what is the answer for 7+n=9+7 in pre-algebra?
143314
7+n=9+7\nfirst subtract 7 from both sides\n7+n=9+7\n-7   -7\nso now we have \nn=9\nThe answer is 9!
373987
of cos!\n\n9*9 + 9 + 9 + 9/9 + (9-9)9
361234
(m-6)(m+7)=0\nm=6 or m=-7\n\n(n-1)(n-7)=0\nn=1 or n=7\n\n(x+1)(x+5)=0\nx=-1 or x=-5\n\n(x-2)(x-7)=0\nx=2 or x=7\n\nas easy as ABC!!!
792643
-9x + 7 = 3x + 19\n\nSubtract 7 on each side...\n\n-9x = 3x + 12\n\nSubtract 3 on each side...\n\n-12x = +12\n\nDivide by -12 on each side...\n\nx = -1
588758
a.  ax + bx\nb.  21x^3 + 24x^2y\nc.  24x^2 + 31x + 10\nd.  24x^2 - 8x - 6\ne.  40x^3 + 66x^2 + 38x + 6  (I assumed that you meant 7x in the first half of the equation.  If not, this is not correct.)\nf.  x^2 -1\ng.  x^3 - 8
16389
x+9 < 16\n\nadd -9 for all\n===============> x+9 -9 < 16 -9    (I)\nx+9 -9 =x (II)\n16 -9 = 7 (III)\n\n(I) & (II) & (III)\n===================> x < 7\n\nif x is a natural number => \nx= 1 or 2 or 3 or 4 or 5 or 6
844276
There is no solution:\n\n6x+7y=7\nMultiply by 2\n12x+14y=14\n\nBut we already have\n12x+14y=5\n\n14 does not equal 5, so there are no common solutions.  What does this mean?  They are parallel lines.\n\nWe can do more:  We can tell the y-intercepts of the lines as well. \n\n6x+7y=7\n7y=7\ny=1\n\n12x+14y=5\n14y=5\ny=5/14\n\nThis means that when the graph of 6x+7y=7 is shifted down 9/14, and then it is the same graph of 12x+14y=5\n\nHope this helps!
822221
Using mods...\n\nn^3 + n - 4 = 0 mod 7\n\nn^3 + n = 4 mod 7\n\nn(n^2+1) = 4 mod 7\n\nIf n=0 mod 7 then n(n^2+1) = 0 mod 7, doesn't work.\nIf n=1 mod 7 then n(n^2+1) = 1(1^2+1) mod 7 = 2 mod 7, doesn't work.\nIf n=2 mod 7 then n(n^2+1) = 2(4+1) mod 7 = 3 mod 7, doesn't work.\nIf n=3 mod 7 then n(n^2+1) = 3(9+1) mod 7 = 2 mod 7, doesn't work.\nIf n=4 mod 7 then n(n^2+1) = 4(16+1) mod 7 = 5 mod 7, doesn't work.\nIf n=5 mod 7 then n(n^2+1) = 5(25+1) mod 7 = 130 mod 7 = 4 mod 7 - YESSSS!\nIf n=6 mod 7 then n(n^2+1) = 6(36+1) mod 7 = 222 mod 7 = 5 mod 7, doesn't work.\n\nSo, n must be equal to 7m+5 for some integer m.  This gives us:\nn^3 + n - 4 =\n(7m+5)^3 + 7m+5 - 4 =\n7^3m^3 + 3*7^2m^2*5 + 3*7m*5^2 + 5^3 + 7m + 1 = \na whole bunch of multiples of 7 + 5^3 + 1 = \nmultiples of 7, plus 126 =\nmultiples of 7, plus 18*7.
1124767
9a*a+6a-4b*b+1\n= (9a^2+6a+1)-4b^2\n=(3a+1)^2-(2b)^2\n=(3a+2b+1)(3a-2b+1)
304425
This way is easier for you to understand...always do the parenthesis first\nin this case 9 + 5 = 14 still got you til there ?\nokay then you will always multiply whatever the parenthesis gave you with the number in front of it in this case...\n7 * 14 =98 \n98 * 66 = 6468\nunderstood ??  Hope so
65662
the above answer is wrong.\n\n( x^2 + 7x - 4 ) + ( x^2 + 9 ) = x^2 + x^2 + 7x -4 +9\n= 2x^2 + 7x + 5
725839
I think it is this:\n\nFirst multiply through the groups..\nso you get:\n6n^7-4n^6-16+4n^6+2n^7+15\n\nThen gather like terms\n\n=8n^7-1\n\n(The -4n^6 and the +4n^6 cancel each other out)\n\nI don't think you can go further than that.\n\nGood luck.
801424
If m equals to 7\n3 (m-5)+1 = 2(m+1)-9\n3m-15+1 = 2m+2-9\n3m-14 = 2m-7 (add 7 to both side)\n3m-7 = 2m (subract 3m from both side)\n-7 = -1m\n-7 = m
973910
17.\n5x+7y=-1  (1)\n-2y+3z=9  (2)\n7x-z=27    (3)\n(3)*3\n21x-3z=81\n-2y+3z=9\nadding\n21x-2y=90 *7\n147x-14y=630\n10x+14y=-2\n157x=628\nx=4\nsub in (1)\ny=-3\nsubin (2)\nz=1\n\n\n18.\nr-s+3t=-8    (1)     \n2s-t=15       (2)\n3r+2t=-7      (3)\n\n(2)*2  4s-2t=30\n          3r+2t=-7\nadding\n4s+3r=23\n(2)*3  6s-3t=45\n       r-s-3t=-8\nadding\n5s+r=37\n4s+3r=23\n15s+3r=111\nsubtracting\n11s=88\ns=8\nsub in (2)\n16-t=15\nt=1\nsub in (3)\n3r+2=-7\nr=-3
935133
Ok, take it back.\n\nFlip the 6, turn it into a 9.  Move it over to the right hand side, move the 2 to an exponent over the 7.\n\n7^2 = 49.\n\nThat's it.
492221
you would multiply the first number of the first by the two in the next and then the second number by the two of the next and add them all.  That is:\n2x x 9x = 18x2  + 2x x 7= 14x then 1 x 9x = 9x and 1 x 7=7\nso you get 18x2 + 14x + 9x + 7 and then you further go to\n18x2 + 23x + 7  is the answer (the x2 is x squared by the way, that's cause it is x times x).
638676
1.)\nstep one: if they are zeroes, use opp. \nexample....\n (x-2) (x+9) (x-7)          if you solve for x it should be opposite of given zeroes.\n\nstep two: FOIL the first two...\nyou'll get   x^ +9x -2x -18 which simplifies to\n\n(x^2 + 7x -18) (x-7) \n\nstep three: FOIL these two together...\nyou'll get     x^3 +7x^2 - 18x -7x^2 -49x + 126\n\nwhich simplifies to  \n   x^3 - 67x +126\n\n\n2.) I don't know about the second one... I could solve it if I spent some time, but start with...\n (x - 4i) (x+5) (x + 2i)\nthan just do what I did in the problem above, factor the first two, than factor your answer with the last one... and remember that i^2 is equal to 1...