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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- jamiebookeater

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- 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

Right..

- Michele_Laino

now we add those two equations together, so we get:
\[\Large - 5x = 10\]

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## More answers

- Michele_Laino

am I right?

- anonymous

Yes

- 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

So it would be - 5x = 10 and that would be an equivalent system of the equation?

- Michele_Laino

yes!

- anonymous

Yay! For Part B, I just check my work an write that as the proof?

- 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

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

which equivalent system do you prefer?

- 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

ok, we can use the first system, namely:
\[\Large \left\{ \begin{gathered}
5x - 5y = 10 \hfill \\
- 5x = 10 \hfill \\
\end{gathered} \right.\]

- Michele_Laino

now, please solve the second equation for, x, what do you get?

- anonymous

You do 15x - 10y and it would equal to 20?

- Michele_Laino

no, it is simple, you have to compute this:
\[\Large x = \frac{{10}}{{ - 5}} = ...?\]

- anonymous

-2?

- Michele_Laino

perfect!

- Michele_Laino

now, we have to substitute x=-2 into the first equation, namely:
\[\Large 5 \times \left( { - 2} \right) - 5y = 10\]

- Michele_Laino

or:
\[ \Large - 10 - 5y = 10\]

- anonymous

Why is there no x?

- Michele_Laino

since I have replaced x with its value, which is -2

- anonymous

OH My bad!

- Michele_Laino

so, what is y?

- anonymous

-2.5

- Michele_Laino

we have these steps:
\[\Large \begin{gathered}
- 10 - 5y = 10 \hfill \\
- 5y = - 20 \hfill \\
y = \frac{{ - 20}}{{ - 5}} \hfill \\
\end{gathered} \]

- anonymous

\[y = 4?\]

- 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

oh -4

- Michele_Laino

ok! so the solution of the equivalent system, is:
\[\Large x = - 2,\quad y = - 4\]

- anonymous

Yes

- 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

values*

- Michele_Laino

so we can write this:
\[\Large 5 \times \left( { - 2} \right) - 5 \times \left( { - 4} \right) = 10\]

- anonymous

Nice!

- anonymous

Thats all I need to do?

- 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

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

so we get:
\[\Large 3 \times \left( { - 2} \right) - 2 \times \left( { - 4} \right) = 2\]

- 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

So we can say the 2 equations are identity then

- 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

Perfect

- Michele_Laino

:)

- anonymous

Thank you so much you have a new fan now haha :)

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