## 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.

1. Michele_Laino

we can multiply the first equation by 2 and the second equation by -5, so we get the subsequent equivalent system: $\Large \left\{ \begin{gathered} 10x - 10y = 20 \hfill \\ - 15x + 10y = - 10 \hfill \\ \end{gathered} \right.$

2. anonymous

Right..

3. Michele_Laino

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

4. Michele_Laino

am I right?

5. anonymous

Yes

6. Michele_Laino

so the requested equivalent system, in part A, can be this: $\Large \left\{ \begin{gathered} 5x - 5y = 10 \hfill \\ - 5x = 10 \hfill \\ \end{gathered} \right.$

7. anonymous

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

8. Michele_Laino

yes!

9. anonymous

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

10. Michele_Laino

another way to get an equivalent system is multipling the firs equation by -2/5, so Iget this: $\left\{ \begin{gathered} - 2x + 2y = - 4 \hfill \\ 3x - 2y = 2 \hfill \\ \end{gathered} \right.$

11. Michele_Laino

then I add the two equation together, so i can write: x=-2 so another equivalent system is: $\Large \left\{ \begin{gathered} x = - 2 \hfill \\ 3x - 2y = 2 \hfill \\ \end{gathered} \right.$

12. Michele_Laino

which equivalent system do you prefer?

13. 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?

14. Michele_Laino

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

15. Michele_Laino

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

16. anonymous

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

17. Michele_Laino

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

18. anonymous

-2?

19. Michele_Laino

perfect!

20. Michele_Laino

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

21. Michele_Laino

or: $\Large - 10 - 5y = 10$

22. anonymous

Why is there no x?

23. Michele_Laino

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

24. anonymous

25. Michele_Laino

so, what is y?

26. anonymous

-2.5

27. Michele_Laino

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

28. anonymous

$y = 4?$

29. Michele_Laino

oops.. I have made an error, here are the right steps: $\Large \begin{gathered} - 10 - 5y = 10 \hfill \\ - 5y = 20 \hfill \\ y = \frac{{20}}{{ - 5}} = ...? \hfill \\ \end{gathered}$

30. anonymous

oh -4

31. Michele_Laino

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

32. anonymous

Yes

33. Michele_Laino

now, we have to substitute those value of x and y into the first equation of the original system, namely: $\Large 5x - 5y = 10$

34. Michele_Laino

values*

35. Michele_Laino

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

36. anonymous

Nice!

37. anonymous

Thats all I need to do?

38. Michele_Laino

we got an identity, since we can write: $\Large \begin{gathered} 5 \times \left( { - 2} \right) - 5 \times \left( { - 4} \right) = 10 \hfill \\ - 10 + 20 = 10 \hfill \\ 10 = 10 \hfill \\ \end{gathered}$

39. Michele_Laino

no, we have to substitute those values of x=-2 and y=-4, into the second equation of your original system, namely: $\Large 3x - 2y = 2$

40. Michele_Laino

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

41. Michele_Laino

again we got an identity, since we can write these steps: $\Large \begin{gathered} 3 \times \left( { - 2} \right) - 2 \times \left( { - 4} \right) = 2 \hfill \\ - 6 + 8 = 2 \hfill \\ 2 = 2 \hfill \\ \end{gathered}$

42. anonymous

So we can say the 2 equations are identity then

43. 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

44. anonymous

Perfect

45. Michele_Laino

:)

46. anonymous

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