Solve for w, x, y, & z. w+x+y+z=18 w=x+y+z w+x-y+z=6 2w=3y 2x+y=8
w+x+y+z=18 -(1) w=x+y+z -(2) w+x-y+z=6 -(3) 2w=3y -(4) 2x+y=8 -(5) Sub. (2) into (1) to solve w. w+w=18 w = ...?
@wendy76 ?
ok hold on
Better off learning linear algebra lol.
@wendy67 What have you got for w?
ok wait sub 2 into 1? on what equation?
@Kainui i'd be glad too but sadly i am taking math 2 but starting with math3 so we took a huge leap foward in math
w+x+y+z=18 -(1) w=x+y+z -(2) w+x-y+z=6 -(3) 2w=3y -(4) 2x+y=8 -(5) Actually, I labelled the equations with (number) So, Sub. (2) into (1) means sub the second equation into the first one. Since w=x+y+z (from the second equation) and w+x+y+z=18 (from the first equation) Sub x+y+z = w into the first equation and you'll get w + w = 18 Understand?
oh im sorry didnt really understand what you meant by the #s on the side and yea i got it now so let me try then..
It's okay. I'm sorry... I should have explained it.
No your fine, and i got 2x+2y+2z=18
Hmm..... How did you get that?
ok so since w=x+y+z (2nd eq.) i plugged into the first: (x+y+z)+x+y+z=18 and got that^
Well.... You can do so... but that doesn't make it easier... Instead you can do it in this way... w=x+y+z (2nd eq.) That means x+y+z =w. Put it into the first one: w + (w) = 18.. Does it make sense to you?
hmm does it mean that x,y,x got canceled and your left with w?
*z
Not really, but substitute x+y+z = w into the equation..
so if i substitude would it give me x+y+z+x+y+z=18?
If you sub w = x+y+z, yes. But you can do it in another way! Since x+y+z = w!
ooooh i get it now so it WOULD be w+(w)=18 2w=18 w=9 ?
Yes!!!
Now, consider the 4th equation. 2w=3y Sub. w = 9 (which you've found) into the fourth equation. You can get the value of y. Can you do it?
yes i already did all of them a got w=9 x=1 y=6 z=2
*and
Yup :)
ok well thank you (:
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