anonymous
  • anonymous
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)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Michele_Laino
  • Michele_Laino
if we multiply the first equation by -3, we get: \[9x - 21y = 48\]
Michele_Laino
  • Michele_Laino
Now if we sum that equation with the second one, what do you get?
anonymous
  • anonymous
x-16y=64?

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More answers

Michele_Laino
  • Michele_Laino
hint: \[9x - 21y + \left( { - 9x + 5y} \right) = 48 + 16\] please simplify
anonymous
  • anonymous
but wouldnt that be x-16y=16?
Michele_Laino
  • Michele_Laino
no, since the next step is: \[9x - 21y - 9x + 5y = 48 + 16\] please simplify
anonymous
  • anonymous
oh you put addition at first, x-25y=64
Michele_Laino
  • Michele_Laino
hint: what is 9 -9 =...
anonymous
  • anonymous
0
Michele_Laino
  • Michele_Laino
ok! so the coefficient of the term with x is zero. Now what is -21 + 5= ...?
anonymous
  • anonymous
-16
Michele_Laino
  • Michele_Laino
ok! so the coefficient of the term with y is -16
Michele_Laino
  • Michele_Laino
then we can write: \[0x - 16y = 64\] and finally: \[ - 16y = 64\]
Michele_Laino
  • Michele_Laino
am I right?
anonymous
  • anonymous
yeah
Michele_Laino
  • 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
  • anonymous
thats it for part A?
Michele_Laino
  • Michele_Laino
yes!
anonymous
  • anonymous
perfect, now what do we do for part B?
Michele_Laino
  • 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
  • Michele_Laino
hint: divide both sides by -16, what do you get?
anonymous
  • anonymous
y=-4
Michele_Laino
  • Michele_Laino
ok!
Michele_Laino
  • 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
  • anonymous
-3x+7-4=16
Michele_Laino
  • Michele_Laino
not exactly, here is the right step: \[ - 3x + 7 \times \left( { - 4} \right) = - 16\] please simplify
Michele_Laino
  • Michele_Laino
what is: \[7 \times \left( { - 4} \right) = ...?\]
anonymous
  • anonymous
-3x-21=-16?
Michele_Laino
  • Michele_Laino
hint: \[7 \times \left( { - 4} \right) = - 28\] am I right?
anonymous
  • anonymous
oops lol, yeah you're right, i accidentally multiplied by -3
anonymous
  • anonymous
sorry its real late over here, cant think properly
Michele_Laino
  • Michele_Laino
ok! So we have: \[ - 3x - 28 = - 16\]
Michele_Laino
  • Michele_Laino
now if we add 28 to both sides, we get: \[ - 3x - 28 + 28 = - 16 + 28\] please simplify
anonymous
  • anonymous
-3x=12
Michele_Laino
  • Michele_Laino
ok! then if we divide both sides by -3, we can write: \[\frac{{ - 3x}}{{ - 3}} = \frac{{12}}{{ - 3}}\] please simplify
anonymous
  • anonymous
-4
anonymous
  • anonymous
x=-4
Michele_Laino
  • 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
  • Michele_Laino
now, please substitute that solution into the first equation of the original system, namely: \[ - 3x + 7y = - 16\] what do you get?
anonymous
  • anonymous
-3-4+7-4=-16
anonymous
  • anonymous
oops
Michele_Laino
  • Michele_Laino
hint: \[ - 3 \times \left( { - 4} \right) + 7 \times \left( { - 4} \right) = - 16\]
anonymous
  • anonymous
gotta multiply
Michele_Laino
  • Michele_Laino
yes!
anonymous
  • anonymous
yeah i meant that
Michele_Laino
  • Michele_Laino
what is (-3) * (-4)=... and 7*(-4)=...?
anonymous
  • anonymous
12 -28
Michele_Laino
  • Michele_Laino
ok! so we can write: \[12 - 28 = - 16\]
Michele_Laino
  • Michele_Laino
now what is 12-28=...?
anonymous
  • anonymous
-16
Michele_Laino
  • Michele_Laino
ok! So we can write: \[ - 16 = - 16\]
anonymous
  • anonymous
ok thanks a lot! appreciate every thing, you taught me better than any teacher has in school!
Michele_Laino
  • Michele_Laino
reassuming the solution of the equivalent system makes the first equation an identity
Michele_Laino
  • 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
  • anonymous
i think we only have to do it with one equation, but i might be wrong
Michele_Laino
  • 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
  • anonymous
-9(-4)+5(-4)=16
Michele_Laino
  • Michele_Laino
ok! \[ - 9 \times \left( { - 4} \right) + 5 \times \left( { - 4} \right) = 16\]
Michele_Laino
  • Michele_Laino
now please simplify: what is (-9)*(-4)=...? and 5*(-4)=...?
anonymous
  • anonymous
36 -20
Michele_Laino
  • Michele_Laino
so we can write: \[36 - 20 = 16\] now what is 36-20=...?
anonymous
  • anonymous
16
anonymous
  • anonymous
16=16
Michele_Laino
  • Michele_Laino
ok! so we got an identity again!
Michele_Laino
  • 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
  • anonymous
ok, thanks for all the help! greatly appreciate it
Michele_Laino
  • Michele_Laino
:)

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