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For what value of the constant $c$ does the system \begin{align*} 3x + 4y &= 7,\\ 6x + 4y &= c- 4y, \end{align*}have infinitely many solutions?

Guest Oct 23, 2017
 #1
avatar+88898 
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3x + 4y = 7

6x + 4y  = c - 4y  →  6x + 8y  =  c

 

The second equation will be a multiple of the first when c  =  14

 

And this value of c will result  in infinite solutions  for the system

 

 

cool cool cool 

CPhill  Oct 23, 2017
 #2
avatar+7451 
+1

My mind is blown

For what value of the constant $c$ does the system \begin{align*} 3x + 4y &= 7,\\ 6x + 4y &= c- 4y, \end{align*}have infinitely many solutions?

 

Hello guest!

 

\( 3x + 4y = 7,\\ 6x + 4y = c- 4y \)

 

\(3x+4y=\frac{c}{2}\\ \frac{c}{2}=7\\ \color{blue} c=14 \)

 

At c = 14 this system of equations has infinitely many results.

laugh  !

asinus  Oct 23, 2017

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