#still need help but diff question
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Multiply #2 by -3 to cancel y
The solution to the system of equations is x = 7/6 and y = 0.@fading gulch
That is wrong though
- 3 (0) + 8 (7/6) = 28/3 ≠ 16
- it's generally considered better to help somebody find the answer, rather than just giving it to them (teach a man to fish...)
If he wants explenation he could ask me
But the thing is wrong
My bad then
what do i do after that
so y become 0 overall
that's not a good way of thinking of it
Your goal is to find what the values of x and y that satify both equation 1 and 2
and -10x=-5?
to do that, you can find the value of x first
by multiplying the second equation by -3 and then adding equation 2 to 1
you'll be left with an equation of only x, then solve for x, plug that x into either equation 1 or 2 and solve for y
when i do -5/-10 it give me -15 that right or am i just gone??
So 3 steps:
- isolate for a single variable (x or y)
- solve for that variable
- plug in that number back into one of the equations and solve for last variable
Do you know how to solve this equation -10x = -5?
divide -5 by -10
yes and that gives you?
-15
on phone calc it does
don't use a calculator.
0.5
so you have x = 0.5 that's correct
yes
the next step is to do what?
go for it.
uhhhhhh
The slope-intercept form of the line whose equation is 3y - 2x = 8 is y = (2/3)x + (8/3).
Yes
3y - x = 8 (1)
5y - 2x = 6 (2)
(1) x 2
-6y + 2x = -16
-6y = -2x - 16
y = (1/3)x + (8/3)
6y - 2x = 16 (3)
What
yikes
my brain
@fading gulch 😭
why are you guys doing his homework for him lol
let the guy solve it himself
you're not helping giving all the answers away
im tryna show him the right working out
innit i got another 4 of the questions after
the 2x
Yea
my signal died
So eliminate that and solve for x and y the rest of the way
ok
hmm
ok
10f - 9g = 12 (1)
2f +3g = 12 (2)
(2) x 3
wait
no
6f + 9g =36 (3)
(3) - (1)
@tropic thorn am i on right track with this question ( diff one tk before)
👍
that can’t be right
someone help
still need help but diff question
$$
10x-9y = 12 \quad (1)\
2x + 3y =12 \quad (2) \
\text{Working with (2) to isolate x} \
2x = 12-3y \
x = 6 - 1.5y \
\text{sub (2) into (1)} \
10(6-1.5y) -9y = 12 \
60-15y - 9y = 12 \
60-12 = 24y \
2 = y
$$
where’d i go wrong?
conquestace
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I don't know.
for one, you're adding equation 1 and 2, but you did 6f-10f
same with 36-12 when its suppose to be 36+12
yeah (3) - (1) is : 6f + 9g - (10f-9g) = 36 - 12
giving you -4f + 18g = 24
you're suppsoe to do (3) + (1) : 6f + 9g + (10f-9g) = 36+12
what is ur goal?
to eliminate a varaible
that's why it's called elimnination method
use your head to determine when to add or subtract
The method I showed was the substitution method, because I isolate for a single variable and then substitute it back into the other equation
o
are you sure what you did was right?
I am pretty sure the answer is f=3 and g =2
if you do -4f = 24 you get f = -6
you plug that back into 2 and you get: 2*(-6) + 3g = 12
g= 8
but when u put in 10(-4) - 9(8) =/= 12
I don't know @tropic thorn is trying to do
troll maybe?
this is right
you continue solving
f = -6
sub f into (1)
10f - 9g = 12 (1)
10 x -6 - 9g = 12
-60 - 9g = 12
-9g = 72
g = -8
@timid otter
That is wrong
Where did you get -10f and -12 from
It works for one of them
this step is wrong
lol mb didn't look at it right
should be 6f + 9g - 10f + 9g = 36 -12 according to (3) - (1)
it should be (1) + (3)