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If the system of equations

\(\begin{align*} 3x+y&=a,\\ 2x+5y&=2a, \end{align*}\)

has a solution  when \($(x,y)$\), compute \(a\)

 Feb 15, 2021
 #1
avatar+120287 
0

Is  this question missing something   ????

 

 

 

cool cool cool

 Feb 15, 2021
 #4
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Yeah sorrry

 

X is 2

Guest Feb 15, 2021
 #2
avatar+32894 
+1

Multiply first equation by -5   to get

-15x -5y = -5a    add to second equation to get

 

-13x  = -3a

x = 3/13 a

 

 

Now multiply second equation by  - 3/2 to get

- 3x - 15/2 y = - 3a      add to first equation to get

-13/2 y = -2a

    y = 4/13 a                      sub red equations into equation one

 

9/13a + 4/13 a = a      check !           so pick any value for a       say  13      then    x = 3    y = 4

 

or if a = 26     then x = 6     y = 8      etc      sooo there is a LOT of answers unless you restrict them somehow !

 Feb 15, 2021
edited by ElectricPavlov  Feb 15, 2021
 #3
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Yeh lel sorry.

 

X is 2

Guest Feb 15, 2021
 #7
avatar+32894 
+1

Welll.... we found that

x = 3/13 a

2 = 3/13 a

a = 26/3

ElectricPavlov  Feb 15, 2021
 #5
avatar+120287 
+2

3(2)  +  y   = a     →    6  + y   =  a      →    y  =  a  - 6      (1) 

2(2)   + 5y   = 2a  →   4   +  5y  = 2a       (2)

 

Sub (1)  into  (2)  for   y

 

4  +  5 ( a  - 6)  =  2a

 

4 + 5a  - 30   = 2a

 

-26  + 5a  =  2a

 

-26  =  2a  - 5a

 

-26  =  -3a

 

a =  26  / 3

 

 

cool cool cool

 Feb 15, 2021
 #6
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0

Thats correct! Thanks so much!

Guest Feb 15, 2021

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