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?
+128475 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
+14915 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.
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