What would be the answer to this system using substitutions?
94721
y= 2x + 11 and y=-x + 5 \n\n2x + 11 = -x + 5\n3x + 11 = 5\n3x = -6\nx = -2\n\ny = -(-2) + 5\ny = 2 + 5\ny = 7\n\n(-2, 7)
107281
Okay, I see your two equations are 5x + 5y = -30 and \n2x - y = 15.  I'm not certain if these are right so I'll explain how to solve systems of equations in general.\nThe way we solve a system of equations by substitution is to solve one of the equations for one of the variables.  We then take what that variable is equal to and substitute it into the equation we did not yet touch.  We then solve the resulting equation.  \nThe explanation is confusing so I will provide an example.\n4x + y = 9\n-x + 3y = 1\nSo I'll solve the second equation for x like so:\n-x + 3y = 1\n-x = 1 - 3y\nx = 3y -1\nThen, I substitute 3y - 1 into the first equation for x, since this expression is equal to x.\n4x + y = 9\n4(3y-1) + y = 9\n12y - 4 + y = 9\nAdding four to both sides and combining like terms I get:\n13y = 13\nTherefore y = 1.\nWe can then solve for x using one of the original equations, we'll pick 4x + y = 9.\n4x + y = 9, and we know y=1 so:\n4x + 1 = 9\nSubtracting 1 and solving gives us\n4x = 8, and therefore x = 2.\nYou can use this same process to solve your problem.  I suggest starting by solving the second equation for y.\nHope this helps, please remember to come back and choose a best answer!  :-)
553253
http://en.wikipedia.org/wiki/Liskov_substitution_principle click the link to know in detail.
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).
408554
yes. take out an x \nyou answer should be x(x+3)
973940
A previous answer was incorrect. It is\n\n-4x(x+3y)(x-2y)
255958
Substitute 2 for x and solve for c.\n2^2-3(2)+c=0\nRemember do powers, then multiplication, then subtraction.
741312
If f(t)=(t-a)*(t-b) = t^2-(a+b)*t+ab; then \nI=(1/3)*t^3 - (a/2+b/2)*t^2 + a*b*t+C
575490
I- + 2e- -------> I2\n\n4P + 3e-  -------------> P4\n\nthere ya go :)
137681
Ok...\nI see what you are doing...\nYou merely substitute the number in for x...\nso the problem becomes 6-2*(-1)\n6+2\n8
434926
2Sb+3 + 3S-2 => Sb_2S_3\n\nquepie's probably right.  I'm not going to look up SB(OH)_3.\n\nThen, it'd be\n\n2Sb(OH)_3 + 3S-2 => Sb_2S_3 + 6OH-
211777
n=3\nl=1\nm sub l =1,0, or -1\nm sub s = + or - 1/2
733305
M - 1 = D + 1\n2 * (D - 1) = M + 1\n\ncan you solve the rest as two equations with two variables?\nM = D + 2\n(2 * D) - 2 = M + 1\n \nM = D + 2 substitute into\n2 * D = M + 3\n\n2 * D = (D + 2) + 3\n\n2 * D = D + 5\n\nLooks like D = 5.\nSo M = D + 2 = 7\n\nDonkey has 5 and Mule has 7.\nCheck it out and see if it works.
898860
First one: d= - 1\n\nEDIT: the rest I'm too lazy to do. Do your own homework! =D
902613
I just plugged this into my calculator.\n\n[1, 0, 2, 0, 2, 0 | 0\n0, 1, -3, 0, -2, 0 | 0\n0, 0, 0, 1, -1, 0 | 0\n0, 0, 0, 0, 0, 1 | 0\n0, 0, 0, 0, 0, 0 | 1 ]\n\nTake it however you want.
417188
y=Asin(2x)+Bcos(2x)+(x/4)\n\nNow you get to determine why.
1090404
look at what is added (in parentheses)\n\n2 (+1) 3 (+2) 5 (+0) 5 (+3) 8 (-1) 7 (+4) 11\n\nStarting with the first and taking every other number you have +1 +0 -1.  The other values are +2 +3 +4.  So you have two patterns interwoven.  Following those patterns will yield:\n\n2, 3, 5, 5, 8, 7, 11, 9, 14, 11, 17, 13, 20