Please Help with System of Equation Problem?
438685
from the last equation: Z = 8 - 2X\nthen replace Z in the second equation = -3X + 2Y -(8 - 2X) = -4\nthis equals to : Y = X/2 +2\nNow that u have Z and Y replace them in the first equation:\nX + 2 (x/2 +2) + 6(8 - 2X) = 4\nthat's X = 24/5\nZ = -8/5\nY = 22/5
247689
Use analytical trig (hint: sin function).  That should give you the answer you are looking for.
469735
The system of equations:\n[a11 a12 a13; a21 a22 a23]\n[1 -3 4; 2 3 1]\n\nPerforming indicated row operation:\n[a11 a12 a13; a21' a22' a23']\n[1 -3 4; -1 -6 3]\n\na21' = a11 - a21 = 1 - 2 = -1\na22' = a12 - a22 = -3 - 3 = -6\na23' = a13 - a23 = 4 - 1 = 3
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
956578
x^2+3x-y=c\n2x +3 -y'=0\ny'= 2x +3\n\n(remember y^0=1)!
736379
for this just substitute for x and y in eqn and see if LHS=RHS\nLHS=4\nRHS=(-1/2)(-3)-5/2\n=3/2-5/2\n=-2/2\n=-1\nnot = RHS\nso pt. not on line
859744
(x + 4y)(x - 4y)\nI answered 2 of your questions. Do the rest of the questions yourself.
1134427
12x + 4y = 8\n(d/dx) * (12x + 4y = 8)\n12 + 4dy/dx = 0\ndy/dx = -12/4\ndy/dx = - 3
901005
Use the following equation and solve:\n\nP1V1/T1=P2V2/T2\n\nT2=P2V2T1/P1V1\n\nalso note the significant figures!!!!
1135408
OH NOES THIS IS LIKE A MATH PROBLEM!!!! i can't handle any more variables :( sry.\n\n\nhahahaha.
210912
Put the left side coefficients into matrix [A],\n[6 2]\n[5 6]\nand the right side values into matrix [B]\n[-16]\n[..4]\nthen tell the calculator to multiply [A] inverse * [B]\nthe result is the solution matrix\n[x]\n[y]\n\nnote, the inverse key looks like x^-1.  \n\nsolve the second system the same way.  coefficient matrix will be 3x3, answer matrix 3x1.
566070
0 = 2y + 2z\n\nz and y must be opposites and equal
929562
∫ax(2 -x)dx\n= ∫(2ax - ax²)dx\n= 2a * ½x² - a * ⅓x³ + c\n= ax² - ⅓ax³ + c
851212
hmm im not quite in college yet, but i hope to be help.\ndo you know the values for f, h, v, and a?\nif you do, id substitute it in.\nthen square both sides. \nand then solve.
978722
You can't, algebraically.  You have two equations and three variables.  Sorry.  You need one more equation, such as a = b.
982897
the 1st one is simple\n(x+y) dy/dx + y =0\nor xdy/dx + ydy/dx + y =0\nor (xdy/dx + y)  + ydy/dx = 0\nxdy/dx + y = d/dx(xy)\n\nand d(y^2/2)/dx \n\nso d/dx(xy+y^2/2) = 0\n\nor xy+y^2/2 = c\nor 2xy+y^2 = 2c or C\n\nproved\nfor the second one\n\ndy/dx -(1/x)y = xy^2\n\nlet y/x = t\ny = xt\ndy/dx -y/x = t+ x dt/dx -t = x dt/dx\nso\nx dt/dx = - x (xt)^2\nor dt/dx = x^2t^2\ndt/t^2 = x^2 dx\nyou should be able to proceed
81494
I'm getting some funky answers on these equations\n\nx= -48/7 and y= -156/7