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What is the answer to 5 -2y equals 3y - 5?
5 - 2y = 3y - 5 10 = 5y So, y = 2.
80 equals 3y plus 2y plus 4 plus 1?
80 = 3y + 2y + 4 + 1 80 = 5y + 5 80 - 5 = 5y + 5 - 5 75 = 5y 75 / 5 = 5y / 5 15 = y
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)
How do you solve 5 plus y plus 2y equal?
5+y+2y? = 5+3y?
What is the value of x and y in this equation 2x plus 3y equals -5 and 5x plus 2y equals 4 The number for x and y stay the same for both parts of the equation?
This is a system of equations, and we can use various methods to solve it. We'll use substitution in this case. We're told: 2x + 3y = -5 5x + 2y = 4 To solve by substitution, what we need to do is take either one of those equations, and solve it for either x or y. Let's take the second one and solve it for x: 5x + 2y = 4 5x = 4 - 2y x = (4 - 2y)/5 Now we can take that solution for x, and substitute it into the other equation: 2x + 3y = -5 2((4 - 2y)/5) + 3y = -5 (8 - 4y) / 5 + 3y = -5 (8 - 4y + 15y) / 5 = -5 8 - 4y + 15y = -25 11y = -33 y = -3 We now have a value for y, and can plug it into either of the original equations to solve for x: 2x + 3y = -5 2x + 3(-3) = -5 2x - 9 = -5 2x = 4 x = 2 To verify our answer, we can plug either x or y into the other of our original equations, and see if we get the same result for the other variable: 5x + 2y = 4 5(2) + 2y = 4 10 + 2y = 4 2y = -6 y = -3 So that confirms our answer, and the two equations intersect at the point (2, -3).
5y plus 4 - 2y equals 9?
5y + (4 - 2y) = 9 5y + 4 - 2y = 9 3y + 4 = 9 3y = 5 y = 5/3
5-2y equals -3y - 2?
5 - 2y = -3y - 2Subtract 5 from each side:-2y = -3y - 7Add 3y to each side:y = -7
-3 plus 3y equals 2y plus 5?
3y - 3 = 2y + 5 3y - 2y = 5 + 3 y = 8
What is the answer to 5 -2y equals 3y - 5?
5 - 2y = 3y - 5 10 = 5y So, y = 2.
-y plus 5 plus 3y - 4?
2y + 1.
How do you solve y2 - 3y plus 2 plus 3y - 4y - 5?
I don't know whether it's y^2-3y+2+3y-4y-5 or 2y-3y+2+3y-4y-5, so I'll do both.y^2-3y+2+3y-4y-5y^2+(-3y+3y-4y)+(2-5); group them togethery^2+(-4y)+(-3)y2-4y-3OR2y-3y+2+3y-4y-5(2y-3y+3y-4y)+(2-5); group them together(-2y)+(-3)-2y-3
What is the equation 80 equals 3y plus 2y plus 4 plus 1?
3y+2y+4+1=80 5y+5=80 5y=80-5 5y=75 y=15
Solve this equation 80 3y plus 2y plus 4 plus 1?
If you mean: 80 = 3y+2y+4+1 then 80 =5y+5 and so y = 15
80 equals 3y plus 2y plus 4 plus 1?
80 = 3y + 2y + 4 + 1 80 = 5y + 5 80 - 5 = 5y + 5 - 5 75 = 5y 75 / 5 = 5y / 5 15 = y
Solve this equation 80 equals 3y plus 2y plus 4 plus 1?
80 = 3y + 2y + 4 + 1 Collect 'like terms'. 80 = 5y + 5 Substract '5' from noth sides 75 = 5y 5y = 75 Divide both sides by '5' y = 15 The answer!!!!!
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)
Ten linear equations?
x + y = 1 x + y = 2 x + 2y = 2 x + 3y = 3 2x + 3y = 4 2x + 5y = 5 x - 2y = 2 x - 3y = 3 2x - 3y = 4 a + b + 2c + 3d +e -6f + 2x - 5y = 5
