anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Michele_Laino
  • Michele_Laino
we can multiply the first equation by 2 and the second equation by -5, so we get the subsequent equivalent system: \[\Large \left\{ \begin{gathered} 10x - 10y = 20 \hfill \\ - 15x + 10y = - 10 \hfill \\ \end{gathered} \right.\]
anonymous
  • anonymous
Right..
Michele_Laino
  • Michele_Laino
now we add those two equations together, so we get: \[\Large - 5x = 10\]

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Michele_Laino
  • Michele_Laino
am I right?
anonymous
  • anonymous
Yes
Michele_Laino
  • Michele_Laino
so 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
  • anonymous
So it would be - 5x = 10 and that would be an equivalent system of the equation?
Michele_Laino
  • Michele_Laino
yes!
anonymous
  • anonymous
Yay! For Part B, I just check my work an write that as the proof?
Michele_Laino
  • Michele_Laino
another 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
  • Michele_Laino
then 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
  • Michele_Laino
which equivalent system do you prefer?
anonymous
  • anonymous
I 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
  • Michele_Laino
ok, we can use the first system, namely: \[\Large \left\{ \begin{gathered} 5x - 5y = 10 \hfill \\ - 5x = 10 \hfill \\ \end{gathered} \right.\]
Michele_Laino
  • Michele_Laino
now, please solve the second equation for, x, what do you get?
anonymous
  • anonymous
You do 15x - 10y and it would equal to 20?
Michele_Laino
  • Michele_Laino
no, it is simple, you have to compute this: \[\Large x = \frac{{10}}{{ - 5}} = ...?\]
anonymous
  • anonymous
-2?
Michele_Laino
  • Michele_Laino
perfect!
Michele_Laino
  • Michele_Laino
now, we have to substitute x=-2 into the first equation, namely: \[\Large 5 \times \left( { - 2} \right) - 5y = 10\]
Michele_Laino
  • Michele_Laino
or: \[ \Large - 10 - 5y = 10\]
anonymous
  • anonymous
Why is there no x?
Michele_Laino
  • Michele_Laino
since I have replaced x with its value, which is -2
anonymous
  • anonymous
OH My bad!
Michele_Laino
  • Michele_Laino
so, what is y?
anonymous
  • anonymous
-2.5
Michele_Laino
  • Michele_Laino
we have these steps: \[\Large \begin{gathered} - 10 - 5y = 10 \hfill \\ - 5y = - 20 \hfill \\ y = \frac{{ - 20}}{{ - 5}} \hfill \\ \end{gathered} \]
anonymous
  • anonymous
\[y = 4?\]
Michele_Laino
  • Michele_Laino
oops.. I have made an error, here are the right steps: \[\Large \begin{gathered} - 10 - 5y = 10 \hfill \\ - 5y = 20 \hfill \\ y = \frac{{20}}{{ - 5}} = ...? \hfill \\ \end{gathered} \]
anonymous
  • anonymous
oh -4
Michele_Laino
  • Michele_Laino
ok! so the solution of the equivalent system, is: \[\Large x = - 2,\quad y = - 4\]
anonymous
  • anonymous
Yes
Michele_Laino
  • Michele_Laino
now, 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
  • Michele_Laino
values*
Michele_Laino
  • Michele_Laino
so we can write this: \[\Large 5 \times \left( { - 2} \right) - 5 \times \left( { - 4} \right) = 10\]
anonymous
  • anonymous
Nice!
anonymous
  • anonymous
Thats all I need to do?
Michele_Laino
  • Michele_Laino
we 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
  • Michele_Laino
no, 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
  • Michele_Laino
so we get: \[\Large 3 \times \left( { - 2} \right) - 2 \times \left( { - 4} \right) = 2\]
Michele_Laino
  • Michele_Laino
again 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
  • anonymous
So we can say the 2 equations are identity then
Michele_Laino
  • Michele_Laino
yes! after our substitution they became 2 identities. That means the solution of the equivalent system is also the solution of your original system
anonymous
  • anonymous
Perfect
Michele_Laino
  • Michele_Laino
:)
anonymous
  • anonymous
Thank you so much you have a new fan now haha :)

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