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Finite Mathematics and Applied Cal...

7th Edition
Stefan Waner + 1 other
ISBN: 9781337274203

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Chapter
Section
BuyFindarrow_forward

Finite Mathematics and Applied Cal...

7th Edition
Stefan Waner + 1 other
ISBN: 9781337274203
Textbook Problem

In Exercises 1-6, find three different particular solutions of the given equation and also its general solution in two forms (if possible): parameterized by x and parameterized by y.

[HINT: See Example 1 and Quick Examples 2-4]

x + 3 y = 3

To determine

To calculate: The general solution in two forms: parameterized by x and parameterized by y

and also three different particular solutions of the equation x+3y=3.

Explanation

Given Information:

The provided equation is x+3y=3.

Formula used:

For any real numbers a, b, and c, the general solution of a linear equation of the form ax+by=c, when parameterized by x has the form (x,caxb) and, when parameterized by y has the form (cbya,y).

For any real numbers a, b, and c, the particular solution of a linear equation of the form ax+by=c is obtained by selecting specific values for x and y which satisfies the equation.

Calculation:

Consider the provided equation,

x+3y=3

To find the general solution of the equation when parameterized by x, express y as a function of x.

x+3y=33y=3xy=3x3

Substitute this expression of y in the solution (x,y). Hence the general solution of the equation x+3y=3 when parameterized by x is (x,3x3).

To find the general solution of the equation when parameterized by y, express x as a function of y.

x+3y=3x=33y

Substitute this expression of x in the solution (x,y). Hence the general solution of the equation x+3y=3 when parameterized by y is (33y,y).

Recall that the particular solution of a linear equation is obtained by selecting specific values for x and y which satisfies the equation x+3y=3

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