Big Ideas Math A Bridge To Success Algebra 2: Student Edition 2015

1st Edition

ISBN: 9781680331165

Author: HOUGHTON MIFFLIN HARCOURT

Publisher: Houghton Mifflin Harcourt

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Question

Chapter 3, Problem 25CR

To determine

To calculate: The solution of the system of equations by any method with explanation.

Expert Solution & Answer

Answer to Problem 25CR

The required solutions are (3−3910,1+33910) and (3+3910,1−33910) .

Explanation of Solution

Given information:

The system of equations:

x2+y2=4

−15x+5=5y

Calculation:

The given system of equation is

x2+y2=4........... (1)

−15x+5=5y..........(2)

Write the equation (2) as:

−315x5+55=5y5−3x+1=y.......... (3)

Substitute −3x+1 for y in equation (1).

x2+(−3x+1)2=4x2+9x2−6x+1=410x2−6x+1=410x2−6x−3=0

Solve above quadratic equation by using the quadratic formula x=−b±b2−4ac2a .

Substitute 10 for a , −6 for b and −3 for c in the above formula and simplify for x .

x=−(−6)±(−6)2−4(10)(−3)2(10)=6±36+12020=6±15620=3±3910

Now, substitute 3+3910 for x in the equation (3) and simplify for y .

−3(3+3910)+1=y−9−339+1010=yy=1−33910

Again, substitute 3−3910 for x in the equation (3) and simplify for y .

−3(3−3910)+1=y−9+339+1010=yy=1+33910

Thus, the solutions are x=3−3910 , y=1+33910 and x=3+3910 , y=1−33910 .

Hence, the required solutions are (3−3910,1+33910) and (3+3910,1−33910) .

Chapter 3, Problem 24CR

Chapter 3, Problem 26CR

Chapter 3 Solutions

Big Ideas Math A Bridge To Success Algebra 2: Student Edition 2015

Ch. 3.1 - Prob. 1E Ch. 3.1 - Prob. 2E Ch. 3.1 - Prob. 3E Ch. 3.1 - Prob. 4E Ch. 3.1 - Prob. 5E Ch. 3.1 - Prob. 6E Ch. 3.1 - Prob. 7E Ch. 3.1 - Prob. 8E Ch. 3.1 - Prob. 9E Ch. 3.1 - Prob. 10E

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