   Chapter 4.1, Problem 3E Finite Mathematics and Applied Cal...

7th Edition
Stefan Waner + 1 other
ISBN: 9781337274203

Solutions

Chapter
Section 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] 3 x + 4 y = 2

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 3x+4y=2.

Explanation

Given Information:

The provided equation is 3x+4y=2.

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,

3x+4y=2

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

3x+4y=24y=23xy=23x4

Substitute this expression of y in the solution (x,y). Hence the general solution of the equation 3x+4y=2 when parameterized by x is (x,23x4).

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

3x+4y=23x=24yx=24y3

Substitute this expression of x in the solution (x,y). Hence the general solution of the equation 3x+4y=2 when parameterized by y is (24y3,y).

The particular solution of a linear equation is obtained by selecting specific values for x and y which satisfies the equation 3x+4y=2

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