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Q: How do you solve 2x subtract y equals 1?

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2x+8=10... Subtract 8 on both sides, then devide by 2... 2x=2... x=1

2x - 1 = 3x + 4 Subtract 2x from both sides: - 1 = x + 4 Subtract 4 from both sides: -5 = x

2x+32 = 1 2x = 1-32 2x = -31 x = -15.5

2x+1 = 9 2x = 9-1 2x = 8 x = 4

get it to y= -2y = 1 - 2x y = x - 0.5

X = 0 (zero) Y = -1 (neg 1)

meant to be e^x = 2e^1-2x

2x + 1 = 9 2x = 9 - 1 2x = 8 2x/2 = 8/2 x=4

If: 2x-1 = 2Then: 2x = 2+1And: 2x = 3So: x = 3/2 = 1.5 or 1 and a 1/2

If: 2x = -2 Then: x = -1 by dividing both sides by 2

x + 8y = 3 therefore 2x + 16y = 6 (a) 2x + 3y = -7 (b) Subtract (b) from (a): 13y = 13 so y = 1 and x = -5

-1-2x-4x=19-6x-1 = -19x = 3

9-2x=-1 subtract the 9 from both sides -2x=-10 divide each side by -2 x=5

If we assume that it equals zero and you wish to solve for x, then the answer is:-x2 + 2x - 3 = 0x2 - 2x + 3 = 0x2 -2x + 1 = -2(x - 1)2 = -2x - 1 = ± √(-2)x = 1 ± i√2

2x + 7 = 1 Therefore, 2x = -6 x = -6/2 x = -3

To solve this equation, we will first subtract 2x from each side:8x=2x+6 (8x minus 2x=6x)6x=6Now, to isolate x, we must divide out the 6:6x=6 (6x divided by 6= x)x=1If 8x=2x+6, x=1

2x + 3 = x + 1 Subtract x from both sides: x + 3 = 1 Subtract 3 from both sides: x = -2

y = -6

4x2 - 2x - 3 = 0 (2x + 1)(2x - 3) = 0 x = -(1/2) or (3/2)

2x + 3y = 9x = 3, y = 1 therefore 6 + 3 = 9

y = 2x -1 y = 5x -13 So: 5x -13 = 2x -1 5x -2x = 13 -1 3x = 12 x = 4 and by substituting y = 7

1. Subtract 2x from each side: x + 2 = -52. Subtract 2 from each side: x = -7

2x + 5x - 1 = 9x 7x - 1 = 9x -1 = 9x - 7x -1 = 2x -1/2=x

-530

For x-2x= 3x-43x+1 x-2x= -x and 3x-43x+1 = -40x+1 so -x= -40x+1 subtract -x from both sides 0= -39x +1 -1 = -39x so 1=39x so x=1/39