Precalculus with Limits
Precalculus with Limits
3rd Edition
ISBN: 9781133947202
Author: Ron Larson
Publisher: Cengage Learning
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Chapter 7.1, Problem 7E
To determine

To find : the solution to the given system of equation using substitution method

Expert Solution & Answer
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Answer to Problem 7E

The solution to the system of equation is (2,  2)

Explanation of Solution

Given information : The system of equation is {2x+y=6−x+y=0   

  Precalculus with Limits, Chapter 7.1, Problem 7E , additional homework tip 1

Concept Involved:

Solution of a system of equation is the point which makes both the equation TRUE.

Graphically the solution to the system of equation is the point where the two lines meet.

Method of Substitution:

1. Solve one of the equations for one variable in terms of the other.

2. Substitute the expression found in Step 1 into the other equation to obtain anequation in one variable.

3. Solve the equation obtained in Step 2.

4. Back-substitute the value obtained in Step 3 into the expression obtained inStep 1 to find the value of the other variable.

5. Check that the solution satisfies each of the original equations.

Calculation:

    DescriptionSteps
    Label the given equation2x+y=6    →1stEquation−x+y=0   →2ndEquation
    Solve the 2nd equation for variable y Add x  on both sides of the equation −x+y=0−x+y+x=0+xy=x
    Substitute x for y in the 1st equation2x+x=6
    Combine like terms in left side of the equation3x=6
    Dividing 3 on both sides3x3=63
    Simplify fraction on both sides of the equationx=2
    Substitute 2 for x in the equation y=xy=2

Conclusion:

The solution to the given system of equation is (2, 2) and it matches with the given graph.

  Precalculus with Limits, Chapter 7.1, Problem 7E , additional homework tip 2

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Chapter 7.1, Problem 6E
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Chapter 7.1, Problem 8E

Chapter 7 Solutions

Precalculus with Limits

Ch. 7.1 - Prob. 1ECh. 7.1 - Prob. 2ECh. 7.1 - Prob. 3ECh. 7.1 - Prob. 4ECh. 7.1 - Prob. 5ECh. 7.1 - Prob. 6ECh. 7.1 - Prob. 7ECh. 7.1 - Prob. 8ECh. 7.1 - Prob. 9ECh. 7.1 - Prob. 10E
Ch. 7.1 - Prob. 11ECh. 7.1 - Prob. 12ECh. 7.1 - Prob. 13ECh. 7.1 - Prob. 14ECh. 7.1 - Prob. 15ECh. 7.1 - Prob. 16ECh. 7.1 - Prob. 17ECh. 7.1 - Prob. 18ECh. 7.1 - Prob. 19ECh. 7.1 - Prob. 20ECh. 7.1 - Prob. 21ECh. 7.1 - Prob. 22ECh. 7.1 - Prob. 23ECh. 7.1 - Prob. 24ECh. 7.1 - Prob. 25ECh. 7.1 - Prob. 26ECh. 7.1 - Prob. 27ECh. 7.1 - Prob. 28ECh. 7.1 - Prob. 29ECh. 7.1 - Prob. 30ECh. 7.1 - Prob. 31ECh. 7.1 - Prob. 32ECh. 7.1 - Prob. 33ECh. 7.1 - Prob. 34ECh. 7.1 - Prob. 35ECh. 7.1 - Prob. 36ECh. 7.1 - Prob. 37ECh. 7.1 - Prob. 38ECh. 7.1 - Prob. 39ECh. 7.1 - Prob. 40ECh. 7.1 - Prob. 41ECh. 7.1 - Prob. 42ECh. 7.1 - Prob. 43ECh. 7.1 - Prob. 44ECh. 7.1 - Prob. 45ECh. 7.1 - Prob. 46ECh. 7.1 - Prob. 47ECh. 7.1 - Prob. 48ECh. 7.1 - Prob. 49ECh. 7.1 - Prob. 50ECh. 7.1 - Prob. 51ECh. 7.1 - Prob. 52ECh. 7.1 - Prob. 53ECh. 7.1 - Prob. 54ECh. 7.1 - Prob. 55ECh. 7.1 - Prob. 56ECh. 7.1 - Prob. 57ECh. 7.1 - Prob. 58ECh. 7.1 - Prob. 59ECh. 7.1 - Prob. 60ECh. 7.1 - Prob. 61ECh. 7.1 - Prob. 62ECh. 7.1 - Prob. 63ECh. 7.1 - 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