In Solving Multi-Step equations?
918315
Distriubutive property:\n3p-6p-24=0\n\nCombine like terms:\n-3p-24=0\n\nAdd 24 on both sides:\n-3p=24\np=-8\n\nCheck:\n3(-8)-(6(-8)+24)=0\n-24-(-48+24)=0\n-24-(-24)=0\n-24+24=0\n0=0
341524
Second order polynomial equation in a single variable is called quadratic equation.\nSo the answer is no.
778162
Step 1: find y '\nStep 2: fill in x=1\nThat gives the right answer 2\n\nTh
408462
to solve such a system in general, what you do is this: take the first equation, solve it for x treating y as you would a constant. In your case, this is x = 4 + y.\n\nThen, take your expression for x and substitute it everywhere you see x in the second equation; in your case you get 3 *(4+y) + 2y = 7.\n\nSolve this now 1-variable equation for y, once you find y (in this case, -1).\n\nSubstitute the value of y into either equation, and you get the value of x (3 in this case).
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
232592
2x = 3y - 4y + 5 - 2\n2x = -y + 3\n2x + y = 3
363348
Are we solving for x?
51216
x = [-(-4) +/- root((-4)^2 - 4x1x(-5))]/2x1\n  = (4 +/- 6)/2\n  = 5 or -1
999333
There are several simple rules for re-arranging and solving equations.  It would be in your best interest to try and learn them, they aren't very hard.
518727
Multiply the first equation by 2.\n\nSubtract the second equation from the new first equation.\n\nSolve for y.\n\nTake your value for y and substitute it into the original first equation.\n\nSolve for x.
932620
solve the equations of any two and make sure that the solution satisfies the equation of the third perpendicular bisector
908374
Okay, the goal for solving simultaneous equations is to isolate one variable, so rearrange one of the equations (it doesn't matter which one) until you get either x or y alone on one side.\n\n5x - 2y = 10\n5x - 2y + 2y = 10 + 2y (if you do something to one side, do it to the other)\n5x = 10 + 2y\n5x/5 = (10 + 2y) / 5\nx = (10+2y)/5\n\nThen, plug this new value for x (the right side of the equation) into the other equation, hence the "substitution."\n3x + 2y = 6\n3 ((10+2y)/5) + 2y = 6\n(30 + 60y)/5 + 2y = 6\n6 + 12y + 2y = 6\n14y = 0\ny = 0\n\nNow, solve for x by substituting this y value back into one of the equations.\n\n5x - 2y = 10\n5x - 2(0) = 10\n5x = 10\nx=2\n\nSo, x=2 and y=0
782468
1.ac\n\n2.-3r\n\n3.3m/-2r\n\n4.(x-4)\n\n5.(a-b)(b-c)
695176
6x^2-3x-4x+2\n3x(2x-1)-2(2x-1)\n(2x-1)(3x-2)
763084
or solve equation for x and y components separately. \nfor x    0 +3+?=0 \nfor y    3+0 +?=0
268600
The addition method for solving simple algebraic simultaneous equations works quite well.  You simply add one equation to the other in order to eliminate a variable.  \n\nWith your example you add two times the first equation to the second (eliminating y) in order to solve for x.\n\n2(-5x + y = 6)\n+ 3x -2y = 3\n----------------\n-7x = 15\n\n=> x = (-15/7)\n\nThen substitue x back into one of the original equations to solve for y:\n\ny = 6-[5*(15/7)]\n\nYou may check your answer by substituting the values you found for x & y into the other equation.