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 23CR

To determine

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

Expert Solution & Answer

Answer to Problem 23CR

The required solutions are (−2,6) and (1,0) .

Explanation of Solution

Given information:

The system of equations:

2x2−2=y

−2x+2=y

Calculation:

The given system of equation is

2x2−2=y..............(1)

−2x+2=y..............(2)

Compare each side of the above equations.

It is seen that the right hand sides of each equation are equal.

So, write the two equations as:

2x2−2=−2x+22x2+2x−2−2=02x2+2x−4=0

Divide each side of the above quadratic equation by 2.

2x22+2x2−422=02x2+x−2=0

Solve above quadratic equation by using the factorization method.

Split the middle term of the above equation.

x2+x−2=0x2+2x−x−2=0x(x+2)−1(x+2)=0(x+2)(x−1)=0

Solve the two factors for x .

x+2=0x=−2

And,

x−1=0x=1

Now, substitute −2 for x in the equation (2) and solve for y .

−2(−2)+2=yy=4+2y=6

Also, substitute 1 for x in the equation (2) and solve for y .

−2(1)+2=yy=−2+2y=0

Thus, the solutions are x=−2,y=6 and x=1,y=0 .

Hence, the required solutions are (−2,6) and (1,0) .

Chapter 3, Problem 22CR

Chapter 3, Problem 24CR

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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