When you have a variable by itself either it be x or y there is always an invisible place holder before it. So a normal "y" would be "1y". So when you put 1y + 1y you get 2y.
1+1+x+x+x+y+y=2+3x+2y
To solve for y in terms of x, divide both sides of the equation by 2: y = x/2.If x=2y then you have already solved for x.
3x + 2y = 5x + 2y = 7Subtract (2) from (1) to give: 3x - x + 2y - 2y = 5 - 7→ 2x = -2→ x = -1Substitute for x in (1)3(-1) + 2y = 5→ -3 + 2y = 5→ 2y = 8→ y = 4Substitute for x and y in (2) to check answer:x + 2y = -1 +2(4) = -1 + 8 = 7 as required. Thus:x = -1y = 4
x=4 x=4
4x + 3y = 1y = 2x + 8 Take the first part: 4x + 3y = 1y 4x = 1y - 3y 4x = -2y x = -y/2 Substitute x in second part: 1y = 2x + 8 1y = -2y/2 + 8 1y = -1y + 8 2y = 8 y = 4 Substitute y into either part: 1y = 2x + 8 1(4) = 2x + 8 4 - 8 = 2x -4 = 2x x = -2 Therefore: x = -2 y = 4
If you mean: 5x+2y+x-3y then it is 6x-1y
When you have a variable by itself either it be x or y there is always an invisible place holder before it. So a normal "y" would be "1y". So when you put 1y + 1y you get 2y.
X 2y-1 4x-3x=18 X 2y-1 7x=18 X 1y 7x=18 X Y 7x=18 X Y X = 2 4/7 and there is ur anser
1+1+x+x+x+y+y=2+3x+2y
To solve for y in terms of x, divide both sides of the equation by 2: y = x/2.If x=2y then you have already solved for x.
3x + 2y = 5x + 2y = 7Subtract (2) from (1) to give: 3x - x + 2y - 2y = 5 - 7→ 2x = -2→ x = -1Substitute for x in (1)3(-1) + 2y = 5→ -3 + 2y = 5→ 2y = 8→ y = 4Substitute for x and y in (2) to check answer:x + 2y = -1 +2(4) = -1 + 8 = 7 as required. Thus:x = -1y = 4
solve for x -x-2y=6 -x=2y+6 x=-2y-6 plug in y=0 x=-6 when y is 0 therefore the x intercept is (-6,0)
x=4 x=4
get it to y= -2y = 1 - 2x y = x - 0.5
X+2y=6 how to solve? 2y=6-x then y=3-x so y=3-x
-2x - 2y = -122x + 2y = 122y = 12 - 2xy = 6 - x