its 2 30 AM need to wake up in 3 hours, please help me finish this last question fast
A system of equations is shown below:
-3x + 7y = -16
-9x + 5y = 16
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. (6 points)
Part B: Show that the equivalent system has the same solution as the original system of equations. (4 points)

- anonymous

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- Michele_Laino

if we multiply the first equation by -3, we get:
\[9x - 21y = 48\]

- Michele_Laino

Now if we sum that equation with the second one, what do you get?

- anonymous

x-16y=64?

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## More answers

- Michele_Laino

hint:
\[9x - 21y + \left( { - 9x + 5y} \right) = 48 + 16\]
please simplify

- anonymous

but wouldnt that be x-16y=16?

- Michele_Laino

no, since the next step is:
\[9x - 21y - 9x + 5y = 48 + 16\]
please simplify

- anonymous

oh you put addition at first,
x-25y=64

- Michele_Laino

hint:
what is 9 -9 =...

- anonymous

0

- Michele_Laino

ok! so the coefficient of the term with x is zero.
Now what is -21 + 5= ...?

- anonymous

-16

- Michele_Laino

ok! so the coefficient of the term with y is -16

- Michele_Laino

then we can write:
\[0x - 16y = 64\]
and finally:
\[ - 16y = 64\]

- Michele_Laino

am I right?

- anonymous

yeah

- Michele_Laino

ok! so the requested system of part A, can be this:
\[\left\{ \begin{gathered}
- 3x + 7y = - 16 \hfill \\
- 16y = 64 \hfill \\
\end{gathered} \right.\]

- anonymous

thats it for part A?

- Michele_Laino

yes!

- anonymous

perfect, now what do we do for part B?

- Michele_Laino

we can consider the last system, namely the equivalent system, and we can solve the second equation for y, namely:
\[ - 16y = 64\]
what is y?

- Michele_Laino

hint:
divide both sides by -16, what do you get?

- anonymous

y=-4

- Michele_Laino

ok!

- Michele_Laino

now substitute that value of y into the first equation of the equivalent system, namely:
\[ - 3x + 7y = - 16\]
what equation do you get?

- anonymous

-3x+7-4=16

- Michele_Laino

not exactly, here is the right step:
\[ - 3x + 7 \times \left( { - 4} \right) = - 16\]
please simplify

- Michele_Laino

what is:
\[7 \times \left( { - 4} \right) = ...?\]

- anonymous

-3x-21=-16?

- Michele_Laino

hint:
\[7 \times \left( { - 4} \right) = - 28\]
am I right?

- anonymous

oops lol, yeah you're right, i accidentally multiplied by -3

- anonymous

sorry its real late over here, cant think properly

- Michele_Laino

ok! So we have:
\[ - 3x - 28 = - 16\]

- Michele_Laino

now if we add 28 to both sides, we get:
\[ - 3x - 28 + 28 = - 16 + 28\]
please simplify

- anonymous

-3x=12

- Michele_Laino

ok! then if we divide both sides by -3, we can write:
\[\frac{{ - 3x}}{{ - 3}} = \frac{{12}}{{ - 3}}\]
please simplify

- anonymous

-4

- anonymous

x=-4

- Michele_Laino

perfect! So the solution of the equivalent system is:
\[\left\{ \begin{gathered}
x = - 4 \hfill \\
y = - 4 \hfill \\
\end{gathered} \right.\]

- Michele_Laino

now, please substitute that solution into the first equation of the original system, namely:
\[ - 3x + 7y = - 16\]
what do you get?

- anonymous

-3-4+7-4=-16

- anonymous

oops

- Michele_Laino

hint:
\[ - 3 \times \left( { - 4} \right) + 7 \times \left( { - 4} \right) = - 16\]

- anonymous

gotta multiply

- Michele_Laino

yes!

- anonymous

yeah i meant that

- Michele_Laino

what is (-3) * (-4)=...
and 7*(-4)=...?

- anonymous

12
-28

- Michele_Laino

ok! so we can write:
\[12 - 28 = - 16\]

- Michele_Laino

now what is 12-28=...?

- anonymous

-16

- Michele_Laino

ok! So we can write:
\[ - 16 = - 16\]

- anonymous

ok thanks a lot! appreciate every thing, you taught me better than any teacher has in school!

- Michele_Laino

reassuming the solution of the equivalent system makes the first equation an identity

- Michele_Laino

now we have to do the same procedure for the second equation of the original system. If the solution of the equivalent system makes an identity the second equation of the original system then we can state that the two systems, namely the original system and the equivalent system have the same solution

- anonymous

i think we only have to do it with one equation, but i might be wrong

- Michele_Laino

so if we substitute the solution of the equivalent system:
\[\left\{ \begin{gathered}
x = - 4 \hfill \\
y = - 4 \hfill \\
\end{gathered} \right.\]
into the second equation of the original system:
\[ - 9x + 5y = 16\]
what do you get?

- anonymous

-9(-4)+5(-4)=16

- Michele_Laino

ok!
\[ - 9 \times \left( { - 4} \right) + 5 \times \left( { - 4} \right) = 16\]

- Michele_Laino

now please simplify:
what is (-9)*(-4)=...?
and
5*(-4)=...?

- anonymous

36
-20

- Michele_Laino

so we can write:
\[36 - 20 = 16\]
now what is 36-20=...?

- anonymous

16

- anonymous

16=16

- Michele_Laino

ok! so we got an identity again!

- Michele_Laino

then we can state that both systems have the same solution, which is:
\[\left\{ \begin{gathered}
x = - 4 \hfill \\
y = - 4 \hfill \\
\end{gathered} \right.\]

- anonymous

ok, thanks for all the help! greatly appreciate it

- Michele_Laino

:)

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