Chapter 4.1, Problem 5E

### Finite Mathematics and Applied Cal...

7th Edition
Stefan Waner + 1 other
ISBN: 9781337274203

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] 4 x = − 5

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 4x=5.

Explanation

Given Information:

The provided equation is 4x=ā5.

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,cāaxb) and, when parameterized by y has the form (cābya,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,

4x=ā5

To find the general solution of the equation when parameterized by x, express y as a function of x in the given equation. Here y cannot be expressed as a function of x, so the general solution cannot be parameterized by x.

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

x=ā54

Substitute this expression of x in the solution (x,y). Hence the general solution of the equation 4x=ā5 when parameterized by y is (ā54,y).

The particular solution of a linear equation is obtained by selecting specific values for x and y which satisfies the equation. Here the given equation can be written as 4x+0y=ā5.

Substitute y=1 in the equation 4x+0y=ā5,

4x+0(1)=ā54x=ā5x=ā54

When y=1,Ā x=ā54

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