substitution problem.?
826575
Use one equation to substitute into the other,then solve for a one variable problem.  Use your new information (the solved variable) to solve for the second variable.\n\nFor the first set:\n\n-3x - y = 2 rearranged becomes:\ny = -3x -2\n\nUsing this "definition" of y, substitute it into the second equation as such:\n\n-6x + 2(-3x -2) = -2\n-6x - 6x - 4 = -2\n-12x = 2\nx = -1/6\n\nSubstitute -(1/6) for x in either equation to solve for y:\n\n-3(-1/6) - y = 2\ny = -3/2\n\nDo the same for the second pair of equations to make sure you got it (the answers for x and y are x=9/7, y = 1/7).
769661
the expression is being cut off.\nPlease break the expression over multiple lines, \nand I'll try to help
1022433
To perhaps elaborate on nardneran's answer:\n\nIn this problem you use the substitution rule. Let u=ln x. Then du=1/x dx and dx=x du. Replacing ln x with u and dx with x du gives ∫u² du (the x's in your equation cancel). You can integrate that to get u³/3 + C and back-substituting gives (ln x)³/3 + C.
874851
2Al (s) + N2(g) ----> 2AlN (s)
642877
int substitute(char c1, char c2, char *s) {\n// init. counter\nint num = 0;\n\nwhile (*s != '&#92;0') {\n\n// compare string char with 1st char\nif (*s == c1) {\n// if equal substitute with 2nd char and incr. counter\n*s = c2;\nnum++;\n} \n// incr. string pointer\ns++;\n}\n\n// return number of occurrences\nreturn num;\n}
1135408
OH NOES THIS IS LIKE A MATH PROBLEM!!!! i can't handle any more variables :( sry.\n\n\nhahahaha.
1149659
substitution:  substitute for I in eq. 2, since you know that I=e/(r+R).  Then you can solve for r.  when you get that, go back to eq. 1 and plug in the numbers
306926
What is wrong with the following list? ABCDEFGHIJKLMNOPQRSTUVWXYZ\n\nIt is a trick.... This question was on here yesterday.
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
1006179
111221, 312211, 13112221, 1113213211, ...\nOR\n1231, 131221, 132231, 232221, 134211, ...\nOR\n3112, 211213, 312213, 212223, 114213, ...\n\nEach is based on the digit makeup of the previous term, but I need one more term in the sequence to be sure of the rule.
746627
Cl2 + 2KI --> 2KCl + I2\n\nsingle replacement
798195
Don't be scared by all the words.  You just have to remove brackets and parentheses from\n2[x+(x+10)+(x+2(10))] \n\nand then collect the terms -- you should get 6x+60.\n\nThe last part just means do it again with 15 replacing 10.
85088
U must have given the name of the variable in the "  ".\nGive it outside double quotes.
157002
71 r 2\n\nor\n\n71 and 2/3\n\nor \n\n215/3
843441
The answer is c.\nC12H22O11+12O2-->12CO2+11H2O
538319
A -> B\nB -> C\n...\nT -> U\nYou (U) -> V\nV -> W\n\nSo, You -> V
387063
X=[-(30y-433)]/1232\nY=[-(1232x-433)]/30