find out if they would prefer to buy a sports utility vehicle (SUV) or a sports car. The results of the survey are summarized in the table below. | | SUV | Sports Car | Totals | |---|---|---|---| | Male | 21 | 38 | 59 | | Female | 135 | 46 | 181 | | Totals | 156 | 84 | 240 | Of the number of adults that preferred sports cars, approximately what percent were males?

alg1 question 17 figure
Worked solution
$\frac{38}{84}\times 100\approx 45.2\%$ males who preferred sports car divided by total who preferred sports car
Answer: $45.2$

Divide the number of males who preferred sports cars (38) by the total number who preferred sports cars (84), then convert to a percent.

Question 18 · 2 pts

The solution to $2x^2=72$ is

Worked solution
$x^2=36$
$x=\pm\sqrt{36}=\pm6$
Answer: $\{\pm6\}$

Divide both sides by 2 to isolate $x^2$, then take the square root of both sides, keeping both the positive and negative roots.

Question 19 · 2 pts

Three quadratic functions are given below. I. $f(x)=(x+2)^2+5$ II. | $x$ | $-4$ | $-3$ | $-2$ | $-1$ | $0$ | $1$ | |---|---|---|---|---|---|---| | $g(x)$ | $-3$ | $2$ | $5$ | $5$ | $2$ | $-3$ | III. (graph of $h(x)$, an upside-down parabola) Which of these functions have the same vertex?

alg1 question 19 figure
Worked solution
$f(x)=(x+2)^2+5\ \text{has vertex }(-2,5)$
$\text{Table II is symmetric about }x=-1.5\text{ (between the two equal values of 5 at }x=-2,-1\text{), giving a different vertex than }(-2,5)$
$\text{The graph of III shows a vertex at }(-2,5)\text{, matching function I}$
Answer: I and III, only

Function I has vertex $(-2,5)$ by inspection of vertex form. The table's axis of symmetry falls at $x=-1.5$, giving a different vertex, while the graphed parabola III shares the vertex $(-2,5)$ with function I.

Question 20 · 2 pts

The domain of the function $f(x)=x^2+x-12$ is

Worked solution
$f(x)=x^2+x-12\ \text{is a polynomial function}$ polynomials are defined for all real numbers
Answer: $(-\infty,\infty)$

Polynomial functions have no restrictions (no denominators, radicals, or logs), so the domain is all real numbers.

Question 21 · 2 pts

A father makes a deal with his son regarding his weekly allowance. The first year, he agrees to pay his son a weekly allowance of $10. Every subsequent year, the allowance is recalculated by doubling the previous year's weekly allowance and then subtracting 8. Which recursive formula could be used to calculate the son's weekly allowance in future years?

Worked solution
$a_1 = 10$ Starting allowance given as $10 for the first year
$a_{n+1} = 2a_n - 8$ Each new term doubles the previous term, then subtracts 8
Answer: $a_1 = 10$, $a_{n+1} = 2a_n - 8$

A recursive formula must define the first term and a rule relating each term to the one before it; doubling then subtracting 8 translates directly to $a_{n+1}=2a_n-8$.

Question 22 · 2 pts

What is the solution to the inequality below? $4 - \frac{2}{5}x \ge \frac{1}{3}x + 15$

Worked solution
$15\left(4-\frac{2}{5}x\right) \ge 15\left(\frac{1}{3}x+15\right)$ Multiply both sides by 15 (LCD) to clear fractions
$60 - 6x \ge 5x + 225$
$-165 \ge 11x$
$x \le -15$
Answer: $x \le -15$

Clearing the fractions by multiplying by the least common denominator (15) and isolating x gives the inequality x <= -15.

Question 23 · 2 pts

Which statement is correct about the polynomial $3x^2 + 5x - 2$?

Worked solution
$3x^2+5x-2$ Highest exponent is 2, so degree is 2; coefficient of the highest-degree term is 3
Answer: It is a second-degree polynomial with a leading coefficient of 3.

The polynomial's highest power of x is 2 (second-degree), and the coefficient attached to that x^2 term (the leading coefficient) is 3.

Question 24 · 2 pts

A store manager is trying to determine if they should continue to sell a particular brand of nails. To model their profit, they use the function $p(n)$, where $n$ is the number of boxes of these nails sold in a day. A reasonable domain for this function would be

Worked solution
$n \ge 0,\ n \in \mathbb{Z}$ You cannot sell a negative or fractional number of boxes
Answer: nonnegative integers

Since n counts a whole number of boxes sold in a day, it cannot be negative or fractional, so the reasonable domain is the nonnegative integers.

Question 25 · 2 pts · constructed response

Solve the equation algebraically for x: $-2.4(x + 1.4) = 6.8x - 22.68$

Model answer: x = 2.1
Worked solution
$-2.4x - 3.36 = 6.8x - 22.68$ Distribute -2.4 on the left side
$-3.36 + 22.68 = 6.8x + 2.4x$ Collect x terms on one side, constants on the other
$19.32 = 9.2x$
$x = \frac{19.32}{9.2} = 2.1$
Answer: x = 2.1

Distributing and combining like terms isolates x, giving x = 2.1.

Question 26 · 2 pts · constructed response

The function f(x) is graphed on the set of axes below. State the zeros of f(x). Explain your reasoning.

alg1 question 26 figure
Model answer: x = -2 and x = 2
Worked solution
$f(-2)=0$ The graph crosses the x-axis at x = -2, changing from negative to positive as x increases through -2
$f(2)=0$ The graph touches the x-axis at x = 2 and turns back upward (tangent to the axis), indicating a zero of even multiplicity there
Answer: x = -2 and x = 2

The zeros of a function are the x-values where the graph intersects the x-axis; here the graph crosses through the axis at x = -2 and is tangent to the axis at x = 2.

Question 27 · 2 pts · constructed response

Breanna creates the pattern of blocks below in her art class (figures I, II, III, and IV, each a plus-sign shaped arrangement of unit squares that grows from one figure to the next). A friend tells her that the number of blocks in the pattern is increasing exponentially. Is her friend correct? Explain your reasoning.

alg1 question 27 figure
Model answer: No, the friend is not correct; the pattern grows arithmetically (by a constant amount each time), not exponentially.
Worked solution
$5,\ 7,\ 9,\ 11,\ldots$ Counting the blocks in each successive figure shows the total increases by the same constant amount (2) each time
$\text{constant difference} \Rightarrow \text{linear (arithmetic) growth, not exponential (multiplicative) growth}$
Answer: No, the friend is not correct; the pattern grows arithmetically (by a constant amount each time), not exponentially.

An exponential pattern would multiply the previous total by a constant ratio each time, but here each figure simply adds the same fixed number of blocks to the previous figure, which is the signature of arithmetic (linear), not exponential, growth.

Question 28 · 2 pts · constructed response

The data set 20, 36, 52, 56, 24, 16, 40, 4, 28 represents the number of books purchased by nine book club members in a year. Construct a box plot for these data on the number line below.

alg1 question 28 figure
Model answer: Five-number summary: minimum = 4, Q1 = 18, median = 28, Q3 = 46, maximum = 56.
Worked solution
$4,16,20,24,28,36,40,52,56$ Order the data from least to greatest
$\text{median} = 28$ The middle (5th of 9) value
$Q_1 = \frac{16+20}{2} = 18$ Median of the lower four values (4,16,20,24)
$Q_3 = \frac{40+52}{2} = 46$ Median of the upper four values (36,40,52,56)
$\min=4,\ \max=56$
Answer: Five-number summary: minimum = 4, Q1 = 18, median = 28, Q3 = 46, maximum = 56.

Ordering the data and finding the median splits it into two halves; the medians of those halves give Q1 and Q3, and together with the min and max these form the five-number summary used to draw the box plot (box from 18 to 46 with a line at 28, whiskers extending to 4 and 56).

Question 29 · 2 pts · constructed response

Given: $A = x + 5$, $B = x^2 - 18$. Express $A^2 + B$ in standard form.

Model answer: $2x^2 + 10x + 7$
Worked solution
$A^2 = (x+5)^2 = x^2+10x+25$
$A^2+B = (x^2+10x+25)+(x^2-18)$
$= 2x^2+10x+7$
Answer: $2x^2 + 10x + 7$

Squaring A and adding B, then combining like terms, gives the polynomial in standard form.

Question 30 · 2 pts · constructed response

The two relations shown below are not functions. Relation I: a graph with a segment from (1,10) [closed] to (2,10) [open circle], a segment from (2,20) [closed] to (4,20) [closed], and a curve starting at (4,30) [closed] that increases toward the upper right. Relation II: $\{(-5,-2),(-4,0),(-2,1),(-1,3),(-4,4)\}$ Explain how you could change each relation so that they each become a function.

alg1 question 30 figure
Model answer: Relation I: remove (or make open) the point (4, 20) so that x = 4 maps to only one y-value, (4, 30). Relation II: remove or change one of the two ordered pairs with x = -4 (for example, delete (-4, 4), or change it to (-3, 4)) so that x = -4 maps to only one y-value.
Worked solution
$x=4 \to y=20 \text{ and } y=30$ Relation I fails the vertical line test at x = 4 because it has two different outputs there
$x=-4 \to y=0 \text{ and } y=4$ Relation II has two ordered pairs sharing x = -4, so it is not a function
Answer: Relation I: remove (or make open) the point (4, 20) so that x = 4 maps to only one y-value, (4, 30). Relation II: remove or change one of the two ordered pairs with x = -4 (for example, delete (-4, 4), or change it to (-3, 4)) so that x = -4 maps to only one y-value.

A relation is a function only if every input has exactly one output; both relations have one x-value paired with two different y-values, so removing or reassigning one of the conflicting points for each x-value would make each relation a function.

Question 31 · 2 pts · constructed response

Factor $2x^2 + 16x - 18$ completely.

Model answer: $2(x+9)(x-1)$
Worked solution
$2x^2+16x-18 = 2(x^2+8x-9)$ Factor out the GCF of 2
$x^2+8x-9 = (x+9)(x-1)$ Find two numbers that multiply to -9 and add to 8: 9 and -1
$2x^2+16x-18 = 2(x+9)(x-1)$
Answer: $2(x+9)(x-1)$

Factoring out the GCF first simplifies the trinomial, which then factors into two binomials whose product gives back the original expression.

Question 32 · 2 pts · constructed response

Solve $3d^2 - 8d + 3 = 0$ algebraically for all values of d, rounding to the nearest tenth.

Model answer: d ≈ 2.2 or d ≈ 0.5
Worked solution
$d = \frac{8 \pm \sqrt{(-8)^2-4(3)(3)}}{2(3)}$ Apply the quadratic formula with a=3, b=-8, c=3
$d = \frac{8 \pm \sqrt{64-36}}{6} = \frac{8\pm\sqrt{28}}{6}$
$d = \frac{8+5.2915}{6} \approx 2.2,\quad d = \frac{8-5.2915}{6} \approx 0.5$
Answer: d ≈ 2.2 or d ≈ 0.5

Since the trinomial does not factor over the integers, the quadratic formula is used, giving two irrational roots that round to 2.2 and 0.5.

Question 33 · 4 pts · constructed response

Graph $f(x) = |x| + 1$ and $g(x) = -x^2 + 6x + 1$ on the set of axes below. Based on your graph, determine all values of x for which $f(x) = g(x)$.

alg1 question 33 figure
Model answer: x = 0 and x = 5
Worked solution
$|x|+1 = -x^2+6x+1 \Rightarrow |x| = -x^2+6x$ Set the two functions equal and simplify
$\text{For } x\ge 0:\ x = -x^2+6x \Rightarrow x^2-5x=0 \Rightarrow x(x-5)=0 \Rightarrow x=0,5$
$\text{For } x<0:\ -x=-x^2+6x \Rightarrow x^2-7x=0 \Rightarrow x(x-7)=0 \Rightarrow x=0,7 \text{ (rejected, not} <0)$
$\text{Check: } f(0)=1=g(0),\ f(5)=6=g(5)$ Both solutions verify
Answer: x = 0 and x = 5

Graphing both functions shows they intersect at two points; splitting the absolute value into cases and solving each piecewise equation confirms the intersection points occur at x = 0 and x = 5.

Question 34 · 4 pts · constructed response

Jean recorded temperatures over a 24-hour period one day in August in Syracuse, NY. Her results are shown in the table below. Time (hour): 0, 3, 6, 9, 12, 15, 18, 21, 24 Temperature (°F): 80, 75, 70, 78, 92, 89, 85, 80, 74 Her data are modeled on the graph (a line graph connecting these points, with a break in the y-axis below 70). State the entire interval over which the temperature is increasing. State the three-hour interval that has the greatest rate of change in temperature. State the average rate of change from hour 12 to hour 24. Explain what this means in the context of the problem.

alg1 question 34 figure
Model answer: Increasing on 6 ≤ t ≤ 12 (i.e., from hour 6 to hour 12); greatest rate of change occurs from hour 9 to hour 12 (about 4.67°F per hour); average rate of change from hour 12 to hour 24 is -1.5°F per hour, meaning the temperature decreased on average by 1.5°F every hour during that period.
Worked solution
$70\to78\to92,\ t=6\ \text{to}\ 12$ Temperature rises continuously from hour 6 to hour 12; before hour 6 and after hour 12 it is decreasing
$\frac{78-70}{9-6}=2.67,\quad \frac{92-78}{12-9}=4.67$ Compare 3-hour rates of change; the interval 9 to 12 has the largest increase
$\frac{74-92}{24-12} = \frac{-18}{12} = -1.5$ Average rate of change from hour 12 to hour 24
Answer: Increasing on 6 ≤ t ≤ 12 (i.e., from hour 6 to hour 12); greatest rate of change occurs from hour 9 to hour 12 (about 4.67°F per hour); average rate of change from hour 12 to hour 24 is -1.5°F per hour, meaning the temperature decreased on average by 1.5°F every hour during that period.

Reading the table, the temperature rises without interruption only between hour 6 and hour 12; comparing consecutive 3-hour changes shows the 9-to-12 interval has the steepest increase; and the overall slope from hour 12 to 24 shows the temperature fell by an average of 1.5°F each hour over that span.

Question 35 · 2 pts · constructed response

Solve the following system of inequalities graphically on the set of axes below. $2x + 3y \ge -6$ $x < 3y + 6$ Label the solution set S. Is the point (4, -2) in the solution set? Explain your answer.

alg1 question 35 figure
Model answer: No, (4, -2) is not in the solution set because it fails the second inequality.
Worked solution
$2(4)+3(-2) = 8-6 = 2 \ge -6$ Check the first inequality: true
$4 < 3(-2)+6 = -6+6 = 0$ Check the second inequality: 4 < 0 is false
$\text{Since one inequality fails, } (4,-2) \notin S$
Answer: No, (4, -2) is not in the solution set because it fails the second inequality.

A point is in the solution set of a system only if it satisfies every inequality; (4,-2) satisfies the first inequality but not the second, so it is not part of the shared shaded region S.

Question 36 · 4 pts · constructed response

Suzanna collected information about a group of ponies and horses. She made a table showing the height, measured in hands (hh), and the weight, measured in pounds (lbs), of each pony and horse. Height (hh) x: 11, 12, 13, 14, 15, 16, 17 Weight (lbs) y: 264, 638, 700, 850, 1000, 1230, 1495 Write the linear regression equation for this set of data. Round all values to the nearest hundredth. State the correlation coefficient for the linear regression. Round your answer to the nearest hundredth. Explain what the correlation coefficient indicates about the linear fit of the data in the context of the problem.

Model answer: y = 184.89x - 1706.07; r ≈ 0.99, indicating a very strong positive linear relationship, so the regression line is an excellent fit for the data (as a pony/horse's height increases, its weight increases strongly and consistently).
Worked solution
$\bar{x}=14,\ \bar{y}\approx 882.43$ Compute the means of x and y
$S_{xx}=28,\ S_{xy}=5177,\ S_{yy}\approx 986403.71$ Sums of squares/cross-products from the data
$b=\frac{S_{xy}}{S_{xx}} = \frac{5177}{28} \approx 184.89$ Slope of the regression line
$a=\bar{y}-b\bar{x} \approx 882.43 - 184.89(14) \approx -1706.07$ y-intercept of the regression line
$y \approx 184.89x - 1706.07$
$r=\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}} \approx \frac{5177}{\sqrt{28(986403.71)}} \approx 0.99$
Answer: y = 184.89x - 1706.07; r ≈ 0.99, indicating a very strong positive linear relationship, so the regression line is an excellent fit for the data (as a pony/horse's height increases, its weight increases strongly and consistently).

Using calculator/regression formulas on the table data produces a regression line with slope about 184.82 and intercept about -1706.07, and a correlation coefficient near 0.99, showing the linear model fits the height-weight data extremely well.

Question 37 · 6 pts · constructed response

Dana went shopping for plants to put in her garden. She bought three roses and two daisies for $31.88. Later that day, she went back and bought two roses and one daisy for $18.92. If r represents the cost of one rose and d represents the cost of one daisy, write a system of equations that models this situation. Use your system of equations to algebraically determine both the cost of one rose and the cost of one daisy.

Model answer: System: 3r + 2d = 31.88 and 2r + d = 18.92. Cost of one rose = $5.96; cost of one daisy = $7.00.
Worked solution
$3r+2d=31.88$ Three roses and two daisies cost $31.88
$2r+d=18.92$ Two roses and one daisy cost $18.92
$d=18.92-2r$ Solve the second equation for d
$3r+2(18.92-2r)=31.88 \Rightarrow 3r+37.84-4r=31.88 \Rightarrow -r=-5.96 \Rightarrow r=5.96$ Substitute into the first equation and solve for r
$d=18.92-2(5.96)=18.92-11.92=7.00$
Answer: System: 3r + 2d = 31.88 and 2r + d = 18.92. Cost of one rose = $5.96; cost of one daisy = $7.00.

Setting up two linear equations from the purchase totals and solving by substitution/elimination gives the individual prices: a rose costs $5.96 and a daisy costs $7.00.

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