Cubic and Quartic and other linear equations?
108399
use binomial theorem, it can be applied to any power.if you don't know this theorem, i'll send it to you.
712758
linear equarion is y=ax+b\n\n1) 10,000=a*100+b\n     b=10,000-100a....(i)\n\n2) 22,000=a*300+b\n    b=22,000-300a.....(ii)\n\nso,\n10,000-100a=22,000-300a\n12,000=200a\na=60, b=4000\n\nthe equation would be\ny=60x+4000
756992
Don't you just love the quadratic equation?\n\nThe first thing you got to do is get the given equation into the correct form, and in order to do that you have to combine all like terms. Since there are no like terms yet you have to do some math first.\n\n-2.0^2 = 4.0because neg * neg = pos\n\nNow you have 2 like terms 4.0 and 0.5 combine to get 4.5. Your eqution should look like this:\n\nf(t)=10t + 4.5 since you need at^2 + bt^2 + c just insert t^2. Now your equation will look like this:\n\nf(t)=t^2 + 10t + 4.5\n\nAnd just in case you need the quadratic equation:\n\nt= -b + or - sqr(b^2 - 4ac) / 2a\n\noh yeah and a = 1
405166
Try this:\nhttp://www.wtamu.edu/academic/anns/mps/math/mathlab/beg_algebra/beg_alg_tut9_bar.htm
783856
First, you should know that you can multiply any one of these equations through by a nonzero constant, without changing the solution. You can also add or subtract any two of the equations term by term without changing the solution.\n\nWe begin by adding equations 1 and 3 together\nx+y+z = -5\nx-y+4z=12\n---------------\n2x + 5z = 7  (notice that the ys cancel out)\n\nNow we triple equation 1, and subtract equation 2 from it\n\n3x+3y+3z= -15\n2x+3y-2z = 8\n--------------------\nx +5z = -23\n\nNow we have two equations, in which we have eliminated y.\nWe subtract them, and solve for x.\n\n2x + 5z = 7 \nx    +5z = -23\n-----------------\nx = 7 - (-23) = 30\n\nIt is easy to find y and z, now that we know x is 30/\n\nx + 5z = -23\n30 + 5z = -23\n5z = -53\nz = -(53/5) or -10.6\n\nx + y + z = -5\n30 + y + (-53/5) = -5\ny = 53/5 - 30 - 5 = -24.4\n\nSo (x,y,z) = (30,-24.4,-10.6)
224079
try this site... it shows step by step how to go about sloving these equations, good luck:
741586
set y=0\nsolve quadratic eqn\nquadratic eqn\n\n-7 +- (49-16)^.5\n-------------------\n2\n= (-7 +- 33^.5)/2
939068
Is this one problem?  If so, you can solve it like this:\n\n(a/b) / (c/d) = (a/b) * (d/c)\n\n~ ~ ~ ~ \n\n( (x² - 4) / x² ) / ( (x²+5x+6) / x ) = \n\n( (x² - 4) / x² ) * ( x / (x²+5x+6) ) = \n\n( (x² - 4) * x) ) / (x² * (x²+5x+6) ) = \n\n( (x² - 4)  / (x * (x²+5x+6) ) = \n\n{{ now notice you can factor your quadratic equations, to get the following }}\n\n(x+2)(x-2)  / (x * (x+2)(x+3) ) = \n\n(x-2) / (x(x+3))\n\nHope that helped!\n\n~ ♥ ~
120172
y-2 = -2(x-3)\ny-2 = -2x + 6 \ny+ 2x = 6+2 \ny+2x  = 8  ( linear equation)\n\nFINDING SOLUTIONS FOR THIS LINEAR EQUATION :\n\ny+2x = 8\n(a)  Assume, x=2 \n     subtitute the assumed value of x in the equation\n     y+ 2 (2) = 8\n     y+4 = 8\n     y = 8-4\n     y = 4\n     HENCE,  when -->   x= 2 then y= 4\n\n   Similarly you can find more solutions for any\n   given linear equation by assuming the value of any one unknown ( x or y ) and then finding the value of the other by substitution in the same given equation.
520168
Try this:\n\nhttp://www.algebra.com/
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.
1160760
what you have is something like\na(b+c)\n\nthe solution is \nab+ac\n\n(x^3)(2x^-2)+(x^3)(4x)\nnow to multiply x^a * x^b you get x^(a+b)\n\n2x^((3+(-2))+4x^(3+1)\n\n2x+4x^4
720929
Let me just work through the steps to solve the problem:\n\nsqrt(2x^2-1)=x   <---Original Equation\n2x^2-1=x         <---Square both sides\n2x^2-x-1=0       <---Subtract x\n\nNow, apply the quadradic equation:\nx=(-b+/-(sqrt(b^2-4ac))/(2a)\n\nFor this equation, the equation is:\nx=(-(-1)+/-(sqrt((-1^2)-4*2*(-1))/(2*(2))\n\nSimplified it is:\nx=(1+/-sqrt(1+8))/(4)\nx=(1+3)/4 or x=(1-3)/4\n\nSo, x=1 or x=-1/2
1007589
1. b^2-4ac>0 : b>2sqrt(ac)\n2. b^2-4ac=0 : b=2sqrt(ac)\n3. b^2-4ac<0 : b<2sqrt(ac)\n\nThe rest should be a piece of cake.
282131
Note that although x1=x2 satisfy the equations but some distinct values for x1 and x2 can satisfy it too: \n\nsuppose x1 and x2 are distinct and so x1-x2 is not zero\nthen eq1 is:\n\n4(x1-x2)(x1+x2)-4(x1-x2)=0\n\n---> 4(x1+x2)-4=0\n---> x1+x2=1\n\nand eq2 is:\n\n-3(x1-x2)(x1+x2)(x1^2+x2^2)+2(x1-x2)(x1+x2)=0\n\nomit (x1-x2)\nand substitute x1 by 1-x2 to get:\n(1-x2)^2+x2^2=2/3\n\n---> x2=(3+-sqrt(3))/6\n\nso we have two values for x1 and x2 which satisfy the equation and x1 and x2 are distinct.
700548
So x4-5*y2*x2+4*y4=0; or (x^2-y^2)*(x^2-4y^2)=0; or (x-y)*(x+y)*(x-2y)*(x+2y)=0\nIt splits into independent linear equations: (a) y=x; (b) y=-x; (c) y=0.5*x; (d) y=-0.5*x;\nNow exclude (a) & (c); thus (x+y)*(x+2y)=0; did you mean it Don?
530596
Given\nax^2 + bx + c = 0\nDivide all terms by a \nx^2 + (b / a) x + c / a = 0 \nSubtract c / a from both sides \nx^2 + (b / a) x + c / a - c / a = - c / a \nand simplify \nx^2 + (b / a) x = - c / a \nAdd (b / 2a)^2 to both sides \nx^2 + (b / a) x + (b / 2a)^2 = - c / a + (b / 2a)^2 \nto complete the square \n[ x + (b / 2a) ]^2 = - c / a + (b / 2a)^2 \nGroup the two terms on the right side of the equation \n[ x + (b / a) ]^2 = [ b^2 - 4a c ] / ( 4a^2 ) \nSolve by taking the square root \nx + (b / a) = ± sqrt { [ b^2 - 4a c ] / ( 4a^2 ) } \nSolve for x to obtain two solutions \nx = - b / a ± sqrt { [ b^2 - 4a c ] / ( 4a^2 ) } \nThe term sqrt { [ b^2 - 4a c ] / ( 4a^2 ) } may be written \nsqrt { [ b^2 - 4a c ] / ( 4a^2 ) } = sqrt(b^2 - 4a c) / 2 | a | \nSince 2 | a | = 2a when a > 0 and 2 | a | = -2a when a < 0, the two solutions to the quadratic equation may be written \nx = [ -b + sqrt( b^2 - 4a c ) ] / 2 a \nx = [ -b - sqrt( b^2 - 4a c ) ] / 2 a \nThe term b^2 - 4a c which is under the square root in both solutions is called the discriminant of the quadratic equation. It can be used to determine the number and nature of the solutions of the quadratic equation. 3 cases are possible \ncase 1: If b 2 - 4a c > 0 , the equation has 2 solutions. \ncase 2: If b 2 - 4a c = 0 , the equation has one solutions of mutliplicity 2. \ncase 3: If b 2 - 4a c < 0 , the equation has 2 imaginary solutions\nps. I did this as quick as possible, hope everything copied correctly.