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Sagot :
We will assume that we want to know if the system of equations is independent or dependent:
[tex]\begin{cases}5x-3y=10\text{ (1)} \\ 6y=kx-42\text{ (2)}\end{cases}[/tex]where k is a real number. We will try to find the solutions to the system, and we will try to give values to k for which the system becomes independent or dependent.
We will use substitution, we solve for the variable x on the first equation to obtain:
[tex]\begin{gathered} 5x-3y=10 \\ 5x=10+3y \\ x=\frac{10+3y}{5} \end{gathered}[/tex]And now we replace it onto the second equation:
[tex]\begin{gathered} 6y=k(\frac{10+3y}{5})-42 \\ 6y=\frac{10k+3ky}{5}-42 \\ 6y=\frac{10k+3ky-210}{5} \\ 30y=10k+3ky-210 \\ 30y-3ky=10k-210 \\ y(30-3k)=10k-210 \\ y=\frac{10k-210}{30-3k} \end{gathered}[/tex]And the value of x will be:
[tex]\begin{gathered} x=\frac{10+3(\frac{10k-210}{30-3k})}{5}=\frac{10+\frac{10k-210}{10-k}}{5} \\ =\frac{10(10-k)+10k-210}{5(10-k)} \\ =\frac{100-10k+10k-210}{50-5k} \\ =-\frac{110}{50-5k} \\ =-\frac{22}{10-k} \end{gathered}[/tex]This means that a solution of the system will be:
[tex]\begin{cases}x=-\frac{55}{10-k} \\ y=\frac{10k-210}{30-3k}\end{cases}[/tex]Now, for finding the values which make the system dependent. This happens when the lines have the same slope, this is, when:
[tex]\begin{gathered} \frac{-5}{-3}=\frac{k}{6} \\ \frac{5}{3}=\frac{k}{6} \\ 30=3k \\ 10=k \end{gathered}[/tex]We did the division of the opposite of the coefficient of x, over the coefficient of y. This means that the system will be independent for each value of k different than 10, and will be dependent for k=10.
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