Use the zero product property to solve: 7y-1+3y^2+y-1.?
549551
The way this is written, there is nothing to solve because it's not an equation.\n\nTo use the zero product property, you want to factor an expression that equals zero. Then you conclude that if a product equals zero, one of its factors must be zero. To factor the expression in your problem, combine any like terms to simplify, then factor the simplified expression.\n\nHope that helps.
982146
Just add and substract terms in both sides.\n\nSince \ny-x^2+8x\nis equal to 7, \nyou can add x^2 and substract 8x to both sides.\n(y-x^2+8x) +x^2-8x = y\n7 +x^2-8x = x^2-8x+7\n\nso y=x^2-8x+7\n\nby the way, if you factorize in terms of x then you'd get\ny=(x-1)(x-7)
701694
7x + y = 1 (subtract 7x from both sides)\ny = 1 - 7x\n\n3x + 2y = -9 (substitute in for y)\n3x + 2(1 - 7x) = -9\n3x + 2 - 14x = -9 (subtract 2 from both sides)\n-11x = -11\nx = 1\n\ny = 1 - 7x (substitute in for x)\ny = 1 - 7(1)\ny = 1 - 7\ny = -6\n\nAnswer: (1, -6)
805693
#1\n4x+1+3=0\n4x+4=0\n4x=-4\nx=-1\n\n#2\n2y+7+4=y\n2y+11=y\ny=-11
197692
y-5-y-5=7y\n-10=7y\ny=-10/7
750522
Begin with where you left of:\n\n2x + 2yy' = 0\n\nDifferentiate again, just like the first time.  Use the chain rule on the yy' term:\n\n2 + 2(y'y' + yy'') = 0\n1 + y'^2 + yy'' = 0\ny'' = (-1-y'^2)/ y\n\nFrom your y' = -x/y\n\ny'' = (-1-(-x/y)^2)/ y = -x^2/y^3 - 1/y\n\nThis isn't the answer you have but see if they are the same:\n\nx^2 = 4 - y^2\n\ny'' = -(4-y^2)/y^3 - 1/y = -4/y^3 + 1/y - 1/y = -4/y^3\n\nSo the two solutions are the same
934203
y + 2y - 10 = 2y + 2\n3y - 10 = 2y + 2\n3y = 2y + 12\ny = 12
1055222
Is that e^y?\n\nCompute the differential of both sides\n\n7dxe^y + 7xe^ydy + 9x^2dx = 20y^3dy\n(20y^3 - 7xe^y)dy = (7e^y + 9x^2)dx\ndy/dx = (7e^y + 9x^2)/(20y^3 - 7xe^y)
141598
If we break down 25x^2, we get 5x.\n\nIf we break down 49y^2, we get 7y.\n\nThe resulting equation is:\n\n(5x + 7y)(5x - 7y)\n\nSolve it from there.
861681
subtract 1st equation from the second:\n  8y^2=32\nso y^2=4, and y = +/- 2\n\nplug back into 1st equation:\n4x^2+4=4\n\n4x^2=0\nx=0
496037
-3x + 5y = 8\nx + 2y = 1\n\n(-3x + 5y) + (x + 2y) = 8 + 1\n-3x + 5y + x + 2y = 9\n-2x + 7y = 9\n\n7y = 2x + 9\ny = (2/7)x + (9/7)\n\nx + 2y = 1\nx + 2((2/7)x + (9/7)) = 1\nx + (4/7)x + (18/7) = 1\n7x + 4x + 18 = 7\n11x + 18 = 7\n11x = -11\nx = -1\n\ny = (2/7)x + (9/7)\ny = (2/7)(-1) + (9/7)\ny = (-2 + 9)/7\ny = (7/7)\ny = 1\n\nx = -1\ny = 1
773396
x+y-z=2---(1)\n2x-y+z=7-(2)\n-x-y+2z=3---(3)\nadding (1) and (2)\n3x=9\nx=3\nadding 1 and 3\nz=5\nsub x=3,z=5 in 1\n3+y-5=2\ny-2=2\ny=4\nthe solution set{3,4,5}\n\n2.x+2y-z=-3\n2x+y+z=6\nadding\n3x+3y=3\n=>x+y=1\neqn. 1*3\n=3x+6y-3z=-9\n-2x+2y+3z=3\nadding\nx+8y=-6\nx+y=1\nsubtracting 7y=-7\ny=-1\nsub in x+y=1,x=2\nsub in x+2y-z=-3\nz=3\nso solution set {2,-1,3}
287717
-4y^+8y+3y^2-6\n-4y(y^2-2)+3(y^2-2)\n(-4y+3)(y^2-2)\n\ndoes that make sense?
642018
Wonderful.\nTHis is some sort of simultaneous equation.\nAccording to the principle of this equation, we try to rationalize one term of the equation so as to get one unknown term. We can either try finding the x term or the y term. For your question as to find the y term.\n\nFirstly, we rationalize the x terms by multipling them by the inverts. then it will result to \n                                    12x + 20y = 32\n                                    12x + 18y = 39\nThe we apply the subtraction rule. ie. we multiply all the terms beneath by the negative (-) sign. This will result to \n                                    12x + 20y = 32\n                                   -12x - 18 = -39\nWe later subtract so as to attain\n                                2y = -7\nWe apply the law of linear equation by dividing both sides by the coefficient which in this case is 2.\n                              2y/2 = -7/2\nThis will eventually cancel the 2 in the first term therby leavin y alone. \n The answer will be in either of the following cases through simplification.  -7/2, -3 1/2 or -3.5
697044
(y+10)^2-72=0 \n(y+10)^2 = 72\ny+10 = +- sqrt(72)\ny= -10 +- sqrt(72)\n-10 + sqrt(72), -10 - sqrt(72)\ns
811856
Okay multiply the top equation by 7 and the bottom equation by negative 3 to obtain 2 new equations with the x variable to eliminate:\n\n7(3x + 7y = -43)\n-3(7x + 6y = -59)\n\nthis becomes \n\n21x + 49y = -301\n-21x - 18y = 177\n\nNow cancel out the 21x's and add down.  Doing that, the equation becomes\n\n31y = -124   this can easily be solved for y\n\ny = -4\n\nNow plug -4 back into one of the original equations\n\n3x + 7(-4) = -43\n\n3x - 28 = -43\n\nadd 28 to both sides to yield\n\n3x = -15   and solve for x\n\nx = -5\n\nNow check x = -5 and y = -4 into both and see that you have solved this system correctly.
683485
It's easy.\nYou have 2 linear equation:\n3x - y = -7  -> equation  1\nx + y = -9   -> equation  2\n\nNote that in equation 2 you can rewrite as follow:\n  x = -9 - y \nBy simply moving the y to the other side.\nSo you know what x is in terms of y.\n\nIn equation 1 you have:\n      3x - y = -7  \nBut you know x now in terms of y and you can SUBSTITUTE to equation 1.\n\n    3(-9 - y) - y = -7\nSolving and expanding the multiplication you have\n    3*(-9) -3*y - y = -7\n    -27 - 4*y = -7\n         -4y  = -7 + 27\n           y  = -5\nSo you know y.\n\nIf you want to know x, you SUBSITUTE back to equation 2, since there you know x in terms of y.\nSo:\n    x = -9 - y \n    x = -9 - (-5)\n    x = -4\n\nChecking the answers by SUBSTITUTING back: \n    x = -4\n    y  = -5\n\n3x - y = -7  -> equation  1\n3(-4) - (-5) = -7  \n-12  + 5 = -7   (correct)\n\nx + y = -9   -> equation  2\n-4 + (-5) = -9 (correct)\n\nHope you understand the process :)