SOLUTION: A) 4x^2 + 3y = 1
3x^2 + 2y = 4
B) w^2/2 + w/4 - z/2 = 3
W^2/3 - 3w/4 + z/6 + 1/3= 0
Algebra.Com
Question 974883: A) 4x^2 + 3y = 1
3x^2 + 2y = 4
B) w^2/2 + w/4 - z/2 = 3
W^2/3 - 3w/4 + z/6 + 1/3= 0
Answer by harpazo(655) (Show Source): You can put this solution on YOUR website!
Use the substitution method to solve both problems.
This is the easiest way to tackle your questions.
Visit youtube.com to learn the substitution method.
You will see it's pretty easy.
You do the math.
RELATED QUESTIONS
Please solve for w, x, y, and z.
1. x + 3w = 4
2y - z -w = 0
3y - 2w = 1
(answered by CharlesG2)
(w+2^2)-(3w-1)-4=0 (answered by Fombitz)
x+2y+0z+3w=0
0x+y+z+w=1
0x+y+0z+w=2... (answered by Alan3354)
(3w^2+2w+3)+(w^2+w+4)+(7w^2+3w+1) (answered by solver91311)
1. ~A ⊃ ~B
2. A ⊃ C
3. Z ⊃ W
4. ~C • ~W /∴~B v W
(answered by RBryant)
what are the values of w,x,y, and z in these equations?
1. w + x + y + z = 2
2. 2w -... (answered by ankor@dixie-net.com)
1. (A v B) ⊃ T
2. Z ⊃ (A v B)
3. T ⊃ W
4. ~ W /∴~Z
(answered by Edwin McCravy)
I need to solve for all variables.
1)2x-x+5y+z=-3
2)3w+2x+2y-6z=-32
3)w+3x+3y-z=-47
(answered by Fombitz)
64. 4w-1/3w+6 – w-1/3 = w-1/w+2
w=2
(answered by Edwin McCravy)