Solving systems of linear equations: 3s+2t-3=c and -7s-5t=4

[Illvoices]

Hello teachers and fellow colleagues U have a question about the linear equation shown below.
3s+2t-3=c
-7s-5t=4

I wanted to know how'd the solution went from -3/-5 to 3/5 in solving the second part of this linear equation.

The rule was that s = -1

Good luck and please get back to me fellow comrades.

 

[JeffM]

Hello teachers and fellow colleagues U have a question about the linear equation shown below.
3s+2t-3=c
-7s-5t=4

I wanted to know how'd the solution went from -3/-5 to 3/5 in solving the second part of this linear equation.

The rule was that s = -1

Good luck and please get back to me fellow comrades.

\(\displaystyle -\ a \equiv (-\ 1) * a \implies\)

\(\displaystyle \dfrac{-\ 3}{-\ 5} = \dfrac{(-\ 1) * 3}{(-\ 1) * 5} = \dfrac{-\ 1}{-\ 1} * \dfrac{3}{5} = 1 * \dfrac{3}{5} = \dfrac{3}{5}.\)

 

[Illvoices]

So what you are trying to implicate is that the negative constant -s is being multiplied by the negative fraction so two negatives equal a positive. I think I get it now

Thank s for your most valuable effort and until next time comrade.

 

[Steven G]

So what you are trying to implicate is that the negative constant -s is being multiplied by the negative fraction so two negatives equal a positive. I think I get it now

Thank s for your most valuable effort and until next time comrade.

JeffM gave an excellent proof to you. Once you understand that proof, you can then remember that a neg divided by a neg is a pos.