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valina [46]
3 years ago
14

Find the general solution of the simple homogeneous "system" below, which consists of a single linear equation. Give your answer

as a linear combination of vectors. Let x2 and x3 be free variables. 3x1 - 6x2 9x3
Mathematics
1 answer:
JulsSmile [24]3 years ago
8 0

Answer:

=  \left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right] = x_2 \left[\begin{array}{ccc}2\\1\\0\end{array}\right] + x_3 \left[\begin{array}{ccc}-3\\0\\1\end{array}\right]

Step-by-step explanation:

Given:  3x1 - 6x2 + 9x3 = 0

x2 and x3 are free variables

We have:

3x1 = 6x2 - 9x3

divide all sides by 3, we have:

x1 = 2x2 - 3x3

Finding the general solution, we have:

\left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right] = \left[\begin{array}{ccc}2x_2 - 3x_3\\x_2\\x_3\end{array}\right]

= \left[\begin{array}{ccc}2x_2\\x_2\\0\end{array}\right] + \left[\begin{array}{ccc}-3x_3\\0\\x_3\end{array}\right]

= x_2 \left[\begin{array}{ccc}2\\1\\0\end{array}\right] + x_3 \left[\begin{array}{ccc}-3\\0\\1\end{array}\right]

The general solution is

=  \left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right] = x_2 \left[\begin{array}{ccc}2\\1\\0\end{array}\right] + x_3 \left[\begin{array}{ccc}-3\\0\\1\end{array}\right]

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Leya [2.2K]

Answer:

      x = 32

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

                    5/8*x+4-(3/8*x+12)=0

Step by step solution :

Step  1  :

           3

Simplify   —

           8

Equation at the end of step  1  :

   5         3

 ((—•x)+4)-((—•x)+12)  = 0

   8         8

Step  2  :

Rewriting the whole as an Equivalent Fraction :

2.1   Adding a whole to a fraction

Rewrite the whole as a fraction using  8  as the denominator :

         12     12 • 8

   12 =  ——  =  ——————

         1        8  

Equivalent fraction : The fraction thus generated looks different but has the same value as the whole

Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator

Adding fractions that have a common denominator :

2.2       Adding up the two equivalent fractions

Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

3x + 12 • 8     3x + 96

———————————  =  ———————

     8             8  

Equation at the end of step  2  :

   5               (3x + 96)

 ((— • x) +  4) -  —————————  = 0

   8                   8    

Step  3  :

           5

Simplify   —

           8

Equation at the end of step  3  :

   5               (3x + 96)

 ((— • x) +  4) -  —————————  = 0

   8                   8    

Step  4  :

Rewriting the whole as an Equivalent Fraction :

4.1   Adding a whole to a fraction

Rewrite the whole as a fraction using  8  as the denominator :

        4     4 • 8

   4 =  —  =  —————

        1       8  

Adding fractions that have a common denominator :

4.2       Adding up the two equivalent fractions

5x + 4 • 8     5x + 32

——————————  =  ———————

    8             8  

Equation at the end of step  4  :

 (5x + 32)    (3x + 96)

 ————————— -  —————————  = 0

     8            8    

Step  5  :

Step  6  :

Pulling out like terms :

6.1     Pull out like factors :

  3x + 96  =   3 • (x + 32)

Adding fractions which have a common denominator :

6.2       Adding fractions which have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

(5x+32) - (3 • (x+32))     2x - 64

——————————————————————  =  ———————

          8                   8  

Step  7  :

Pulling out like terms :

7.1     Pull out like factors :

  2x - 64  =   2 • (x - 32)

Equation at the end of step  7  :

 2 • (x - 32)

 ————————————  = 0

      8      

Step  8  :

When a fraction equals zero :

8.1    When a fraction equals zero ...

Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.

Now,to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator.

Here's how:

 2•(x-32)

 ———————— • 8 = 0 • 8

    8    

Now, on the left hand side, the  8  cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :

  2  •  (x-32)  = 0

Equations which are never true :

8.2      Solve :    2   =  0

This equation has no solution.

A a non-zero constant never equals zero.

Solving a Single Variable Equation :

8.3      Solve  :    x-32 = 0

Add  32  to both sides of the equation :

                     x = 32

One solution was found :

                  x = 32

Step-by-step explanation:

6 0
3 years ago
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