First move the -3 to the other side so you have: y^2+8y+3=0My suggestion we be to apply the quadratic formula, cause good luck finding two numbers that add to 3 and multiply to 8. FYI Quadratic formula is -8+-sqr((-8)^2-(4)(1)(3)/2(1)= -8+-sqr(64-12)/2=-8+-sqr(52)/2... keep going.
y2 +8y + 16 = 0 can factor to (y+4) (y+4) = 0 so y+4 = 0 so y = -4
It goes to (x-5)(x-3)
y2 + 8y + 16 = y2 + 4y + 4y + 16 = y(y + 4) + 4(y + 4) = (y + 4)(y + 4) or (y + 4)2
If you mean y2+8y+7=0 then it is a quadratic. y2+8y+7+9-9=0 y2+8y+16-9=0 (y+4)2=9 y+4=3 , y+4=-3 y=-1 , y=-7 So y can equal -1 or -7.
Let's first reformat it: x2 + y2 - 12x - 8y - 43 = 0 x2 - 12x + 36 + y2 - 8y + 16 = 43 + 36 + 16 (x - 6)2 + (y - 4)2 = 95 So this equation defines a circle, with a center point at (6, 4) and a perimeter (i.e. circumference) of √95.
88 + 5y - y2 66 - 3y + y2 Subtract: 22 + 8y -2y2
The standard equation of a circle with center C(h,k) and radius r is as follows:(x - h)2 + (y - k)2 = r2(x2 -2hx +h2) + (y2 - 2ky + k2)We have...x2 + y2 - 10x + 8y + 5 =0x2 - 10x + y2 + 8y + 5 =0add h2 and k2 to both sides(x2 - 10x + h2) + (y2 + 8y + k2) = -5 + h2 + k2Using the above framework we can see that 10 = 2h, h = 5Using the above framework we can see that 8 = 2k, k = 4So the formula is(x2 - 10x + 25) + (y2 + 8y + 16) = -5 + 25 + 16(x - 5)2 + (y - 4)2 = 36the center of the circle is C(5,4) and radius 6
x2 + 2y - 6x + 8y - 1 = x2 - y2 + 4x + 6y - 1 y2 - 10x + 4y = 0 y(y+4) = 10x It cannot be solved completely because with two variables (x and y) you need two independent equations for a full solution.
(x + y)(x - y) = 16
It is: y2
Yes, as x-y2=0
multiplying the first equation by the power of route 7, you will work out a formula to give you and equal equation to the one above, thus allowing you to work out the solutions to this eqation as being y= 14 and x= 21Improved Answer:-x = 2y-2 and x2 = y2+7(2y-2)2 = y2+7(2y-2)(2y-2) = y2+74y2-8y+4 = y2+73y2-8y-3 = 0(3y+1)(y-3) = 0Therefore y = -1/3 or y = 3So when x = 4 then y = 3 and when x = -8/3 then y = -1/3