find all the solutions of the system of equations{ x^2 + y^2=100,3x+y+10}?
439038
for first equation: x^2 and y^2 have to be 50. So the ansawer is &#92;/50 or radical 50 for both of them or it could be radical 51 and radical 49 and so on. For the second equation there is no solution.
197840
a^2 + b^2 = c^2, looks like a right triangle to me, 3,4,5, is the "obvious" solution.\n\nIf you don't recongnize that you have\n\nx^2 + (x+1)^2 = (x+2)^2\n\nx^2 + x^2 +2x + 1 = x^2 +4x +4\nSimplify\nx^2 - 2x - 3 = 0\n(x-3)(x+1) = 0\nx = 3, and x = -1 are the zeros so you have\n\n{3,4,5} and {-1, 0, 1} for solutions.
572161
a^2 + b^2 = c^2\n8^2 + 6^2 = c^2\n64 + 36 = c^2\n100 = c^2\nc = 10\n\nThis only work when you have a right triangle.
869879
Complete the square:\nx^2 - 4x + 4 + y^2 + 6y + 9 = 12+4+9\n(x-2)^2 + (y+3)^2 = 25 = 5^2\nRadius 5, centre (2,-3).
86546
(3y+1)(-2)+y\n-6y-3+y\n-5y-3\n\n12s+(7-s)2\n12s+14-2s\n10s+14\n\n7x-3x(x+1)\n7x-3x^2-3x\n-3x^2+4x\n\n3t(t-5)+6t^2\n3t^2-15t+6t^2\n9t^2-15t\n\n4w^2-w(2w-3)\n4w^2-2w^2+3w\n2w^2+3w
504285
It has been a long time, here are suggestions:\n1. 5x^3 + 11x^2 + 3x + 1  \n2. 4x^3 - 13x^2 - 8x + 7 \n3. 2x^4 - x^3\n4. 21x^2 + 26 x + 8
776012
(x - 99)(x -101)(x - 103)(x - 105) + 16 = 0\n(x - 102 + 3)(x -102 + 1)(x - 102 - 1)(x - 102 - 3) + 16 = 0\n[(x - 102)² - 3² ] [ (x - 102)² - 1] + 16 = 0\nLet (x - 102)² = y \n(y - 5 - 4)(y - 5 + 4) + 16 = 0\n(y - 5)² - 4² + 16 = 0\n(y - 5)² =0 --> y = 5\n(x - 102)² = 5\n\nx = 102 + sqrt(5)\nx = 102 - sqrt(5)
801726
You solve one of the equations for one of the variables, and put the solution into the other equation to solve the other variable:\n\n3x+y=4\ny=4-3x  then go to the other equation, and put this in for y:\n\n-2x-10(4-3x)=-12\n-2x-40+30x=12\n28x=28\nx=1     Now drop this answer into one of the original equations:\n\n3(1)+y=4\ny=4-3=1     So both x & y =1
798648
(x+3)(x^2+2x+4)\n\nStart by taking the first term in the first parenthesis and multiplying it by each term in the second parenthesis:\n\nx*x^2 + x*2x + x*4 = x^3 + 2x^2 + 4x\n\nNow take the second term and multiply it by each term in the second parenthesis:\n\n3*x^2 + 3*2x + 3*4 = 3x^2 + 6x +12\n\nNow add all of that together\n\nx^3 + 2x^2 + 4x + 3x^2 + 6x +12\n\nand combine terms\n\nx^3 + 5x^2 + 10x +12
665576
x^2<4x\n\ncase 1: x<=0\nno solution\n\ncase 2: x>0\ndivide by x\nx<4\n\ntherfore the solutions are 0<x<4
703163
6n + 2 + 4n = 102\n10n + 2 = 102\n10n = 100\nn = 10
698387
y= -(2/3)x^2 + (2/3)x + 4\n\n\n(P): y = ax^2 + bx + c\n\nPoints (0;4)  (-2;0)  (3;0) are element of (P)\n-->  4 = 0a + 0b + c --> c = 4\n0 = 4a - 2b + c\n0 = 9a + 3b + c\n\n--> 5a + 5b = 0 --> b = -a\n--> 4a + 2a + 4 = 0 --> a = -2/3,  b=2/3
1082168
y=x^2+10x+8\ny=(x+5)^2 - 17\n\nVertex form: (5,-17)
805031
1)  √10 · ³√100\n= 10^(1/2) · 10^(2/3)\n= 10^(7/6)\n\n2)  (6x)^(1/2) · (36x²)^(1/4)\n= (6x)^(1/2) · (6x)²^(1/4)\n= (6x)^(1/2) · (6x)^(2/4)\n= 6x\n\n...and so on.  Change the radicands to rational exponents with common bases, and follow the rules from there.  Good luck!
966925
There are 3 solutions for ABCDEFGHJ -\n\n2 6 7 8 3 5 1 4 9\n2 6 7 9 5 3 1 4 8\n6 7 8 4 1 3 2 5 9
753426
X2   5x - 4y = 1\n10x -8y =2\n-10x + 8y = -3\n-------------------------\n  0=-1                NO SOLUTION
1103883
Y=X^5-X^-3+X^2\nY'=5X^4+3X^-4+2X\nY''=20X^3-12X^-5+2