-(3*y/(y+2) = - (18/ y^2 - 3y - 10)?
966529
I just typed this all out and then lost it. Here I go again.\n\nfirst I would multiply both sides by -1/3 because i want to get rid of the neg and i know 3 goes into 18 and 3 and I want to work with smaller numbers.Also factor the denominator.\n\ny/(y+2)= 6/(y+2)(y-5)   Now mult. by the LCD (y+2)(y+5)\n\ny(y+5)=6   remove () and sub 6\n\ny^2+5x-6=0  factor\n(y+1)(y-6)=0  set each factor =0\ny+1=0  or  y-6=0  solve\ny=-1  or  y=6\n\nalways remember to check your answers in teh original denominators to make sue you won't get zero.
495630
your answer is incorrect.\n\nthis is the correct answer:\n\n18x+9y=27 ----->  y= -2x+3; intercept at y= 3 and x= 3/2\n\nhow do i get it?\n\nwhen intercept at y, the x value must be zero, so put 0 into x, you get y= -(2 * 0) + 3 = 3\n\nwhen intercept at x, the y value must be 0, so put 0 into y, you get 0 = -2x + 3 -----> -2x = -3 ----> 2x = 3 ----> x= 3/2\n\ni got trouble also when  i was in school but i keep on practising\n\ngood luck
831999
y = (x+3)(x-1)(x-2) = x^3 - 7x + 6
169494
The answer is:\n2y^3 - y^2\nOR\n(2y-1)*y^2
844586
9x^2+y^2=9......(eqn no 1)\n3x-y=3 .............(eqn. no. 2)\nFrom equn. no. 2,we get y=3x-3=3(x-1)\nPutting the value of y in eqn.no.1,w get\n9x^2+{3(x-1)}^2=9\n=> 9x^2+9(x^2-2x+1)=9\n=>9x^2+9x^2-18x+9=9\n=>18x^2-18x=0\n=>18x(x-1)=0\nTherefore either 18x=0 or x-1=0\ntherefore x=0 or 1\nPutting the value of x in eqn no 2,we get 3X0-y=3 or y=-3  or 3X1-y=3,in which case y=0 \nTherefore x=0,y=-3 or x=1 and y=0
758594
2. The equation is (y-y1)/(y1-y2)=(x-x1)/(x1-x2)\n=>(y+2)(-2-0)=(x+1)(-1-0)\n=>(y+2)(-2)=(x+1)(-1)\n=>-y-2=-2x-2\n=>-y+2x=0=> 2x-y=0 which is the required equn.\n3.The equation is (y-4)(4+4)=(x-0)/(0-4)\n=>y-4/8 =x/-4\n=>8x=-4y+16  or,8x+4y-16=0  o4 2x+y-4=0 is the reqd.equation\n4.the eqn. is (y+3)/(-3-3)= (x+2)/(-2-0)\n=>(y+3)/-6=(x+2)/-2\n=>-6x-12=-2y-6\n=>-8x+2y-6=0 or,-4x+y-4=0 which is the equired equation
70526
=((-4y)^2)/(10y-5)= (16y^2)/(10y-5)\nand if the equation is equal to zero then y=4i
934203
y + 2y - 10 = 2y + 2\n3y - 10 = 2y + 2\n3y = 2y + 12\ny = 12
973940
A previous answer was incorrect. It is\n\n-4x(x+3y)(x-2y)
360927
y'=dy/dx=\n\n(6√x)' + (3/x^2)' + (10)'=\n\n[6*x^(1/2)]' + [3x^(-2)]' + 0= \nnow use the power rule\n\n6*(1/2)* x^[(1/2)-1] + 3*(-2)*x^(-2-1)=\n\n3*x^(-1/2)-6x^(-3)=\n\n3/x^(1/2) -6/x^3=\n\n3/√x - 6/x^3
809908
2(y-3) = y-6\n\nThis is what you are supposed to calculate : a value for the y, such that the equation holds.\n\nfor instance try y = 1\n\n2(1-3) = 1-6\n2 - 6 = 1- 6\nand you see that this is not so, thus y=1 is not a solution.\n\nthe standard way to solve is : move all y''s to the left of the=, and all numbers to the right of the =\n\n2(y-3) = y - 6\n2y - 6 = y - 6\n2y -y - 6 = -6 ( move left y to the right, sign changes )\n2y - y = -6 + 6 moved left -6 to right, change sign\n\ny = 0. ( adding the 2y - y together, and the -6 + 6\n\nso the solution is y = o
728357
3x^2y^2(2y+3y)/3x^2y^2\n=5y
446101
switch the y and x\n\nx = 2y - 3\n\nThen solve for y\n\nx + 3 = 2y\n\n(x+3)/2 = y\n\nnow you have the inverse.
641389
8Y(2Y-1)=0\n16Y^2-8Y=0\n16Y^2=8Y\n16Y^2/y=8Y/y\n16Y=8\nY=8/16 or 1/2
864521
y = 3/(1 - x)²\n\n1, (1 - x)² ≥ 0 for all x and therefore 3/(1 - x)² ≥ 0 for all x\n\n2. 3/(1 - x)² is undefined when x = 0. Therefore 3/(1 - x)² (ie y) → +∞ when x → 1 from the left of 1 and when x → 1 from the right of 1\n\n3. As x → ±∞, 3/(1 - x)² (ie y) → 0\n\n4. Plot a few other points and Bob's your uncle!!
823232
y = (3x - 2)/(5x² + 4)\n\nW.w.t.f dy/dx.  You can use the formula:\nd/dx (u/v) = (v du/dx - u dv/dx)/v²\n\nNow, we have u = 3x - 2, v = 5x² + 4.  du/dx = 3 and dv/dx = 10x. We can now substitute them to the formula:\ndy/dx = [(5x² + 4)(3) - (3x - 2)(10x)]/(5x² + 4)²\n\nNow,\ndy/dx = (15x² + 12 - 30x² + 20x)/(5x² + 4)²\n\nThus,\ndy/dx = (-15x² + 20x + 12)/(5x² + 4)²\n\ntherefore,\ndy/dx = -(15x² - 20x - 12)/(5x² + 4)²\n\n^_^
683580
y^2 + 7y + 3y + 21\ny^2 + (7 + 3)y + 21\ny^2 + 10y + 21\n(y + 7)*(y + 3)