1. Solve the system of equations.
y = –3x + 4
x + 4y = –6
A. x = –2,y = –1
B. x = –2,y = 10
C. x = 2,y = –2
D. x = 3,y = –5
E. x = 4,y = –8
2. John can mow his lawn in 3 hours and his sister, Julie, can mow it in 2 hours. How long will it take them to mow their lawn if they work together?
A. 1 hour 12 minutes
B. 1 hour 15 minutes
C. 1 hour 20 minutes
D. 1 hour 30 minutes
E. 1 hour 35 minutes
3. Solve the equation.
B. –5 and 2
D. 2 and 4
4. Factor the expression completely.
6×3- 4×2 – 16x
B. 2x(3×2 – 2x – 8)
C. 2x(3x + 4)(x – 2)
D. 4x(2x + 1)(x – 4)
E. 2x(2×2 + 7x – 4)
5. Solve the equation for x.
5×2 + 6x = 3
6. What should be added to both sides of the equation x2 – 12x = 5 in order to solve it by completing the square?
7. Find the vertical asymptotes of the function.
A. x = –5 and x = –3
B. x = –5, x = –3, and x = 6
C. x = 3 and x = 5
D. x = 3 and x = 6
E. x = 6
8. Two cars are traveling north along a highway. The first drives at 40 mph, and the second, which leaves 3 hours later, travels at 60 mph. How long after the second car leaves will it take for the second car to catch the first?
A. 1 hour 12 minutes
B. 2 hours
C. 5 hours
D. 6 hours
E. 6 hours 40 minutes
9. What is the domain of the function f(x) = 2x – 4?
10. Solve the equation for x.
1. C. Notice that the given system has two equations, and each equation has two variables, x and y. Therefore, the solution of the system of equations will have values for each variable. Since the first equation already has y isolated on the left side, it will be easier to use the substitution method than the elimination method to solve the system of equations. To begin, substitute the left side of the first equation, –3x+4, for y into the second equation, and then solve for x.
x + 4y = –6
x + 4(–3x + 4) = –6
x – 12x + 16 = –6
–11x = –22
x = 2
To find the value of y, substitute 2 for x in the first equation.
y = –3(2) + 4
= –6 + 4
Therefore, the solution of the given system of equations is x = 2,y = –2. Check this solution by substituting the values into the second equation and making sure the resulting equality is true.
2. A. Using, rate x time = amount, determine the rate at which John and Julie each mows the lawn if they work separately. To make the calculation easier, rewrite the formula as .
Next, if John and Julie work together, their total rate can be found by adding the individual rates together. In other words, their total rate working together is lawns per hour. Substitute this value into the original rate formula and solve for t, the variable that represents time spent mowing.
rate x time = amount
Therefore, it will take John and Julie hours, or 1 hour 12 minutes, to mow the lawn if they work together.
3. B. First, eliminate the denominators by multiplying both sides by x(x + 4). Then, simplify the result.
The result is a quadratic equation. Solve it by factoring the left side and setting each factor equal to zero.
(x + 5)(x – 2) = 0
x = –5, x = 2
Therefore, the possible solutions are x = –5 and x = 2. Unfortunately, there is a risk of finding an incorrect solution when solving rational equations in this manner. Consequently, the two possible solutions must be verified. To do this, substitute them into the given equations and make sure that the result is a true statement.
Therefore, the solutions are x = –5 and x = 2.
4. C. First, factor out any common factors from each of the three terms, 6×3, -4×2, and -16x. Notice that the greatest common factor (GCF) of the coefficients is 2, and each term is divisible by x. Therefore, factor out 2x.
6×3 – 4×2 – 16x = 2x(3×2 – 2x – 8)
Finally, factor the trinomial, 3×2 – 2x – 8, into two binomials.
2x(3×2 – 2x – 8) = 2x(3x + 4)(x – 2)
5. C. To begin, rewrite the equation in the form ax2 + bx + c = 0 by subtracting 3 from both sides of the equation.
5×2 + 6x = 3
5×2 + 6x – 3 = 0
Since the left side cannot be factored, use the quadratic formula to solve the equation, which is written in the form ax2 + bx + c = 0. For this equation, a = 5, b = 6, and c =–3.
6. E. To solve an equation by completing the square, manipulate it algebraically so that one side (in this case, the left side) is a perfect square trinomial and the other side (the right side) is a constant. Recall that a perfect square trinomial is a trinomial that can be factored as (ax + b)2or (ax – b)2. In the given equation, the left side only has two terms, an x2-term and an x-term; a constant term needed to make the expression a perfect square trinomial.
To calculate that constant, divide the coefficient of the x-term (which is –12) by 2 (giving –6) and square the result ( (-6)2 = 36). Adding 36 to both sides of the original equation will complete the square: notice that x2 – 12x + 36 is indeed a perfect square trinomial because it can be factored as (x – 6)2.
7. C. An asymptote is a line that the graph of a function approaches but never touches. For a rational function like the given one, the vertical asymptotes are vertical lines that occur at every x-value for which the denominator is zero. For the given function, set up and solve an equation to determine when x2 – 8x + 15 is zero.
x2 – 8x + 15 = 0
(x – 3)(x – 5) = 0
x = 3, x = 5
Therefore, the vertical asymptotes are the lines x = 3 and x = 5.
8. D. Use the rate formula, , to find the answer. Let t represent the number of hours the second car travels.
40(t + 3) = distance that first car travels
60t = distance that second car travels
The second car will catch the first car when both have traveled the same distance. So, set the two formulas equal to each other and solve for t.
40(t + 3) = 60t
40t + 120 = 60t
120 = 20t
t = 6
Therefore, the second car catches the first car six hours after it leaves.
9. E. The domain of a function is the set of all possible input values for the function. In this case, the input for f(x) is x. The easiest way to find the domain of f(x) is to figure out which x-values do not work (that is, give undefined output values) in the function. Then you can exclude those values from the domain.
Examine the given function. Notice that it does not contain a square-root sign, a logarithm, or a fraction with x in the denominator. Therefore, the function is defined for all real numbers. Notice from the graph below that the function can go to the left and right without end.
10. D. The logarithm of a number is the exponent to which the base must be raised to in order to get that number. For example, since 23 = 8 , it is also true that log2 8 = 3. Therefore, the equation can be rewritten as
Simplify the equation and solve for x.
Thus, the solution is x = 3. Check this solution by substituting 3 into the original equation and making sure that the result is a true statement.
In addition, Algebra 2 is the first math class in a student's math career that introduces topics that are more complex and less concrete, like complex numbers or logarithms, which makes Algebra 2 harder to grasp than other math classes whose concepts are more straight forward and easier to visualize.How many questions do I need to get right to pass the algebra 2 Regents? ›
The difficulty of the Algebra 2 Regents exam really depends on your goal for taking it. If you are taking the Algebra 2 exam in order to satisfy the graduation requirements for a Regents diploma, you need to answer 33% of the questions on the exam correctly to pass.Is it hard to pass algebra 2? ›
Algebra II is sometimes perceived as a rather hard subject by some students, and we agree that it is a bit difficult to understand, but only if you lack the area of knowledge related to previous math classes and some basic concepts, including geometry and algebra I.Is algebra 2 harder than Geometry? ›
Let's begin with the “why” question. Geometry is simpler than algebra 2. So if you want to look at these three courses in order of difficulty, it would be algebra 1, geometry, then algebra 2. Geometry does not use any math more complicated than the concepts learned in algebra 1.Is algebra 2 easier than Calculus? ›
We often consider calculus to be more difficult than algebra. Algebra courses explore the many operations, properties, and rules that can be used to manipulate equations. Calculus courses apply algebraic operations to functions in a more complex way.Can I skip algebra 2 in high school? ›
In general, skipping Algebra II is a bad idea if you plan on going to a college or university since they have the choice to reject you, There could also be other hidden consequences for not completing Algebra successfully.Do colleges look at Regents? ›
College Admissions: Many colleges and universities consider Regents Exam scores as part of the admissions process. Scoring well on these exams can improve your chances of being admitted to your desired school.What happens if you fail the Regents but pass the class? ›
Students may earn diploma credit for successfully completing a course but fail to meet the corresponding Regents assessment requirement; or • Students can fail a course and receive no academic credit but fulfill an assessment requirement if they pass a Regents examination in that course.How do I study for the Algebra 2 exam? ›
Consider forming a study group. Working with your fellow students to solve problems and going over algebraic concepts is a great way to succeed in an Algebra 2 class. You can also find out if your school offers a math study lab or tutors. Taking advantage of these resources can make passing Algebra 2 a lot easier.Why is algebra 2 so hard for me? ›
It's not only learning new things, but combining old things together in ways that students may have never experienced. One thing I discovered with students who struggle with Algebra II is that there is something from past math courses that they never quite got, especially working with decimals and fractions.
Because most colleges do require 3-4 years of math, including an algebra and a geometry for admission, almost all schools require that a student passes algebra 2 in order to meet that standard.Is it normal for a freshman to take algebra 2? ›
Traditionally, freshmen enroll in Algebra, Geometry, Honors Geometry, or Honors Algebra II. Though you are advanced in your course subject, the regular Algebra II does not prepare you for future honors math classes.Should I skip geometry and go to algebra 2? ›
Geometry is typically taken before algebra 2 and after algebra 1. Whether or not a student can take algebra 2 before Geometry depends on each student's school policies. However, I would recommend taking the traditional order of math classes. Some schools allow their students to place out of certain math concepts.Is it OK to take algebra 2 before geometry? ›
Order of math classes:
As a general rule, HSML strongly recommends that a student take Geometry prior to Algebra II. Although it is certainly possible to have a successful year in Algebra II before taking Geometry, here are some thoughts to consider as you decide what is right for your student.
There is also the argument as to what makes sense in terms of course sequence, especially if a student struggles with math. If retaining algebra concepts is a challenge then taking Algebra 1 and Algebra 2 back-to-back makes the most sense.Is Algebra 2 or Algebra 1 harder? ›
Now Algebra 1 is easier than Algebra 2. However for some Geometry is much harder than Algebra.What math class is hardest? ›
What is the Hardest Math Class in High School? In most cases, you'll find that AP Calculus BC or IB Math HL is the most difficult math course your school offers. Note that AP Calculus BC covers the material in AP Calculus AB but also continues the curriculum, addressing more challenging and advanced concepts.What level of math is Algebra 2? ›
Students typically learn Algebra II in 11th grade. An Algebra II curriculum usually builds on knowledge and skills that are gained in Algebra I and reinforced in Geometry, including relationships between quantities through equations and inequalities, graphing of functions, and trigonometry.Is Algebra 2 higher than algebra? ›
Students will likely take algebra 1 before taking geometry, and will then follow up these two topics with algebra 2. Algebra 1 focuses on solving and graphing equations and inequalities while algebra 2 covers new functions like exponential and logarithmic equations.