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anonymous
 one year ago
I need help.
A system of equations is shown below:
5x  5y = 10
3x  2y = 2
Part A: Create an equivalent system of equations by replacing one equation with the sum of that equation and a multiple of the other. Show the steps to do this.
Part B: Show that the equivalent system has the same solution as the original system of equations.
anonymous
 one year ago
I need help. A system of equations is shown below: 5x  5y = 10 3x  2y = 2 Part A: Create an equivalent system of equations by replacing one equation with the sum of that equation and a multiple of the other. Show the steps to do this. Part B: Show that the equivalent system has the same solution as the original system of equations.

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we can multiply the first equation by 2 and the second equation by 5, so we get the subsequent equivalent system: \[\Large \left\{ \begin{gathered} 10x  10y = 20 \hfill \\  15x + 10y =  10 \hfill \\ \end{gathered} \right.\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now we add those two equations together, so we get: \[\Large  5x = 10\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1so the requested equivalent system, in part A, can be this: \[\Large \left\{ \begin{gathered} 5x  5y = 10 \hfill \\  5x = 10 \hfill \\ \end{gathered} \right.\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So it would be  5x = 10 and that would be an equivalent system of the equation?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yay! For Part B, I just check my work an write that as the proof?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1another way to get an equivalent system is multipling the firs equation by 2/5, so Iget this: \[\left\{ \begin{gathered}  2x + 2y =  4 \hfill \\ 3x  2y = 2 \hfill \\ \end{gathered} \right.\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1then I add the two equation together, so i can write: x=2 so another equivalent system is: \[\Large \left\{ \begin{gathered} x =  2 \hfill \\ 3x  2y = 2 \hfill \\ \end{gathered} \right.\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1which equivalent system do you prefer?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I prefer the first way you did, the 2nd one was kind of confusing but I could probably understand it if I put the time. But either way is great, thank you so much! Thats for Part A right? Then for Part B I use the way you used but I like do the opposite of everything and divide to get the equation again? Would that work or no?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1ok, we can use the first system, namely: \[\Large \left\{ \begin{gathered} 5x  5y = 10 \hfill \\  5x = 10 \hfill \\ \end{gathered} \right.\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now, please solve the second equation for, x, what do you get?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You do 15x  10y and it would equal to 20?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1no, it is simple, you have to compute this: \[\Large x = \frac{{10}}{{  5}} = ...?\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now, we have to substitute x=2 into the first equation, namely: \[\Large 5 \times \left( {  2} \right)  5y = 10\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1or: \[ \Large  10  5y = 10\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1since I have replaced x with its value, which is 2

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we have these steps: \[\Large \begin{gathered}  10  5y = 10 \hfill \\  5y =  20 \hfill \\ y = \frac{{  20}}{{  5}} \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1oops.. I have made an error, here are the right steps: \[\Large \begin{gathered}  10  5y = 10 \hfill \\  5y = 20 \hfill \\ y = \frac{{20}}{{  5}} = ...? \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1ok! so the solution of the equivalent system, is: \[\Large x =  2,\quad y =  4\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now, we have to substitute those value of x and y into the first equation of the original system, namely: \[\Large 5x  5y = 10\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1so we can write this: \[\Large 5 \times \left( {  2} \right)  5 \times \left( {  4} \right) = 10\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thats all I need to do?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we got an identity, since we can write: \[\Large \begin{gathered} 5 \times \left( {  2} \right)  5 \times \left( {  4} \right) = 10 \hfill \\  10 + 20 = 10 \hfill \\ 10 = 10 \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1no, we have to substitute those values of x=2 and y=4, into the second equation of your original system, namely: \[\Large 3x  2y = 2\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1so we get: \[\Large 3 \times \left( {  2} \right)  2 \times \left( {  4} \right) = 2\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1again we got an identity, since we can write these steps: \[\Large \begin{gathered} 3 \times \left( {  2} \right)  2 \times \left( {  4} \right) = 2 \hfill \\  6 + 8 = 2 \hfill \\ 2 = 2 \hfill \\ \end{gathered} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So we can say the 2 equations are identity then

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! after our substitution they became 2 identities. That means the solution of the equivalent system is also the solution of your original system

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so much you have a new fan now haha :)
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