2x+5=-91-10x 12x+5=-91 12x=-96 x=-8
10x + 5 which is also equal to 5(2x + 1)
If the line y = 2x+1.25 is a tangent to the curve y^2 = 10x then it works out that when x = 5/8 then y = 5/2
4x2 + 12x + 5 = 4x2 + 2x + 10x + 5 = 2x(2x+1) + 5(2x+1) = (2x+1)(2x+5)
12x + 5 - 2x =(12x - 2x) + 5 =10x + 5 =5 (2x + 1)
2x+5=-91-10x 12x+5=-91 12x=-96 x=-8
10x + 5 which is also equal to 5(2x + 1)
x=5. 10x-4x (6x) +10-6x (=10) If 10 is 2x, x=5.
13 + 2x = 32x = 3 - 132x = -10x = -5
3x-2x-3x+6 = 5-10x+9x-5 Clean up both sides (the fives on the right cancel out) -2x + 6 = -1x add 2x to both sides 6 = x ■
10x+5y = 25 5y = -10x+25 y = -2x+5 Therefore: the slope is -2 and the intercept is 5
6x2 + 10x = 2x*(3x + 5)
If the line y = 2x+1.25 is a tangent to the curve y^2 = 10x then it works out that when x = 5/8 then y = 5/2
6x2 + 10x = 2x(3x + 5)
4x2 + 12x + 5 = 4x2 + 2x + 10x + 5 = 2x(2x+1) + 5(2x+1) = (2x+1)(2x+5)
6x-8+2x-5 = 7-2x 6x+2x+2x = 7+8+5 10x = 20 x = 20/10 x = 2
If y = 2x +10 and y^2 = 10x then by forming a single quadratic equation and solving it the point of contact is made at (5/8, 5/2)