2x-1=9 2x=10 x=5
10x2
y = 2x-5 4x-y = 7 => y = 4x-7 2x-5 = 4x-7 2x-4x = 5-7 -2x = -2 x = 1 Therefore by substituting: x = 1 and y = -3
(2x + 1) + (x*x - 2x + 1) = x^2 + 2x - 2x + 1 + 1 = x^2 + 2
3
2x-6=x+5 2x -x=6+5 x=11 Check the math.
2x2 + 3x - 5 = 02x2 - 2x + 5x - 5 = 02x(x - 1) + 5(x - 1) = 0(x - 1)(2x + 5) = 0x - 1 = 0 or 2x + 5 = 0x = 1 or x = -5/2
2x - 1 = x + 15 Add 1 to both sides: 2x = x + 16 Subtract x from both sides: x = 16
2x-5<7 2x<12 x<6 -3x+5>11 -3x>6 x<-2
2x+5+x-1=3x+4
2x+5+x-1=3x+4
x + 5 = 2x - 35Subtract 2x from both sides of the equationx - 2x + 5 = -35Subtract 5 from both sides of the equationx - 2x = -35 - 5Simplify-x = -40Multiply both sides of the equation by -1-x (-1) = -40 (-1)x = 40
Two solutions: One with math, one with common sense. 1). 5 + x = 2x. Subtract x from both sides of the equation: 5 = x2). If adding 5 to something made it double, what do you think it was before you added the 5 ?
2x-1=9 2x=10 x=5
4x + 1 = 2x + 5 Subtract '1' frp, bpth sides Hence 4x = 2x + 4 Subtract '2x' from botbh sides 2x = 4 Divide both sides by '2' x = 2 The answer!!!!!!
x2 - 2x - 5 = 0 x2 - 2x + 1 = 6 (x - 1)2 = 6 x - 1 = ± √6 x = 1 ± √6
6-2kx=5-2x 2x-2kx=-1 x(2-2k)=-1 x=-1/(2-2k)