#Need help with Solving Two Variable Systems with Substitution
65 messages · Page 1 of 1 (latest)
3x+y=3
y=3-3x
substitute for y in the 2nd equation
2x - 5(3-3x) = 6
2x - 15 - 15x = 6
-15 - 13x = 6
- 13x = 21
x = -21/13
So is the solution -21/13 or is it another equation, because I believe on another similar question I put in the value of x but it was wrong and it was instead something like y=something + something
Thanks for the step by step help.
wait it is wrong
3x+y=3
y=3-3x
substitute for y in the 2nd equation
2x - 5(3-3x) = 6
2x - 15 + 15x = 6
-15 + 17x = 6
17x = 21
x = 21/17
this is correct
its simple
I'm not good at math
and x = (3+y)/3 = (y/3) + 1
oops minus
sry
no solution
because the x and y have ration but their constant have no ratio
I will pretend I understand
wait
you can write first equation as 6x -2y -10=0
and second equation as 6x -2y +15=0 (multiplying by 2)
from here you can see that whatever value you put there will be no solution
I need to get a math solver/tutor bro
heheh
Hey can you help me with one last question?
ok
in which grade are you?
dropout?
I got a whole backstory,
And idk if I wanna post it here
Maybe in DMs
Just not here for the whole public to see
Essentially
can you send friend request
Sure
option c
.close
Post marked as solved by @winged badger.
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