0
What else can I help you with?
How do you solve y-2y'-3y equals 0?
For y - 2y - 3y equals 0, y equals 0.
What is the locus of points that satisfies both equations 2x-3y equals 5 and x-2y equals 1?
2x-3y=5 and x-2y=12x-3y+2y=5 and x-2y+2y=12x-y=5 and x=12(1)-y=52-y=52-y-2=5-2-y=3y=-3(x,y) is (1,-3)
2y plus 3y-7y plus 4 equals 16?
2y+3y-7y+4=16 -2y=12y=-6Check it-12-18+42+4=16-30+46=1616=16
9y-3y plus 2y equals?
9y-3y=6y; 6y+2y=8y; so your answer is 8y
What is the answer to 5 -2y equals 3y - 5?
5 - 2y = 3y - 5 10 = 5y So, y = 2.
What does 3y equals 12 equals what does 2y equal?
If 3y = 12, 2y = 8
How do you solve the equation 5y equals 3Y plus 12?
5y = 3y + 12 5y - 3y = (3y - 3y) + 12 2y = 12 y = 6
X - 3y equals -3 and 6x plus 2y equals -12?
x = 3y - 3 so 6 (3y - 3) + 2y = -12 ie 18y - 18 + 2y = -12 ie 20y = 6 so y = 0.3 and x = -2.1
How do you solve y-2y'-3y equals 0?
For y - 2y - 3y equals 0, y equals 0.
5y - 2y equals 3y plus 2 what is the answer?
5y-2y= 3y+2 5y-2y-3y= 2 0y= 2
5-2y equals -3y - 2?
5 - 2y = -3y - 2Subtract 5 from each side:-2y = -3y - 7Add 3y to each side:y = -7
What type of lines are these x-3y equals -3 and 6x plus 2y equals -12?
perpendicular
What is the locus of points that satisfies both equations 2x-3y equals 5 and x-2y equals 1?
2x-3y=5 and x-2y=12x-3y+2y=5 and x-2y+2y=12x-y=5 and x=12(1)-y=52-y=52-y-2=5-2-y=3y=-3(x,y) is (1,-3)
2y plus 3y-7y plus 4 equals 16?
2y+3y-7y+4=16 -2y=12y=-6Check it-12-18+42+4=16-30+46=1616=16
9y-3y plus 2y equals?
9y-3y=6y; 6y+2y=8y; so your answer is 8y
2y plus 3 equals 3y plus 9?
2y + 3 = 3y + 9 2y - 3y = 9 - 3 or -y = 6 so y = -6
-3 plus 3y equals 2y plus 5?
-3 + 3y = 2y + 5 Add '3' to both sides Hence 3y = 2y + 8 Subtract '2y' from both sides. y = 8 The answer!!!!!
