Call Now to Set Up Tutoring:

(888) 888-0446

ISEE Upper Level Math : Algebraic Concepts

Study concepts, example questions & explanations for ISEE Upper Level Math

Create An Account

All ISEE Upper Level Math Resources

6 Diagnostic Tests 244 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

← Previous 1 2 3 4 5 6 7 8 9 … 32 33 Next →

Example Question #1 : How To Find The Solution To An Equation

Solve for x:

$\dpi{100} 3x^{2}+4=31$

Possible Answers:

$\dpi{100} \pm 1$

$\dpi{100} \pm 9$

$\dpi{100} \pm 3$

$\dpi{100} 9$

Correct answer:

$\dpi{100} \pm 3$

Explanation:

First, subtract 4 from both sides:

$\dpi{100} 3x^{2}+4-4=31-4$

$\dpi{100} 3x^{2}=27$

Next, divide both sides by 3:

$\dpi{100} \frac{3x^{2}}{3}=\frac{27}{3}$

$\dpi{100} x^{2}=9$

Now take the square root of both sides:

$\dpi{100} \sqrt{x^{2}}=\sqrt{9}$

$\dpi{100} x=\pm 3$

Report an Error

Example Question #1 : How To Find The Solution To An Equation

Solve for $\dpi{100} x$ :

$\dpi{100} \frac{x}{3}+22 = 45$

Possible Answers:

$\dpi{100} 66$

$\dpi{100} 69$

$\dpi{100} 3$

$\dpi{100} 135$

Correct answer:

$\dpi{100} 69$

Explanation:

$\dpi{100} \frac{x}{3}+22 = 45$

$\dpi{100} \frac{x}{3}+22-22 = 45-22$

$\dpi{100} \frac{x}{3} = 23$

$\dpi{100} (3)\cdot \frac{x}{3} = 23\cdot (3)$

$\dpi{100} x=69$

Report an Error

Example Question #3 : How To Find The Solution To An Equation

Solve for :

| 3x + 9| = 24

Possible Answers:

$x = -5 \textrm{ or }x = 11$

$x = -11 \textrm{ or }x = 11$

$x = -11 \textrm{ or }x = 5$

$x = 5 \textrm{ or }x = 11$

$x = -5\textrm{ or }x = 5$

Correct answer:

$x = -11 \textrm{ or }x = 5$

Explanation:

Rewrite this as a compound statement and solve each separately:

| 3x + 9| = 24

$3x + 9 = - 24 \textrm{ or }3x + 9 = 24$

3x + 9 = - 24

3x + 9 -9= - 24-9

3x = - 33

$3x\div 3 = - 33 \div 3$

x = -11

3x + 9 = 24

3x + 9 -9= 24-9

3x = 15

$3x\div 3 = 15 \div 3$

x = 5

Report an Error

Example Question #1 : Algebraic Concepts

Solve for :

$\left (x -4 \right )\left( x +3 \right )= \left (x-2 \right) \left ( x-1\right )$

Possible Answers:

$x = 1\textrm{ or }x = 4$

x = 12

x = 7

$x = 1\textrm{ or }x = 7$

$x = 3\textrm{ or }x = 4$

Correct answer:

x = 7

Explanation:

FOIL each of the two expressions, then solve:

$\left (x -4 \right )\left( x +3 \right )= \left (x-2 \right) \left ( x-1\right )$

$x ^{2} + 3 \cdot x -4\cdot x -4 \cdot 3 = x^{2} -1 \cdot x - 2 \cdot x + 2\cdot1$

$x ^{2} + 3 x -4 x -12 = x^{2} -1 x - 2x + 2$

$x ^{2} - x -12 = x^{2} - 3x + 2$

$x ^{2} - x -12 - x^{2} = x^{2} - 3x + 2 - x^{2}$

- x -12= - 3x + 2

Solve the resulting linear equation:

- x -12 +3x + 12 = - 3x + 2+3x+12

2x =14

$2x\div 2 =14 \div 2$

x = 7

Report an Error

Example Question #5 : How To Find The Solution To An Equation

Solve for :

$x^{2} -10x = 24$

Possible Answers:

$x=-6\textrm{ or }x=4$

$x=-12 \textrm{ or }x=2$

$x=-6\textrm{ or }x=-4$

$x=-2 \textrm{ or }x=12$

$x=-4\textrm{ or }x=6$

Correct answer:

$x=-2 \textrm{ or }x=12$

Explanation:

First, rewrite the quadratic equation in standard form by moving all nonzero terms to the left:

$x^{2} -10x = 24$

$x^{2} -10x -24= 24-24$

$x^{2} -10x -24= 0$

Now factor the quadratic expression $x^{2} -10x -24$ into two binomial factors (x + ?)(x + ?) , replacing the question marks with two integers whose product is and whose sum is . These numbers are -12,2 , so:

(x -12)(x + 2) = 0

$x-12 = 0 \Rightarrow x = 12$

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

The solution set is $\left \{ -2,12\right \}$ .

Report an Error

Example Question #6 : How To Find The Solution To An Equation

Solve for :

$\left ( x-3\right )(2x-1) = 18$

Possible Answers:

$x = 9\frac{1}{2} \textrm{ or }x=19$

$x= - 1 \frac{1}{2} \textrm{ or } x= 5$

$x= - 5\textrm{ or } x= 1 \frac{1}{2}$

$x= -3 \textrm{ or } x= 5$

$x= -5 \textrm{ or } x= 3$

Correct answer:

$x= - 1 \frac{1}{2} \textrm{ or } x= 5$

Explanation:

First, rewrite the quadratic equation in standard form by FOILing out the product on the left, then collecting all of the terms on the left side:

$\left ( x-3\right )(2x-1) = 18$

$x\cdot 2x -x\cdot 1 -3 \cdot 2x +3\cdot 1= 18$

$2x^{2} -x -6x +3= 18$

$2x^{2} -7x +3= 18$

$2x^{2} -7x +3-18= 18 -18$

$2x^{2} -7x -15= 0$

Use the method to factor the quadratic expression $2x^{2} -7x -15$ ; we are looking to split the linear term by finding two integers whose sum is and whose product is 2 (-15) = -30 . These integers are -10,3 , so:

$(2x^{2} -10x)+(3x -15)= 0$

2x (x -5)+3(x -5)= 0

$\left (2x +3 \right) \left ( x-5 \right )= 0$

Set each expression equal to 0 and solve:

2x +3= 0

2x = -3

$x = -\frac{3}{2} = -1\frac{1}{2}$

x-5= 0

x=5

The solution set is $\left \{ - 1 \frac{1}{2}, 5\right \}$ .

Report an Error

Example Question #7 : How To Find The Solution To An Equation

Solve for :

$x^{2} + 12 = 8x$

Give all solutions.

Possible Answers:

$x = 6 \textrm{ or }x = 8$

$x = -6 \textrm{ or }x =6$

$x = 2 \textrm{ or }x = 6$

$x = -8 \textrm{ or }x = 8$

$x = -6 \textrm{ or }x =- 2$

Correct answer:

$x = 2 \textrm{ or }x = 6$

Explanation:

Rewrite this quadratic equation in standard form:

$x^{2} + 12 = 8x$

$x^{2} -8x + 12 = 8x-8x$

$x^{2} -8x + 12 = 0$

Factor the expression on the left. We want two integers whose sum is and whose product is . These numbers are -6,-2 , so the equation becomes

(x-6)(x-2)=0 .

Set each factor equal to 0 separately, then solve:

$x - 6 = 0 \Rightarrow x = 6$

$x - 2 = 0 \Rightarrow x = 2$

Report an Error

Example Question #1 : How To Find The Solution To An Equation

Solve for :

4 (x + 6) - 3 (x + 8) = 2 (x + 4) + 2 (x - 4)

Possible Answers:

x =4

The equation has the set of all real numbers as its solution set.

x = 0

x = 2

The equation has no solution.

Correct answer:

x = 0

Explanation:

Simplify both sides, then solve:

4 (x + 6) - 3 (x + 8) = 2 (x + 4) + 2 (x - 4)

$4 \cdot x + 4 \cdot 6 - 3 \cdot x - 3 \cdot 8 = 2 \cdot x + 2 \cdot 4 +2 \cdot x - 2 \cdot 4$

4x + 24 - 3 x - 24 = 2x +8 +2 x -8

4x - 3 x+ 24 - 24 = 2x +2 x +8-8

$\left ( 4 - 3 \right ) x= \left ( 2 +2 \right ) x$

x= 4 x

x - 4x = 4 x - 4x

$\left (1 - 4 \right ) x = 0$

-3 x = 0

x = 0

Report an Error

Example Question #9 : How To Find The Solution To An Equation

3 (x + 2) + 4x = 5x + 2 (x +3)

Possible Answers:

x = -2

The equation has the set of all real numbers as its solution set.

x = 0

x = 6

The equation has no solution.

Correct answer:

The equation has the set of all real numbers as its solution set.

Explanation:

Simplify both sides, then solve:

3 (x + 2) + 4x = 5x + 2 (x +3)

$3 \cdot x + 3 \cdot 2 + 4x = 5x + 2 \cdot x + 2 \cdot 3$

3x + 6 + 4x = 5x + 2x + 6

3x + 4x + 6 = 5x + 2x + 6

$\left ( 3 + 4 \right ) x + 6 = \left ( 5 + 2 \right ) x + 6$

7x + 6 =7 x + 6

This is an identically true statement, so the original equation has the set of all real numbers as its solution set.

Report an Error

Example Question #10 : How To Find The Solution To An Equation

Solve for :

W = 5xp + 3q

Possible Answers:

$p = \frac{W-3q}{5x}$

$p = \frac{W- 5x }{3q}$

$p = \frac{W}{3q }-5x$

$p = \frac{W}{5x}-3q$

$p = W-\frac{3q}{5x}$

Correct answer:

$p = \frac{W-3q}{5x}$

Explanation:

5xp + 3q = W

5xp + 3q -3q = W-3q

5xp = W-3q

$5xp \div 5x =\left ( W-3q \right ) \div 5x$

$p =\frac{ W-3q }{5x}$

Report an Error

← Previous 1 2 3 4 5 6 7 8 9 … 32 33 Next →

View Tutors

Jennifer
Certified Tutor

McKendree University, Bachelor in Business Administration, Accounting and Business Management. Lindenwood University, Masters...

View Tutors

Myung Hee
Certified Tutor

Pukyung National University, Bachelor of Science, Japanese Studies. University of Padova, Master of Science, International Bu...

View Tutors

Gregory
Certified Tutor

Pennsylvania State University-Main Campus, Bachelor of Science, Chemical Engineering. Carnegie Mellon University, Doctor of P...

All ISEE Upper Level Math Resources

6 Diagnostic Tests 244 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Calculus Tutors in Los Angeles, Reading Tutors in Phoenix, SAT Tutors in Chicago, GMAT Tutors in Los Angeles, SSAT Tutors in Dallas Fort Worth, English Tutors in Los Angeles, SSAT Tutors in Seattle, Reading Tutors in Los Angeles, Reading Tutors in Washington DC, SSAT Tutors in San Francisco-Bay Area

Popular Courses & Classes

ACT Courses & Classes in Seattle, MCAT Courses & Classes in Seattle, GRE Courses & Classes in Miami, SAT Courses & Classes in Atlanta, Spanish Courses & Classes in Seattle, GMAT Courses & Classes in Washington DC, ACT Courses & Classes in Phoenix, Spanish Courses & Classes in Miami, SSAT Courses & Classes in Seattle, SSAT Courses & Classes in Washington DC

Popular Test Prep

ACT Test Prep in Denver, LSAT Test Prep in Washington DC, GRE Test Prep in Phoenix, MCAT Test Prep in Denver, MCAT Test Prep in Atlanta, GMAT Test Prep in Washington DC, LSAT Test Prep in New York City, GRE Test Prep in Washington DC, ACT Test Prep in Seattle, MCAT Test Prep in Houston

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

ISEE Upper Level Math : Algebraic Concepts

Study concepts, example questions & explanations for ISEE Upper Level Math

All ISEE Upper Level Math Resources

Example Questions

Example Question #1 : How To Find The Solution To An Equation

Example Question #1 : How To Find The Solution To An Equation

Example Question #3 : How To Find The Solution To An Equation

Example Question #1 : Algebraic Concepts

Example Question #5 : How To Find The Solution To An Equation

Example Question #6 : How To Find The Solution To An Equation

Example Question #7 : How To Find The Solution To An Equation

Example Question #1 : How To Find The Solution To An Equation

Example Question #9 : How To Find The Solution To An Equation

Example Question #10 : How To Find The Solution To An Equation

All ISEE Upper Level Math Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: