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avatar+681 

Solve the system of equations 
10b+5v+5z=55
3b+5v+5z=34
5b+6v+3z=42 
I have gotten b=3 but I am stuck and Do not know how to proceed. 

 Sep 15, 2019
 #1
avatar+106539 
+2

You're on the right track  !!!!

 

10b+5v+5z=55   ⇒  5v + 5z  =  55 - 10b    (1)
3b+5v+5z=34      ⇒  5v + 5z  = 34  - 3b    (2)
5b+6v+3z=42         (3)

 

Note that   (1)  must = (2)....therefore

 

55 - 10b  =  34 - 3b   rearrange as

55-34  = 7b

21 = 7b 

b = 3

 

Knowing this  (1)  becomes 

10(3) + 5v + 5z  = 55

30 + 5v + 5z = 55

5v + 5z  = 25    [divide through by 5 ]

v + z =  5     →    z = 5 - v     (4)

 

Sub (4)  into (3)  and sub 3 for b   and we have that

 

5(3) + 6v + 3(5 - v )  = 42

15 + 6v + 15 - 3v  = 42

30 + 3v  = 42

3v = 12

v = 4

 

So  z = 5 - v  =  5 - 4   = 1

 

So    { b, v , z}   =  ( 3, 4 , 1}

 

 

cool cool cool

 Sep 15, 2019
 #2
avatar+681 
+1

thanks !! :)

Nirvana  Sep 16, 2019

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