3x = 7-4x
3x+4x = 7
7x = 7
x = 1
If: 3x = 15 Then: x = 5
2x+3x = 5x
y = 3x 3x + 4y = 30 3x + 4(3x) = 30 3x + 12x = 30 15x = 30 x = 2 y = 3x y = 3(2) y = 6 (2, 6)
3x+7=37 -7 -7 3x=30 /3 /3 x=10
3x = 18divide both sides by 3x = 6
3x+12 = 18 3x = 18-12 3x = 6 x = 2
no
y = 3x = 12y = 12
3x-3x equal = 0
If that's 3x2 + 7x + 2, the answer is (x + 2)(3x + 1)
Algebraically rearrange both lines to ' y = ''' Hence y = 3x & 3y = x => y = (1/3)x Since the two lines are now y = 3x & y = (1/3)x The lines y = 3x has the greater/steeper slope at '3' , Whereas the other line has a slope of ' 1/3 '. (much less steep).
If x-3x=18 then x equals -9
3x + 2 = 3x + 6 This is not possible.
3x=81 3x/3=81/3x=27
3x-5=3x+5 There is no solution.
(2, -1)
y = 15 (by transitivity)