We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

College Algebra : Non Linear Systems

Study concepts, example questions & explanations for College Algebra

Create An Account

All College Algebra Resources

5 Diagnostic Tests 84 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Non Linear Systems

Solve the following system of equations

$\\1: x^2+y^2=9\\2: 3x+y=1$

Possible Answers:

$x=\frac{3-\sqrt{89}}{10}$

$y=\frac{1+3\sqrt{89}}{10}$

$x=\frac{3+\sqrt{89}}{10}$

$x=\frac{3-\sqrt{89}}{10}$

$y=\frac{1+3\sqrt{89}}{10}$

$y=\frac{1-3\sqrt{89}}{10}$

There are no Real Solutions

$x=\frac{3+\sqrt{89}}{10}$

$x=\frac{3-\sqrt{89}}{10}$

$x=\frac{3+\sqrt{89}}{10}$

$y=\frac{1-3\sqrt{89}}{10}$

Correct answer:

$x=\frac{3+\sqrt{89}}{10}$

$x=\frac{3-\sqrt{89}}{10}$

$y=\frac{1+3\sqrt{89}}{10}$

$y=\frac{1-3\sqrt{89}}{10}$

Explanation:

First lets solve equation 2 for y.

$\\3x+y=1\\y=1-3x$

Now we plug in y=1-3x into equation 1.

x^2+(1-3x)^2=9

FOIL the left hand side, simplify and then subtract from each side.

$\\x^2+(1-3x)^2=9\\x^2+1-6x+9x^2=9\\10x^2-6x-8=0$

Now use the quadratic equation in order to solve for .

Recall the quadratic formula

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ , where a,b,c refer to the coefficients in the quadratic equation ax^2+bx+c=0 .

$x=\frac{6\pm\sqrt{(-6)^2-4(10)(-8)}}{2(10)}$

$x=\frac{6\pm\sqrt{36+320}}{20}$

$x=\frac{6\pm\sqrt{356}}{20}$

$x=\frac{6\pm\sqrt{4\cdot 89}}{20}$

$x=\frac{6\pm2\sqrt{89}}{20}$

So the solutions for x are as follows

$x=\frac{6+2\sqrt{89}}{20}=\frac{3+\sqrt{89}}{10}$

$x=\frac{6-2\sqrt{89}}{20}=\frac{3-\sqrt{89}}{10}$

Now we just plug these values of x into equation 2 to solve for y.

y=1-3x

$y=1-3(\frac{3+\sqrt{89}}{10})$

$y=\frac{10}{10}-(\frac{9+3\sqrt{89}}{10})=\frac{1+3\sqrt{89}}{10}$

y=1-3x

$y=1-3(\frac{3-\sqrt{89}}{10})$

$y=\frac{10}{10}-(\frac{9-3\sqrt{89}}{10})=\frac{1-3\sqrt{89}}{10}$

So the solutions to these pair of equations are:

$x=\frac{3+\sqrt{89}}{10}$

$x=\frac{3-\sqrt{89}}{10}$

$y=\frac{1+3\sqrt{89}}{10}$

$y=\frac{1-3\sqrt{89}}{10}$

Report an Error

Example Question #2 : Non Linear Systems

Solve the system of equations to find the points where the graphs of the equations intersect.

x+3=4y

$\frac{1}{4}(x^2+1)=y$

Possible Answers:

$\left(-1, 2\right)$ , $\left(\frac{1}{2}, \frac{5}{4} \right )$

$\left(\frac{3}{2}, -1\right)$ , $\left( 2, \frac{11}{2} \right )$

$\left(\frac{3}{7}, 1\right)$ , $\left( 2, \frac{3}{4} \right )$

$\left(-1, \frac{1}{2}\right)$ , $\left(2, \frac{5}{4} \right )$

$\left(2, 1\right)$ , $\left( 1, \frac{3}{4} \right )$

Correct answer:

$\left(-1, \frac{1}{2}\right)$ , $\left(2, \frac{5}{4} \right )$

Explanation:

Solve the system of equations,

x+3=4y (1)

$\frac{1}{4}(x^2+1)=y$ (2)

Notice that we can multiply both sides of equation (2) by to obtain,

x^2 +1 = 4y (3)

Now equate equations (3) and (4) to obtain,

x+3 = x^2 +1

This is a quadratic equation, rearrange to find the roots by setting to zero and factoring,

x^2-x-2=0

(x+1)(x-2)=0

$x = \left \{ -1, 2\right \}$

Use these values to find the corresponding values of by substituting into either of the original equations. Using equation (1),

$y_1 = \frac{1}{4}(-1+3)=\frac{1}{2}$

$y_2 =\frac{1}{4}(2+3)=\frac{5}{4}$

$y = \left \{ \frac{1}{2}, \frac{5}{4}\right \}$

The points $\left(-1, \frac{1}{2}\right)$ and $\left(2, \frac{5}{4} \right )$ are the intersection points of the two functions,

$y=\frac{1}{4}(x+3)$

$y=\frac{1}{4}(x^2+1)$

Intersection

Report an Error

Example Question #3 : Non Linear Systems

Solve the following system of equations:

$y=x^{2}+1$

y=2x+1

Possible Answers:

(0,1)(5,2)

(0,1)(2,5)

(1,5)(2,1)

(0,5)(1,2)

Correct answer:

(0,1)(2,5)

Explanation:

$y=x^{2}+1$

y=2x+1

Subtract the second equation from the first, to eliminate the y terms:

$y=x^{2}+1$

- y=2x+1

This yields the following: $0=x^{2}-2x$

Factor out an x from both terms: 0=x(x-2)

Solve for x: x=0,2

Plug in the values for x into the first equation:

$y=(0)^{2}+1=1$

$y=(2)^{2}+1=4+1=5$

Therefore, when x=0 , y=1 , and when x=2 , y=5

Plug in the values for x and y to make sure they satisfy the second equation too:

1=2(0)+1=0+1=1

5=2(2)+1=4+1=5

Solution: (0,1)(2,5)

Report an Error

Example Question #2 : Non Linear Systems

Solve the following system of equations:

y=x-1

$x=y^{2}$

Possible Answers:

$(\frac{3+\sqrt{5}}{2}, \frac{1-\sqrt{5}}{2})(\frac{3-\sqrt{5}}{2},\frac{1+\sqrt{5}}{2})$

$(\frac{-3+\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2})(\frac{-3-\sqrt{5}}{2},\frac{1-\sqrt{5}}{2})$

$(\frac{3+\sqrt{5}}{3}, \frac{1+\sqrt{5}}{3})(\frac{3-\sqrt{5}}{3},\frac{1-\sqrt{5}}{3})$

$(\frac{3+\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2})(\frac{3-\sqrt{5}}{2},\frac{1-\sqrt{5}}{2})$

Correct answer:

$(\frac{3+\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2})(\frac{3-\sqrt{5}}{2},\frac{1-\sqrt{5}}{2})$

Explanation:

y=x-1 $x=y^{2}$

Substitute the value of x from the second equation, into the first equation: $y=(y^{2})-1$

Subtract y from both sides of the equation: $0=y^{2}-1-y$

Use the quadratic formula to solve for y: $y=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}=\frac{1\pm \sqrt{1-4(1)(-1)}}{2}=\frac{1\pm \sqrt{5}}{2}$

$y=\frac{1+\sqrt{5}}{2}$ or $\frac{1-\sqrt{5}}{2}$

Plug the first value of y into the first equation: $\frac{1+\sqrt{5}}{2}=x-1$

Re-write the equation to make it easier to solve for x: $\frac{1+\sqrt{5}}{2}=x-\frac{2}{2}$

Add $\frac{2}{2}$ to each side of the equation: $x=\frac{3+\sqrt{5}}{2}$

The second value of y into the first equation: $\frac{1-\sqrt{5}}{2}=x-1$

Re-write the equation to make it easier to solve for x: $\frac{1-\sqrt{5}}{2}=x-\frac{2}{2}$

Add $\frac{2}{2}$ to each side of the equation: $x=\frac{3-\sqrt{5}}{2}$ .

Plug the values of x and y into the second equation to make sure they satisfy both of them:

$(\frac{1+\sqrt{5}}{2})^{2}=\frac{1+2\sqrt{5}+5}{4}=\frac{6+2\sqrt{5}}{4}=\frac{3+\sqrt{5}}{2}$

$(\frac{1-\sqrt{5}}{2})^{2}=\frac{1-2\sqrt{5}}{4}=\frac{1-2\sqrt{5}+5}{4}=\frac{6-2\sqrt{5}}{4}=\frac{3-\sqrt{5}}{2}$

Solution: $(\frac{3+\sqrt{5}}{2},\frac{1+\sqrt{5}}{2})(\frac{3-\sqrt{5}}{2},\frac{1-\sqrt{5}}{2})$

Report an Error

Example Question #3 : Non Linear Systems

Solve the following system of equations:

y=x-1

$\frac{x^{2}}{3}+\frac{y^{2}}{2}=1$

Possible Answers:

$(\frac{2\sqrt{6}-3}{5},\frac{2\sqrt{6}-2}{5})(\frac{3-2\sqrt{6}}{5},\frac{-2\sqrt{6}-2}{5})$

$(\frac{5+2\sqrt{6}}{5},\frac{2\sqrt{6}-2}{5})(\frac{3-2\sqrt{6}}{5},\frac{-2\sqrt{6}-2}{5})$

$(\frac{3+2\sqrt{6}}{5},\frac{2\sqrt{6}-2}{5})(\frac{3-2\sqrt{6}}{5},\frac{-2\sqrt{6}-2}{5})$

$(\frac{3+2\sqrt{6}}{5},\frac{3\sqrt{6}-2}{5})(\frac{3-2\sqrt{6}}{5},\frac{-2\sqrt{6}-2}{5})$

Correct answer:

$(\frac{3+2\sqrt{6}}{5},\frac{2\sqrt{6}-2}{5})(\frac{3-2\sqrt{6}}{5},\frac{-2\sqrt{6}-2}{5})$

Explanation:

y=x-1 $\frac{x^{2}}{3}+\frac{y^{2}}{2}=1$

Multiply the second equation by 6, to eliminate the denominator: $6(\frac{x^{2}}{3}+\frac{y^{2}}{2}=1)6$

Simplify: $2x^{2}+3y^{2}=6$

Plug the value of y, from the first equation, into the second equation: $2x^{2}+3(x-1)^{2}=6$

Simplify: $2x^{2}+3x^{2}-6x+3=6$

Simplify further: $5x^{2}-6x-3=0$

Use the quadratic formula to solve for x: $x=\frac{6\pm \sqrt{36-4(5)(-3)}}{10}=\frac{6\pm \sqrt{96}}{10}=\frac{6\pm 4\sqrt{6}}{10}=\frac{3\pm 2\sqrt{6}}{5}$

$x=\frac{3+2\sqrt{6}}{5}$ or $x=\frac{3-2\sqrt{6}}{5}$

Plug the first value of x into the first equation: $y=\frac{3+2\sqrt{6}}{5}-1=\frac{3+2\sqrt{6}}{5}-\frac{5}{5}=\frac{2\sqrt{6}-2}{5}$

Plug the second value of x into the first equation: $y=\frac{3-2\sqrt{6}}{5}-1=\frac{3-2\sqrt{6}}{5}-\frac{5}{5}=\frac{-2\sqrt{6}-2}{5}$

Plug the values of x and y into the second equation to make sure they satisfy both of them:

$\frac{(\frac{3+2\sqrt{6}}{5})^{2}}{3}+\frac{(\frac{2\sqrt{6}-2}{5})^{2}}{2}=(\frac{9+12\sqrt{6}+24}{25})(\frac{1}{3})+(\frac{24-8\sqrt{6}+4}{5})(\frac{1}{2})=$

$\frac{11+4\sqrt{6}}{25}+\frac{14-4\sqrt{6}}{25}=\frac{25}{25}=1$

$\frac{(\frac{3-2\sqrt{6}}{5})^{2}}{3}+\frac{(\frac{-2\sqrt{6}-2}{5})^{2}}{2}=(\frac{9-12\sqrt{6}+24}{25})(\frac{1}{3})+(\frac{24+8\sqrt{6}+4}{25})(\frac{1}{2})=\frac{11-4\sqrt{6}}{25}+\frac{14+4\sqrt{6}}{25}=\frac{25}{25}=1$

Solution: $(\frac{3+2\sqrt{6}}{5},\frac{2\sqrt{6}-2}{5})(\frac{3-2\sqrt{6}}{5},\frac{-2\sqrt{6}-2}{5})$

Report an Error

Example Question #2 : Non Linear Systems

Solve the following system of equations:

$x=2y^{2}$

y=-x+1

Possible Answers:

$(2,-1)(\frac{-1}{2},\frac{1}{2})$

$(-2,-1)(\frac{1}{2},\frac{1}{2})$

$(2,-1)(\frac{1}{2},\frac{1}{2})$

$(2,1)(\frac{1}{2},\frac{1}{2})$

Correct answer:

$(2,-1)(\frac{1}{2},\frac{1}{2})$

Explanation:

$x=2y^{2}$

y=-x+1

Substitute the value of y from the second equation, into the first equation:

$x=2y^{2}=2(-x+1)^{2}=2(x^{2}-2x+1)=2x^{2}-4x+2$

Subtract x from both sides of the equation:

$2x^{2}-5x+2=0$

Use the quadratic formula to solve for x: $x=\frac{5\pm \sqrt{25-4(2)(2)}}{4}=\frac{5\pm \sqrt{25-16}}{4}=\frac{5\pm \sqrt{9}}{4}=\frac{5\pm 3}{4}$

x=2 or $x=\frac{1}{2}$

Plug the first value of x into the second equation: y=-(2)+1=-1

Plug the second value of x into the second equation: $y=-(\frac{1}{2})+1=\frac{1}{2}$

x=2,y=-1 or $x=\frac{1}{2},y=\frac{1}{2}$

Verify the first solution by plugging the first values of x and y into both equations:

$(2)=2(-1)^{2}=2$

(-1)=-(2)+1=-2+1=-1

Verify the second solution by plugging the second values of x and y into both equations:

$(\frac{1}{2})=2(\frac{1}{2})^{2}=2(\frac{1}{4})=\frac{1}{2}$

$(\frac{1}{2})=-(\frac{1}{2})+1=\frac{1}{2}$

Solution:

$(2,-1)(\frac{1}{2},\frac{1}{2})$

Report an Error

View College Algebra Tutors

Philipp
Certified Tutor

University of Nevada-Las Vegas, Bachelor of Science, Mechanical Engineering.

View College Algebra Tutors

Ezam
Certified Tutor

University of the West Indies, Bachelor of Science, Mathematics and Computer Science. University of Phoenix-Central Valley Ca...

View College Algebra Tutors

Yucheng
Certified Tutor

The University of Texas at Austin, Bachelor of Science, Mechanical Engineering.

All College Algebra Resources

5 Diagnostic Tests 84 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

ISEE Tutors in Philadelphia, LSAT Tutors in New York City, Math Tutors in San Diego, Biology Tutors in Phoenix, English Tutors in Philadelphia, Computer Science Tutors in San Diego, English Tutors in Los Angeles, GRE Tutors in Philadelphia, ISEE Tutors in San Diego, ACT Tutors in Washington DC

Popular Courses & Classes

SAT Courses & Classes in New York City, SSAT Courses & Classes in Washington DC, ACT Courses & Classes in Washington DC, GRE Courses & Classes in Washington DC, GMAT Courses & Classes in Atlanta, SAT Courses & Classes in Denver, ISEE Courses & Classes in Seattle, SSAT Courses & Classes in New York City, SAT Courses & Classes in Houston, SSAT Courses & Classes in Los Angeles

Popular Test Prep

MCAT Test Prep in San Francisco-Bay Area, GRE Test Prep in Washington DC, SSAT Test Prep in Seattle, LSAT Test Prep in Dallas Fort Worth, SSAT Test Prep in Denver, SSAT Test Prep in San Diego, SAT Test Prep in San Francisco-Bay Area, ACT Test Prep in Seattle, SSAT Test Prep in Dallas Fort Worth, GRE 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

College Algebra : Non Linear Systems

Study concepts, example questions & explanations for College Algebra

All College Algebra Resources

Example Questions

Example Question #1 : Non Linear Systems

Example Question #2 : Non Linear Systems

Example Question #3 : Non Linear Systems

Example Question #2 : Non Linear Systems

Example Question #3 : Non Linear Systems

Example Question #2 : Non Linear Systems

All College Algebra Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: