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

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

x-16y=64?

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

but wouldnt that be x-16y=16?

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

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

hint:
what is 9 -9 =...

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

-16

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

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

am I right?

yeah

thats it for part A?

yes!

perfect, now what do we do for part B?

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

y=-4

ok!

-3x+7-4=16

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

-3x-21=-16?

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

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

sorry its real late over here, cant think properly

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

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

-3x=12

-4

x=-4

-3-4+7-4=-16

oops

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

gotta multiply

yes!

yeah i meant that

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

12
-28

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

now what is 12-28=...?

-16

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

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

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

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

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

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

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

36
-20

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

16

16=16

ok! so we got an identity again!

ok, thanks for all the help! greatly appreciate it