Page 1
Question ID: d1b66ae6
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Hard
Question
If satisfies the system of equations above, what is the value of y ?
Rationale
The correct answer is . One method for solving the system of equations for y is to add corresponding sides of the two equations. Adding the
left-hand sides gives , or 4y. Adding the right-hand sides yields . It follows that . Finally, dividing
both sides of by 4 yields or . Note that 3/2 and 1.5 are examples of ways to enter a correct answer.
Page 2
Question ID: cb8f449f
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Medium
Question
The system of equations above has solution (x, y). What is the value of x ?
Answer
A. 3
B.
C. 4
D. 6
Correct Answer: D
Rationale
Choice D is correct. Adding the corresponding sides of the two equations eliminates y and yields , as shown.
If (x, y) is a solution to the system, then (x, y) satisfies both equations in the system and any equation derived from them. Therefore, .
Choices A, B, and C are incorrect and may be the result of errors when solving the system.
Page 3
Question ID: ff501705
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Hard
Question
In the given system of equations, is a constant. If the system has no solution, what is the value of ?
Correct Answer: 6
Rationale
The correct answer is . A system of two linear equations in two variables, and , has no solution if the lines represented by the equations in
the xy-plane are parallel and distinct. Lines represented by equations in standard form, and , are parallel if the
coefficients for and in one equation are proportional to the corresponding coefficients in the other equation, meaning ;îš•and the lines
are distinct if the constants are not proportional, meaning is not equal toîš• or . The first equation in the given system is
. Multiplying each side of this equation by yields . Adding to each side of this equation yields ,
or . The second equation in the given system is . Multiplying each side of this equation by yields
. Subtracting from each side of this equation yields . Subtracting from each side of this equation yields
. Therefore, the two equations in the given system, written in standard form, are
îš•
and . As previously
stated, if this system has no solution, the lines represented by the equations in the xy-plane are parallel and distinct, meaning the proportion
îš•
, or
îš•
, is true and the proportion is not true. The proportion is not true. Multiplying each side of the true
proportion,
,îš• by yields . Therefore, if the system has no solution, then the value of is .
îš•
Page 4
Question ID: b86123af
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Easy
Question
Hiro and Sofia purchased shirts and pants from a store. The price of each shirt purchased was the same and the price of each pair of pants
purchased was the same. Hiro purchased 4 shirts and 2 pairs of pants for $86, and Sofia purchased 3 shirts and 5 pairs of pants for $166. Which
of the following systems of linear equations represents the situation, if x represents the price, in dollars, of each shirt and y represents the price,
in dollars, of each pair of pants?
Answer
A.
B.
C.
D.
Correct Answer: A
Rationale
Choice A is correct. Hiro purchased 4 shirtsîš•and each shirt cost xîš•dollars, so heîš•spent aîš•total ofîš•4xîš•dollars on shirts. Likewise, Hiro purchased 2
pairs of pants, and each pair of pants cost yîš•dollars, so heîš•spent a total of 2yîš•dollars on pants. Therefore, the total amount that Hiro spentîš•was 4x
+ 2y. Since Hiro spent $86 in total, this can be modeled by the equation 4x + 2y = 86. Using the same reasoning, Sofia bought 3 shirts at xîš•dollars
each and 5 pairs of pants at yîš•dollars each, so she spent a total ofîš•3x + 5yîš•dollars on shirts and pants. Since Sofia spent $166 in total, this can be
modeled by the equation 3x + 5y = 166.
Choice B is incorrect and may be the result of switching the number ofîš•shirts Sofia purchased with the number of pairs of pants Hiro purchased.
Choice C is incorrect and may be the result of switchingîš•the total price each person paid. Choice D is incorrect and may be the result of
switchingîš•the total price each person paid as well as switching the number of shirts Sofia purchased with the number of pairs of pants Hiro
purchased.
Page 5
Question ID: 608eeb6e
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Easy
Question
The solution to the given system of equations is . What is the value of ?
Answer
A.
B.
C.
D.
Correct Answer: C
Rationale
Choice C is correct. Adding the second equation of the given system to the first equation yields
îš•
, which is
equivalent to
îš•
. So the value of
îš•
is .
Choice A is incorrect and may result from conceptual or calculation errors.
Choice B is incorrect. This is the value of .
Choice D is incorrect and may result from conceptual or calculation errors.
Page 6
Question ID: b0fc3166
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Easy
Question
The graph of a system of linear equations is shown. What is the solution
îš•
to the system?
Answer
A.
B.
C.
D.
Correct Answer: C
Rationale
Choice C is correct. The solution to this system of linear equations is represented by the point that lies on both lines shown, or the point of
intersection of the two lines. According to the graph, the point of intersection occurs when and , or at the point . Therefore, the
solution to the system is .
Choice A is incorrect and may result from conceptual or calculation errors.
Choice B is incorrect and may result from conceptual or calculation errors.
Choice D is incorrect and may result from conceptual or calculation errors.
Page 7
Question ID: 0dd6227f
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Medium
Question
At how many points do the graphs of the equations and intersect in the xy-plane?
Answer
A.
B.
C.
D.
Correct Answer: B
Rationale
Choice B is correct. Each given equation is written in slope-intercept form, , where is the slope and
îš•
is the y-intercept of the
graph of the equation in the xy-plane. The graphs of two lines that have different slopes will intersect at exactly one point. The graph of the first
equation is a line with slope . The graph of the second equation is a line with slope . Since the graphs are lines with different slopes, they will
intersect at exactly one point.
Choice A is incorrect because two graphs of linear equations have intersection points only if they are parallel and therefore have the same
slope.
Choice C is incorrect because two graphs of linear equations in the xy-plane can have only , , or infinitely many points of intersection.
Choice D is incorrect because two graphs of linear equations in the xy-plane can have only , , or infinitely many points of intersection.
Page 8
Question ID: 7efe5495
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Medium
Question
The solution to the given system of equations is . What is the value of ?
Answer
A.
B.
C.
D.
Correct Answer: C
Rationale
Choice C is correct. It's given by the first equation in the system that . Substituting for in the equation yields
, or
îš•
.
îš•
Choice A is incorrect and may result from conceptual or calculation errors.
Choice B is incorrect and may result from conceptual or calculation errors.
Choice D is incorrect and may result from conceptual or calculation errors.
Page 9
Question ID: 71189542
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Medium
Question
A group of 202îš•people went on an overnight camping trip, taking 60îš•tents with them. Some of the tents held 2îš•people each, and the rest held
4îš•people each. Assuming all the tents were filled to capacity and every person got to sleep in a tent, exactly how many of the tents were 2-person
tents?
Answer
A. 30
B. 20
C. 19
D. 18
Correct Answer: C
Rationale
Choice C is correct. Let x represent the number of 2-person tents and let y represent the number of 4-person tents. It is given that the total
number of tents was 60 and the total number of people in the group was 202. This situation can be expressed as a system of two equations,
and . The first equation can be rewritten as . Substituting for y in the equation
yields . Distributing and combining like terms gives . Subtracting 240 from
both sides of and then dividing both sides by gives . Therefore, the number of 2-person tents is 19.
Alternate approach: If each of the 60 tents held 4 people, the total number of people that could be accommodated in tents would be 240.
However, the actual number of people who slept in tents was 202. The difference of 38 accounts for the 2-person tents. Since each of these
tents holds 2 people fewer than a 4-person tent, gives the number of 2-person tents.
Choice A is incorrect. This choice may result from assuming exactly half of the tents hold 2 people. If that were true, then the total number of
people who slept in tents would be ; however, the total number of people who slept in tents was 202, not 180. Choice B
is incorrect. If 20 tents were 2-person tents, then the remaining 40 tents would be 4-person tents. Since all the tents were filled to capacity, the
total number of people who slept in tents would be ; however, the total number of people who slept in
tents was 202, not 200. Choice D is incorrect. If 18 tents were 2-person tents, then the remaining 42 tents would be 4-person tents. Since all the
tents were filled to capacity, the total number of people who slept in tents would be ; however, the total
number of people who slept in tents was 202, not 204.
Page 10
Question ID: dba8d38a
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Easy
Question
A petting zoo sells two types of tickets. The standard ticket, for admission only, costs $5. The premium ticket, which includes admission and
food to give to the animals, costs $12. One Saturday, the petting zoo sold a total of 250 tickets and collected a total of $2,300 from ticket sales.
Which of the following systems of equations can be used to find the number of standard tickets, s, and premium tickets, p, sold on that
Saturday?
Answer
A.
B.
C.
D.
Correct Answer: A
Rationale
Choice A is correct. It’s given that the petting zoo sells two types of tickets, standard and premium, and that s represents the number of standard
tickets sold and p represents the number of premium tickets sold. It’s also given that the petting zoo sold 250 tickets on one Saturday; thus,
. It’s also given that each standard ticket costs $5 and each premium ticket costs $12. Thus, the amount collected in ticket sales
can be represented by for standard tickets and for premium tickets. On that Saturday the petting zoo collected a total of $2,300 from
ticket sales; thus, . These two equations are correctly represented in choice A.
Choice B is incorrect. The second equation in the system represents the cost per standard ticket as $12, not $5, and the cost per premium ticket
as $5, not $12. Choices C and D are incorrect. The equations represent the total collected from standard and premium ticket sales as $250, not
$2,300, and the total number of standard and premium tickets sold as $2,300, not $250. Additionally, the first equation in choice D represents the
cost per standard ticket as $12, not $5, and the cost per premium ticket as $5, not $12.
Page 11
Question ID: f75bd744
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Hard
Question
In the given system of equations, is a constant. If the system has no solution, what is the value of ?
Correct Answer: 8
Rationale
The correct answer is . The given system of equations can be solved using the elimination method. Multiplying both sides of the second
equation in the given system by yields , or . Adding this equation to the first equation in the given
system, , yields , or . Subtracting from
both sides of this equation yields , or . If the given system has no
solution, then the equation has no solution. If this equation has no solution, the coefficients of on each side of the
equation,
îš•
and , must be equal, which yields the equation . Dividing both sides of this equation by îš•yields . Thus, if
the system has no solution, the value of is .
Alternate approach: A system of two linear equations in two variables, and , has no solution if the lines represented by the equations in the xy-
plane are parallel and distinct. Lines represented by equations in the form , where , , and are constant terms, are parallel if
the ratio of the x-coefficients is equal to the ratio of the y-coefficients, and distinct if the ratio of theîš•x
-coefficients are not equal to the ratio of the
constant terms. Subtracting from both sides of the first equation in the given system yields , or
. Subtracting from both sides of the second equation in the given system yields , or
. The ratio of the x-coefficients for these equations is , or . The ratio of the y-coefficients for these equations is . The ratio
of the constant terms for these equations is
, or . Since the ratio of the x-coefficients, , is not equal to the ratio of the constants, , the
lines represented by the equations are distinct. Setting the ratio of the
x
-coefficients equal to the ratio of the y-coefficients yields .
Multiplying both sides of this equation by
îš•
yields , or . Therefore, when
îš•
, the lines represented by
these equations are parallel. Thus, if the system has no solution, the value of is .
Page 12
Question ID: 6e6a3241
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Medium
Question
Which of the following graphs in the xy-plane could be used to solve the system of equations above?
Answer
A.
B.
C.
D.
Correct Answer: C
Rationale
Choice C is correct. The graph of a system of equations is the graph that shows the lines represented by each of the equations in the system.
The x-intercept of the graph of each given equation can be found by substituting 0 for y in each equation: , or , and
, or . The y-intercept of the graph of each equation can be found by substituting 0 for x in each equation: ,
or , and or . Using these x- and y- intercept values, the line that has equation passes through the
points and , and the line that has equation passes through the points and . Only the lines in choice C
pass through these points and can be used to solve the given system of equations.
Choices A, B, and D are incorrect. In choices A and B, neither line passes through and or and . In choice D, although
one line passes through and the other line doesn’t pass through and .
Page 13
Page 14
Question ID: 8abed0fb
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Easy
Question
What is the solution to the given system of equations?
Answer
A.
B.
C.
D.
Correct Answer: B
Rationale
Choice B is correct. Since it’s given that , substituting 1 for x in the first equation yields . Simplifying the right-hand side of
this equation yields , or . Therefore, the ordered pair is a solution to the given system of equations.
Choice A is incorrect and may result from a calculation error when substituting 1 for x in the first equation. Choices C and D are incorrect.
Because it’s given that , x cannot equal 2 as stated in these ordered pairs.
Page 15
Question ID: e1259a5a
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Easy
Question
A system of two linear equations is graphed in the xy-plane below.
Which of the following points is the solution to the system of equations?
Answer
A.
B.
C.
D.
Correct Answer: A
Rationale
Choice A is correct. The solution to this system of linear equations is the point that lies on both lines graphed, or the point of intersection of the
two lines. According to the graphs, the point of intersection occurs when and , or at the point .
Choices B and D are incorrect. Each of these points lies on one line, but not on both lines in the xy-plane. Choice C is incorrect. This point doesn’t
lie on either of the lines graphed in the xy-plane.
Page 16
Question ID: 70feb725
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Hard
Question
During a month, Morgan ran rîš•
miles at 5îš•miles per hour and biked bîš•
miles at 10îš•miles per hour. She ran and biked a total of 200îš•miles that month,
and she biked for twice as many hours as she ran. What is the total number of miles that Morgan biked during the month?
Answer
A. 80
B. 100
C. 120
D. 160
Correct Answer: D
Rationale
Choice D is correct. The number of hours Morgan spent running or biking can be calculated by dividing the distance she traveled during that
activity by her speed, in miles per hour, for that activity. So the number of hours she ran can be represented by the expression , and the
number of hours she biked can be represented by the expression . It’s given that she biked for twice as many hours as she ran, so this can
be represented by the equation , which can be rewritten as . It’s also given that she ran r miles and biked b miles, and
that she ran and biked a total of 200 miles. This can be represented by the equation . Substituting for b in this equation yields
, or . Solving for rîš• yields . Determining the number of miles she biked, b, can be found by substituting 40 for r in
, which yields . Solving for b yields .
Choices A, B, and C are incorrect because they don’t satisfy that Morgan biked for twice as many hours as she ran. In choice A, if she biked 80
miles, then she ran 120 miles, which means she biked for 8 hours and ran for 24 hours. In choice B, if she biked 100 miles, then she ran 100
miles, which means she biked for 10 hours and ran for 20 hours. In choice C, if she biked 120 miles, then she ran for 80 miles, which means she
biked for 12 hours and ran for 16 hours.
Page 17
Question ID: ed92fb68
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Medium
Question
If the system of equations above has solution , what is the value of ?
Answer
A. 0
B. 9
C. 18
D. 38
Correct Answer: C
Rationale
Choice C is correct. Adding the given equations yields 9x + 9y = 162. Dividing each side of the equation 9x + 9y = 162 by 9 gives x + y = 18.
Choice A is incorrect and may result from incorrectly adding the equations. Choiceîš•B is incorrect and may result from conceptual or
computational errors. Choice D is incorrect. This value is equivalent to y – x.
Page 18
Question ID: 19fdf387
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Medium
Question
In the xy-plane, the graph of intersects the graph of at the point . What is the value of a ?
Answer
A. 3
B. 6
C. 9
D. 12
Correct Answer: C
Rationale
Choice C is correct. Since the graph of intersects the graph of at the point , the ordered pair is the solution
to the system of linear equations consisting of and , and the value of a is the value of x in the solution of this system.
Since both and are equal to y, it follows that . Subtracting x from and adding 6 to both sides of the equation
yields . Therefore, the value of a is 9.
Choices A and B are incorrect and may result from a calculation or conceptual error in solving the system of equations consisting of
and . Choice D is incorrect. This is the value of b, not a.
Page 19
Question ID: e1248a5c
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Hard
Question
In the system of equations below, a and c are constants.
If the system of equations has an infinite number of solutions , what is the value of a ?
Answer
A.
B. 0
C.
D.
Correct Answer: D
Rationale
Choice D is correct. A system of two linear equations has infinitely many solutions if one equation is equivalent to the other. This means that
when the two equations are written in the same form, each coefficient or constant in one equation is equal to the corresponding coefficient or
constant in the other equation multiplied by the same number. The equations in the given system of equations are written in the same form, with
x and y on the left-hand side and a constant on the right-hand side of the equation. The coefficient of y in the second equation is equal to the
coefficient of y in the first equation multiplied by 3. Therefore, a, the coefficient of x in the second equation, must be equal to 3 times the
coefficient of x in the first equation: , or .
Choices A, B, and C are incorrect. When , , or , the given system of equations has one solution.
Page 20
Question ID: 52cb8ea4
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Hard
Question
If is the solution to the system of equations above, what is the value of
îš•
?
Answer
A.
B.
C. 5
D. 13
Correct Answer: B
Rationale
Choice B is correct. Subtracting the second equation, , from the first equation, , results in
, or . Combining like terms on the left-hand side of this equation yields
.
Choice A is incorrect and may result from miscalculating as . Choice C is incorrect and may result from miscalculating as 5.
Choice D is incorrect and may result from adding 9 to 4 instead of subtracting 9 from 4.
Page 21
Question ID: c5082ce3
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Medium
Question
The score on a trivia game is obtained by subtracting the number of incorrect answers from twice the number of correct answers. If a player
answered 40îš•questions and obtained a score of 50, how many questions did the player answer correctly?
Rationale
The correct answer is 30. Let x represent the number of correct answers from the player and y represent the number of incorrect answers from
the player. Since the player answered 40 questions in total, the equation represents this situation. Also, since the score is found by
subtracting the number of incorrect answers from twice the number of correct answers and the player received a score of 50, the equation
represents this situation. Adding the equations in the system of two equations together yields .
This can be rewritten as . Finally, solving for x by dividing both sides of the equation by 3 yields .
Page 22
Question ID: d7bf55e1
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Hard
Question
A movie theater sells two types of tickets, adult tickets for $12 and child tickets for $8. If the theater sold 30îš•tickets for a totalîš•of $300, how
much, in dollars, was spent on adult tickets? (Disregard the $ sign when gridding your answer.)
Rationale
The correct answer is 180. Let a be the number of adult tickets sold and c be the number of child tickets sold. Since the theater sold a total of 30
tickets, a + c = 30. The price per adult ticket is $12, and the price per child ticket is $8. Since the theater received a total of $300 for the 30 tickets
sold, it follows that 12a + 8c = 300. To eliminate c, the first equation can be multiplied by 8 and then subtracted from the second equation:
Because the question asks for the amount spent on adult tickets, which is 12a dollars, the resulting equation can be multiplied by 3 to give 3(4a)
= 3(60) = 180. Therefore, $180 was spent on adult tickets.
Alternate approach: If all the 30 tickets sold were child tickets, their total price would be 30($8) = $240. Since the actual total price of the 30
tickets was $300, the extra $60 indicates that a certain number of adult tickets, a, were sold. Since the price of each adult ticket is $4 more than
each child ticket, 4a = 60, and it follows that 12a = 180.
Page 23
Question ID: bd45df49
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Medium
Question
The solution to the given system of equations is
îš•
. What is the value of ?
Answer
A.
B.
C.
D.
Correct Answer: D
Rationale
Choice D is correct. The first equation in the given system of equations defines as . Substituting for in the second equation in
the given system of equations yields . Applying the distributive property on the left-hand side of this equation yields
. Subtracting from both sides of this equation yields . Subtracting from both sides of this equation yields
. Substituting for in the first equation of the given system of equations yields ,îš•or . Substituting for
and for into the expression yields , or .
Choice A is incorrect. This is the value of , not .
Choice B is incorrect. This is the value of , not .
Choice C is incorrect and may result from conceptual or calculation errors.
Page 24
Question ID: f718c9cf
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Hard
Question
The solution to the given system of equations is
îš•
. What is the value of ?
Correct Answer: 1.8, 9/5
Rationale
The correct answer is . Multiplying the first equation in the given system by yields . Subtracting the second equation in the
given system, , from yields , which is equivalent to
, or . Dividing both sides of this equation by yields . The value of can be found by substituting for in
either of the two given equations. Substituting for in the equation yields , or . Subtracting
from both sides of this equation yields . Dividing both sides of this equation by yields , or . Therefore, the value of
is , or . Note that 9/5 and 1.8 are examples of ways to enter a correct answer.
Page 25
Question ID: 6e50ce28
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Medium
Question
The sum of a number and is twice as large as a number . The number is less than the number . Which system of equations describes
this situation?
Answer
A.
B.
C.
D.
Correct Answer: A
Rationale
Choice A is correct. It's given that the sum of a number and is twice as large as a number . This can be described by the equation
. It’s also given that the number is less than the number . This can be described by the equation . Therefore, the system
consisting of the equations and describes this situation.
Choice B is incorrect. The equation describes a situation where the number is less than .
Choice C is incorrect. The equation describes a situation where the number is twice the sum of a number and .
Choice D is incorrect. The equation describes a situation where the number is twice the sum of a number and , and the
equation describes a situation where a number is less than .
Page 26
Question ID: 2875ba81
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Medium
Question
The solution to the given system of equations is . What is the value of y?
Answer
A.
B.
C.
D.
Correct Answer: A
Rationale
Choice A is correct. The given system of linear equations can be solved by the elimination method. Multiplying each side of the second equation
in the given system by yields , or . Subtracting this equation from the first equation in the given
system yields , which is equivalent to
îš•
, or .
îš•
Choice B is incorrect. This is the value of , not the value of .
Choice C is incorrect and may result from conceptual or calculation errors.
Choice D is incorrect and may result from conceptual or calculation errors.
Page 27
Question ID: ee031767
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Easy
Question
A dance teacher ordered outfits for students for a dance recital. Outfits for boys cost $26, and outfits for girls cost $35. The dance teacher
ordered a total of 28 outfits and spent $881. If b represents the number of outfits the dance teacher ordered for boys and g represents the
number of outfits the dance teacher ordered for girls, which of the following systems of equations can be solved to find b and g ?
Answer
A.
B.
C.
D.
Correct Answer: B
Rationale
Choice B is correct. Outfits for boys cost $26 each and the teacher ordered b outfits for boys, so the teacher spent 26b dollars on outfits for
boys. Similarly, outfits for girls cost $35 each and the teacher ordered g outfits for girls, so the teacher spent 35g dollars on outfits for girls. Since
the teacher spent a total of $881 on outfits for boys and girls, the equation 26b + 35g = 881 must be true. And since the teacher ordered a total
of 28 outfits, the equation b + g = 28 must also be true.
Choice A is incorrect and may result from switching the constraint on the total number of outfits with the constraint on the cost of the outfits.
Choice C is incorrect and may result from switching the constraint on the total number of outfits with the constraint on the cost of the outfits, as
well as switching the cost of the outfits for boys with the cost of the outfits for girls. Choice D is incorrect and may result from switching the cost
of the outfits for boys with the cost of the outfits for girls.
Page 28
Question ID: dcc4886a
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Medium
Question
One of the two equations in a system of linear equations is given. The system has infinitely many solutions. If the second equation in the system
is , where and are constants, what is the value of ?
Answer
A.
B.
C.
D.
Correct Answer: D
Rationale
Choice D is correct. It’s given that the system has infinitely many solutions. The graphs of two lines in the xy-plane represented by equations in
slope-intercept form, , whereîš• andîš• are constants, have infinitely many solutions if their slopes, , are the same and if their y-
coordinates of the y-intercepts, , are also the same. The first equation in the given system is . For this equation, the slope is and
the y-coordinate of the y-intercept is . If the second equation is in the form , then for the two equations to be equivalent, the values
ofîš• andîš• in the second equation must equal the corresponding values in the first equation. Therefore, the second equation must have a slope,
, of , and a y-coordinate of the y-intercept, , of . Thus, the value ofîš• is .
Choice A is incorrect and may result from conceptual errors.
Choice B is incorrect and may result from conceptual errors.
Choice C is incorrect and may result from conceptual errors.
Page 29
Question ID: 466b87e3
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Hard
Question
In the system of equations above, c is a constant. If the system has no solution, what is the value of c ?
Rationale
The correct answer is . A system of two linear equations has no solution when the graphs of the equations have the same slope and
different y-intercepts. Each of the given linear equations is written in the slope-intercept form, , where m is the slope and b is the
y-coordinate of the y-intercept of the graph of the equation. For these two linear equations, the y-intercepts are and . Thus, if the
system of equations has no solution, the slopes of the graphs of the two linear equations must be the same. The slope of the graph of the first
linear equation is . Therefore, for the system of equations to have no solution, the value of c must be . Note that 1/2 and .5 are examples
of ways to enter a correct answer.
Page 30
Question ID: cd33b015
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Easy
Question
If is the solution to the given system of equations, what is the value of y ?
Answer
A. 10
B. 15
C. 60
D. 65
Correct Answer: B
Rationale
Choice B is correct. Substituting 20 for in the second equation in the system yields , or . Subtracting
40 from both sides of this equation yields . Dividing both sides of this equation by 3 yields .
Choice A is incorrect. If , then since . However, substituting 10 for both x and y in the second equation yields
, which is a false statement. Choice C is incorrect. If , then since . However, substituting these values
for x and y in the second equation yields , which is a false statement. Choice D is incorrect. If , then since
. However, substituting these values for x and y in the second equation yields , which is a false statement.
Page 31
Question ID: e2e3942f
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Hard
Question
In the system of equations above, aîš•is a constant. If the system of equations has no solution, what is the valueîš•ofîš•aîš•?
Answer
A.
B. 0
C. 1
D. 2
Correct Answer: D
Rationale
Choice D is correct. A system of two linear equations has no solution when the graphs of the equations have the same slope and different y-
coordinates of the y-intercepts. Each of the given equations is written in the slope-intercept form of a linear equation, , where m is
the slope and b is the y-coordinate of the y-intercept of the graph of the equation. For these two linear equations, the y-coordinates of the y-
intercepts are different: and . Thus, if the system of equations has no solution, the slopes of the two linear equations must be the same.
The slope of the first linear equation is 2. Therefore, for the system of equations to have no solution, the value of a must be 2.
Choices A, B, and C are incorrect and may result from conceptual and computational errors.
Page 32
Question ID: 4fb8adf7
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Easy
Question
What is the solution
îš•
to the given system of equations?
Answer
A.
B.
C.
D.
Correct Answer: A
Rationale
Choice A is correct. The second equation in the given system is . Substituting for in the first equation in the given system yields
, or . Subtracting from both sides of this equation yields
îš•
. Dividing both sides of this equation by
yields . Therefore, the solution to the given system of equations is .
Choice B is incorrect and may result from conceptual or calculation errors.
Choice C is incorrect and may result from conceptual or calculation errors.
Choice D is incorrect and may result from conceptual or calculation errors.
Page 33
Question ID: 1e0a46e4
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Hard
Question
Which system of linear equations has no solution?
Answer
A.
B.
C.
D.
Correct Answer: D
Rationale
Choice D is correct. A system of linear equations can be solved by the elimination method. Multiplying the equation
îš•
by îš•yields
. Adding this equation to the equation
îš•
yields , which has no solution. It follows that the system of linear
equations consisting of
îš•
and has no solution.
Choice A is incorrect. This system of linear equations has infinitely many solutions, rather than no solution.
Choice B is incorrect. This system of linear equations has one solution, rather than no solution.
Choice C is incorrect. This system of linear equations has infinitely
many solutions, rather than no solution.
Page 34
Question ID: 1e11190a
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Hard
Question
Store A sells raspberries for
îš•
per pint and blackberries for per pint. Store B sells raspberries for per pint and blackberries for
per pint. A certain purchase of raspberries and blackberries would cost at Store A or at Store B. How many pints of
blackberries are in this purchase?
Answer
A.
B.
C.
D.
Correct Answer: B
Rationale
Choice C is correct. It’s given that store A sells raspberries for
îš•
per pint and blackberries for
îš•
per pint, and a certain purchase of
raspberries and blackberries at store A would cost . It’s also given that store B sells raspberries for
îš•
per pint and blackberries for
îš•
per pint, and this purchase of raspberries and blackberries at store B would cost . Let represent the number of pints of
raspberries and represent the number of pints of blackberries in this purchase. The equation
îš• îš•
represents this purchase
of raspberries and blackberries from store A and the equation
îš•
represents this purchase of raspberries and blackberries
from store B. Solving the system of equations by elimination gives the value of and the value of that make the system of equations true.
Multiplying both sides of the equation for store A by yields
îš•
, or
îš•
.
Multiplying both sides of the equation for store B by yields
îš•
, or .
Subtracting both sides of the equation for store A,
îš•
, from the corresponding sides of the equation for store B,
îš•
, yields
îš•
, or
îš•
. Dividing both sides of this equationby
yields
îš•
. Thus, pints of blackberries are in
this purchase.
Choices A and B are incorrect and may result from conceptual or calculation errors. Choice D is incorrect. This is the number of pints of
raspberries, not blackberries, in the purchase.
Page 35
Question ID: e77a76ce
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Medium
Question
Which of the following systems of linear equations has no solution?
Answer
A.
B.
C.
D.
Correct Answer: A
Rationale
Choice A is correct. A system of two linear equations in two variables, and , has no solution if the graphs of the lines represented by the
equations in the xy-plane are distinct and parallel. The graphs of two lines in theîš•xy
-plane represented by equations in slope-intercept form,
, where and are constants, are parallel if their slopes, , are the same and are distinct if their y-coordinates of the y-intercepts, ,
are different. In the equations
îš•
and
îš•
, the values of are each , and the values of are and , respectively. Since the
slopes of these lines are the same and the y-coordinates of the y-intercepts are different, it follows that the system of linear equations in choice
A has no solution.
Choice B is incorrect. The two lines represented by these equations are a horizontal line and a line with a slope of that have the same y-
coordinate of the y-intercept. Therefore, this system has a solution, , rather than no solution.
Choice C is incorrect. The two lines represented by these equations have different slopes and the same y-coordinate of theîš•y
-intercept. Therefore,
this system has a solution, , rather than no solution.
Choice D is incorrect. The two lines represented by these equations are a vertical line and a horizontal line. Therefore, this system has a solution,
, rather than no solution.
Page 36
Question ID: fb5e7f59
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Hard
Question
In the given system of equations, is a constant. In the xy-plane, the graphs of these equations intersect at the point , where is a
constant. What is the value of ?
Correct Answer: 11
Rationale
The correct answer is . It’s given that the graphs of the equations in the given system intersect at the point , where is a constant.
Therefore, the coordinates of this point must satisfy both equations. Substituting the point into the first equation, ,
yields . Adding
îš•
to both sides of this equation yields , which is equivalent to .
Substituting the point
îš•
into the second equation yields . Substituting in place of in the equation
yields . Applying the distributive property to the left-hand side of this equation yields
. Combining like terms on the left-hand side of this equation yields . Subtracting from both sides of this
equation yields . Dividing both sides of this equation by yields .
Page 37
Question ID: 5e422ff9
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Medium
Question
In the solution to the system of equations above, what is the value of y ?
Answer
A.
B.
C. 9
D. 15
Correct Answer: D
Rationale
Choice D is correct. Multiplying both sides of by 5 results in . Multiplying both sides of by 2 results in
. Subtracting the resulting equations yields , which simplifies to . Dividing both sides
of by results in .
Choices A and B are incorrect and may result from incorrectly subtracting the transformed equation. Choice C is incorrect and may result from
finding the value of x instead of the value of y.
Page 38
Question ID: 567ac7ab
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Hard
Question
One of the two equations in a linear system is . The system has no solution. Which of the following could be the other equation
in the system?
Answer
A.
B.
C.
D.
Correct Answer: B
Rationale
Choice B is correct. A system of two linear equations written in standard form has no solution when the equations are distinct and the ratio of
the x-coefficient to the y-coefficient for one equation is equivalent to the ratio of the x-coefficient to the y-coefficient for the other equation. This
ratio for the given equation is 2 to 6, or 1 to 3. Only choice B is an equation that isn’t equivalent to the given equation and whose ratio of the x-
coefficient to the y-coefficient is 1 to 3.
Choice A is incorrect. Multiplying each of the terms in this equation by 2 yields an equation that is equivalent to the given equation. This system
would have infinitely many solutions. Choices C and D are incorrect. The ratio of the x-coefficient to the y-coefficient in (choice C)
is to 2, or to 1. This ratio in (choice D) is 6 to 2, or 3 to 1. Since neither of these ratios is equivalent to that for the given
equation, these systems would have exactly one solution.
Page 39
Question ID: 2704399f
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Medium
Question
Which of the following systems of equations has the same solution as the system of equations graphed above?
Answer
A.
B.
C.
D.
Correct Answer: A
Rationale
Choice A is correct. The solution to a system of equations is the coordinates of the intersection point of the graphs of the equations in the xy-
plane. Based on the graph, the solution to the given system of equations is best approximated as . In the xy-plane, the graph of is
a horizontal line on which every y-coordinate is 0, and the graph of is a ver tical line on which every x-coordinate is . These graphs
intersect at the point . Therefore, the system of equations in choice A has the same solution as the given system.
Page 40
Choices B, C, and D are incorrect. If graphed in the xy-plane, these choices would intersect at the points , , and , respectively,
not .
Page 41
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Easy
Question
If is the solution to the given system of equations, what is the value of y ?
Rationale
The correct answer is 23. Since it’s given that , the value of y can be found by substituting 2 for x in the first equation and solving for y.
Substituting 2 for x yields , or . Subtracting 6 from both sides of this equation yields .
Page 42
Question ID: 0df106df
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Easy
Question
An online bookstore sells novels and magazines. Each novel sells for $4, and each magazine sells for $1. If Sadie purchased a total of 11îš•novels
and magazines that have a combined selling price of $20, how many novels did she purchase?
Answer
A. 2
B. 3
C. 4
D. 5
Correct Answer: B
Rationale
Choice B is correct. Let n be the number of novels and m be the number of magazines that Sadie purchased. If Sadie purchased a total of 11
novels and magazines, then . It is given that the combined price of 11 novels and magazines is $20. Since each novel sells for $4
and each magazine sells for $1, it follows that . So the system of equations below must hold.
Subtracting corresponding sides of the second equation from the first equation yields ,îš•so . Therefore, Sadie purchased 3 novels.
Choice A is incorrect. If 2 novels were purchased, then a total of $8 was spent on novels. That leaves $12 to be spent on magazines, which
means that 12 magazines would have been purchased. However, Sadie purchased a total of 11 novels and magazines. Choices C and D are
incorrect. If 4 novels were purchased, then a total of $16 was spent on novels. That leaves $4 to be spent on magazines, which means that 4
magazines would have been purchased. By the same logic, if Sadie purchased 5 novels, she would have no money at all ($0) to buy magazines.
However, Sadie purchased a total of 11 novels and magazines.
Page 43
Question ID: b544a348
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Medium
Question
In the solution to the system of equations above, what is the value of x ?
Rationale
The correct answer is 7. Subtracting the second equation from the first equation eliminates the variable y.
Dividing both sides of the resulting equation by 4 yields x = 7.
Page 44
Question ID: 7d89376f
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Easy
Question
A discount airline sells a certain number of tickets, x, for a flight for $90 each. It sells the number of remaining tickets, y, for $250 each. For a
particular flight, the airline sold 120 tickets and collected a total of $27,600 from the sale of those tickets. Which system of equations represents
this relationship between x and y ?
Answer
A.
B.
C.
D.
Correct Answer: A
Rationale
Choice A is correct. The airline sold two types of tickets for this flight: x tickets at $90 each and the remainingîš• tickets, y, at $250 each. Because
the airline sold a total of 120 tickets for this flight, it must be true that x + y = 120. The amount, in dollars, collected from the sale of x tickets at
$90 each is represented by 90x.îš•The amount, in dollars, collected from the sale of the remaining y tickets at $250 each is represented by 250y. It
is given that a total of $27,600 was collected from the sale of all tickets. Therefore, it must also be true that 90x + 250y = 27,600.
Choice B is incorrect. The total number of tickets sold is represented correctly as x + y = 120. The total amount, in dollars, collected from the sale
of the x tickets at $90 each and the remaining tickets, y, at $250 has been correctly represented as 90x + 250y. However, according to the
information given, this total should be equal to 27,600, not 120(27,600) dollars. Choice C is incorrect. The total number of tickets sold has been
correctly represented as x + y. However, according to the information given, this total should be equal to 120, not 27,600, as shown in choice C.
The total amount, in dollars, collected from the sale of the x tickets at $90 each and the remaining tickets, y, at $250 has been correctly
represented as 90x + 250y. However, according to the information given, this total should be equal to 27,600, not 120(27,600) dollars. Choice D is
incorrect. The two equations given in choice D have no meaning in this context.
Page 45
Question ID: 17f176ec
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Easy
Question
A movie theater charges $11 for each full-price ticket and $8.25 for each reduced-price ticket. For one movie showing, the theater sold a total of
214îš•full-price and reduced-price tickets for $2,145. Which of the following systems of equations could be used to determine the number of full-
price tickets, f, and the number of reduced-price tickets, r, sold?
Answer
A.
B.
C.
D.
Correct Answer: B
Rationale
Choice B is correct. The movie theater sells f full-price tickets and r reduced-price tickets, so the total number of tickets sold is f + r. Since the
movie theater sold a total of 214 full-price and reduced-price tickets for one movie showing, it follows that f + r = 214. The movie theater charges
$11 for each full-price ticket; thus, the sales for full-price tickets, in dollars, is given by 11f. The movie theater charges $8.25 for each reduced-
price ticket; thus, the sales for reduced-price tickets, in dollars, is given by 8.25r. Therefore, the total sales, in dollars, for the movie showing is
given by 11f + 8.25r. Since the total sales for all full-price and reduced-price tickets is $2,145, it follows that 11f + 8.25r = 2,145.
Choice A is incorrect. This system of equations suggests that the movie theater sold a total of 2,145 full-price and reduced-price tickets for a
total of $214. Choice C is incorrect. This system suggests that the movie theater charges $8.25 for each full-price ticket and $11 for each
reduced-price ticket. Choice D is incorrect. This system suggests that the movie theater charges $8.25 for each full-price ticket and $11 for each
reduced-price ticket and sold a total of 2,145 tickets for a total of $214.
Page 46
Question ID: 65833256
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Medium
Question
What is the solution
îš•
to the given system of equations?
Answer
A.
B.
C.
D.
Correct Answer: A
Rationale
Choice A is correct. The given system of linear equations can be solved by the substitution method. The first equation in the given system of
equations defines as . Substituting
îš•
for in the second equation of the given system of equations yields
. Applying the distributive property on the left-hand side of this equation yields , or . Adding to
both sides of this equation yields . Dividing both sides of this equation by yields . Substituting for in the first
equation of the given system of equations, , yields ,îš•or . Therefore, the solution to the given system
of equations is .
Choice B is incorrect and may result from conceptual or calculation errors.
Choice C is incorrect and may result from conceptual or calculation errors.
Choice D is incorrect and may result from conceptual or calculation errors.
Page 47
Question ID: 44d65912
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Easy
Question
Angela is playing a video game. In this game, players can score points only by collecting coins and stars. Each coin is worth c points, and each
star is worth s points.
The first time she played, Angela scored 700 points. She collected 20 coins and 10 stars.
The second time she played, Angela scored 850 points. She collected 25 coins and 12 stars.
Which system of equations can be used to correctly determine the values of c and s ?
Answer
A.
B.
C.
D.
Correct Answer: B
Rationale
Choice B is correct. The number of coins collected can be multiplied by c to give the score from the points earned from coins. Similarly, the
number of stars collected can be multiplied by s to give the score from the points earned from the stars. Therefore, the total score each time
Angela played is , and the total score the second time she played is .
Choices A, C, and D are incorrect and may result from misidentifying the terms of the equation. Choice A switches coins and stars, choice C
switches stars and points, and choice D misidentifies coins, stars, and points.
Page 48
Question ID: 4b76c7f1
Assessment Test Domain Skill Difficulty
SAT Math Algebra Systems of two linear
equations in two
variables
Easy
Question
In the given system of equations, a is a constant. If the system has infinitely many solutions, what is the value of a ?
Answer
A. 4
B. 9
C. 36
D. 54
Correct Answer: C
Rationale
Choice C is correct. A system of two linear equations has infinitely many solutions if one equation is equivalent to the other. This means that
when the two equations are written in the same form, each coefficient or constant in one equation is equal to the corresponding coefficient or
constant in the other equation multiplied by the same number. The equations in the given system of equations are written in the same form, with
x and y on the left-hand side of the equation and a constant on the right-hand side of the equation. The coefficients of x and y in the second
equation are equal to the coefficients of x and y, respectively, in the first equation multiplied by 4: and . Therefore, the
constant in the second equation must be equal to 4 times the constant in the first equation: , or .
Choices A, B, and D are incorrect. When , , or , the given system of equations has no solution.
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