Find Constants A, B, and C so that y = A(x^3) + Bx + C satisfies the equation y''' + 2y'' - 3y' + y = x
92557
Derivate y = A(x^3) + Bx + C to obtain:\n\ny'''=6A\ny''=6Ax\ny'=3Ax^2+B\n\n\nThen substitute in the ODE:\n\n---> 6A+12Ax-9Ax^2-3B+Ax^3+Bx+C=x\n\nArrange for x:\n\nx^3(A)+x^2(-9A)+x(12A+B-1)+(6A-3B+C)=0\n\nThe left hand side the equation above must be identically zero, so any parentheses must be zero:\n\n----> A=0 , B=1 , C=1/3\n\n---> y=x+1/3
108152
(1/3)*e^(3x)-4sinx + c, where c is any arbitrary constant.
886265
gotta put the equation into y=mx+b first\ny= 1/3x - 1\n\nperpendicular would be y = -3x (+ or - any #)
1031270
x^2 + 5 + 5x + x y +5y\n\nx^2 + xy + 5x + 5y + 5\n\nx(x+y) + 5(x+y) + 5\n\nx(x+y) + 5(x+y+1)       OR     (x+5)(x+y)+5\n\nCynthia
994889
Start with a general cubic equation:\nx^3 +ax^2 +bx+c=0.\nFirst, we want to get rid of the x^2 term, so let \nx=y-(a/3).\nIf you plug this in and expand, the y^2 terms will drop out and you will get an equation of the form\ny^3 +Cy+D=0.\nNow, if you let y=u+v (to get an extra bit of freedom), and expand, you get\nu^3 +v^3 +(3uv+C)(u+v)+D=0.\nChoose v so that 3uv+C=0, i.e. v=-C/(3u). Plug this into\nu^3 +v^3 +D=0\nand multiply through by u^3 to get a quadratic in u^3. Solve\nthis by the quadratic formula and take a cube root to find\nu. There will be three cube roots in the complex plane, so find all of them. For each u, find v=-C/(3u), then y=u+v, and finally x=y-(a/3).\n\nThis is often a rather cumbersome proceedure, even if you do some of it ahead of time. Also, the roots may not 'look' like what you expect. If you apply this to an equation that has a root of, say, 4, you will obtain 4 as a sum of two cube roots of quadratic expressions. To recognize the answer as 4 is a challenge in and of itself.
1169467
y = x^2 + 5x - 3\n2y=3x - 2\n----------------\n2y = 2x^2 + 10x - 6\n2y = 3x - 2\n-----------------------\n    0 = 2x^2 + 7x - 4\n=>2x^2 + 7x - 4 =0\n=>2x^2 + 8x - x - 4 = 0\n=>2x(x + 4) -1(x+4) = 0\n=> (2x - 1)(x + 4) = 0\ntherefore 2x - 1 = 0 => x = 1/2 or \nx + 4 = 0 or x = -4\nx = - 4 or 1/2  \ngot it?
859744
(x + 4y)(x - 4y)\nI answered 2 of your questions. Do the rest of the questions yourself.
913056
The trick here is to find three equations where three of the variables are expressed functions of another variable. \nSo for example we could express x, y and w functions of z\nIf you add the first equation to the third and substract the second, you'll find: y= -3-z. We'll call this new equation A\nIf you add the second and the fourth equation and replace y by its value found in (A), you'll have: w= 7/4 + 1/4 z, which we'll call (B)\nNow take the first equation and replace w and y by their value from (A) and (B): \nx= -17/2 -7/2 z, this is the equation (C).\nNow that you have your three equations A,B and C you can substitute them in one of the four initial equations, for example the third one. You'll find:\nz=1\ny= -4\nw=2\nx= -12\nThis is not the entire solution (you have to detail how to get to A, B and C) but you should be able to do it (I don't want to be the only one working here!)
699742
-(3x+y)(3x-y)\n-(9x^2-y^2)\ny^2-9x^2
394265
Factor it out-- A(B+C)=AB+AC.  So, let X=A, B=1, C=Y.  You get X+(X*Y)=X(1+y)=Z
913768
y = ax² + bx + c\n\nWhen x = 2 y = 30 so 4a + 2b + c = 30 ...... (1)\nWhen x = 3 y = 15 so 9a + 3b + c = 15 ...... (2)\nWhen x = 4 y = 6 so 16a + 4b + c = 6 ........ (3)\n\n(2) - (1) 5a + b = -15 ....... (4)\n(3) - (2) 7a + b = -9 ......... (5)\n(5) - (4) 2a = 6\na = 3 so b = - 30 and c = 78\n\nie y = 3x² - 30x + 78
932620
solve the equations of any two and make sure that the solution satisfies the equation of the third perpendicular bisector
697874
Multiply or divide all components in one question to set the variables in the opposite questions to cancel out if you added the two problems together by combining like terms. For example:\n2x + 3y = -13  and 4x + y = -1\n\nTo keep this simple I will multiply all the components of the 2x + 3y = -13 equation by negative 2, resulting in:\n\n-4x - 6y = 26\n\nThen add the new equation to the other equation 4x + y = -1\n\n-4x -6y = 26\n4x + y = -1\n----------------\n0x -5y = 25  which simplifies to y = -5\n\nInsert the value found for y back into one of the original eqations and solve for x\n\n2x + 3(-5) = -13\n\n2x + -15 = -13\n\n2x = 2\n\nx = 1\n\nYour end result is x =1 and y = -5
11448
Substitution Method\nSolve this system of equations:\n3y - 2x = 11\ny + 2x = 9\n \n\n1.  Solve one of the equations for either "x =" or "y =".\nThis example solves the second equation for "y ="\n \n\n3y - 2x = 11\ny = 9 - 2x\n \n\n2.  Replace the "y" value in the first equation by what "y" now equals.  Grab the "y" value and plug it into the other equation. \n\n \n\n3(9 - 2x) - 2x = 11\n \n3.  Solve this new equation for "x".\n \n\n(27 - 6x) - 2x = 11\n27 - 6x - 2x = 11\n27 - 8x = 11\n-8x = -16\nx = 2\n4.  Place this new "x" value into either of the ORIGINAL equations in order to solve for "y".  Pick the easier one to work with!  \n\n \n \n\ny = 9 - 2(2)\ny = 9 - 4\ny = 5\n \n\n5.  Check:  substitute x = 2 and y = 5 into BOTH ORIGINAL equations.  If these answers are correct, BOTH equations will be TRUE!\n   \n\n3y - 2x = 11\n3(5) - 2(2) = 11\n15 - 4 = 11\n11 = 11\n\ny + 2x = 9\n5 + 2(2) = 9\n5 + 4 = 9\n9 = 9 \n\nElimination:\n x - 2y = 14\nx + 3y = 9\n \n\na.  First, be sure that the variables are "lined up" under one another.  In this problem, they are already "lined up".\n \n\nx - 2y = 14\nx + 3y = 9\n \nb.  Decide which variable ("x" or "y") will be easier to eliminate.  In order to eliminate a variable, the numbers in front of them (the coefficients) must be the same or negatives of one another.  Looks like "x" is the easier variable to eliminate in this problem.\n  \n\nx - 2y = 14\nx + 3y = 9\n \n\nc.  Now, subtract to eliminate the "x" variable.  \n(Remember:  when you subtract signed numbers, you change the signs and follow the rules for adding signed numbers.)\n \n x - 2y = 14\n-x - 3y = - 9    \n\n    - 5y = 5  \nd.  Solve this simple equation.   \n\n-5y = 5\n   y = -1\n \n\ne.  Plug "y = -1" into either of the ORIGINAL equations to get the value for "x".\n   \n\nx - 2y = 14\nx - 2(-1) = 14\nx + 2 = 14\nx = 12\n \n\nf.  Check:  substitute x = 12 and y = -1 into BOTH ORIGINAL equations.  If these answers are correct, BOTH equations will be TRUE!\n   \n\nx - 2y = 14\n12 - 2(-1) = 14\n12 + 2 = 14\n14 = 14  \n\nx + 3y = 9\n12 + 3(-1) = 9\n12 - 3 = 9\n9 = 9
23387
aY + b = c\n=> aY = c-b\n=> Y = (c-b)/a\n\nin this case:\n5 - 2Y = 7\n\na= -2\nb= 5\nc= 7\n\n=> Y = (7-5)/(-2)\n=> Y = -1
121228
here is a simple program algo\n\nprogram quad\nbegin\nvar\n     declare variables;\nget variable values of a,b,c for equation ax2+bx+c=0\nd=)b*b)-4*a*c;\nif(d<0)\nbegin\n\n   root are imaginary;\nend\nelse\n\n    root1=b(sqrt(d))/2a;\n    root2=-b(sqrt(d))/2a;\n ( note : formula i am not sure make changes as per required)\nprint the value of root1 and root2;\nend
612379
1. Move the non variable constant to the opposite side\n3x^2 - 8x = 2\n\n2. Factor out the 'a' term\n3(x^2 - 8/3x) = 2\n\n3. Take 1/2 the middle term, square it, and add it to both sides\n1/2 of 8/3 = 8/6\n3(x^2 - 8/3x + (8/6)^2) = 2 + (8/6)^2\n\n4. Factor and simplify\n3(x - 8/6)^2 = 2 + 64/36\n3(x - 8/6)^2 = 72/36 + 64/36\n3(x - 8/6)^2 = 136/36\n(x - 8/6)^2 = 136/108\n(x - 8/6)^2 = 68/54\n(x - 8/6)^2 = 34/27\nx - 8/6 = sqrt(34/27)\n\nx = 8/6 + sqrt(34/27)\nand\nx = 8/6 - sqrt(34/27)