Can someone help me solve this problem by substitution?
930601
put -2y=5-3x call it equation {1}\ncall 4x-5y=86 {2}\n\nmultiply all of the terms i {1} by 5\nmultiply all ot the terms in {2) by 2\nthis gives -10y=25-15x {3} &  8x-10y=172 {4}\nyou've got -10y in terms of x in {3}, so just put that into {4}  8x+25-15x=172, -7x=157, \nx = -172/7 sub in to {1} to get y \nman that was ablast from the past, haven't done that for years.
511948
There are an infinite number of ordered pairs that would satisfy this equation. If you want to find a few of them, assign a value to x and then solve for y.
958467
Replace x with 2:\nf(2)=2^3-4(2)^2+14(2)-20\nf(2)=8-16+28-20\nf(2)=-8+8\nf(2)=0
851212
hmm im not quite in college yet, but i hope to be help.\ndo you know the values for f, h, v, and a?\nif you do, id substitute it in.\nthen square both sides. \nand then solve.
518727
Multiply the first equation by 2.\n\nSubtract the second equation from the new first equation.\n\nSolve for y.\n\nTake your value for y and substitute it into the original first equation.\n\nSolve for x.
1017997
The equations are wrongor insufficient. Post the correct data if you want me to help/ Alternately, you could also message me.
920579
use synthetic  divsion method \ni think this link would help\nhttp://www.purplemath.com/modules/synthdiv.htm
808688
the 1st three terms means n=1, n=2, n=3 so substitute\n\nThis gives, 3*1+2,3*2+2,3*3+2\n\n=5,8,11
51150
UGH!!!  Thx for the effort (I was the one to post that question, as a part of a solution to another problem I found here on Yahoo! Answers.)  Incidentally, I made a mistake when simplifying the other problem and it should be xln(2)-2ln(x)-ln(86400) = 0\n\nI don't know any way to get rid of the ln(x) unless maybe it's using calculus techniques?  I'm just glad this problem is making other ppl trouble about it as much as I did :-).
934633
By substitution, the first equation when substituting in y=2+x becomes:\n4x - 4(2+x) = -8 ==> -8 = -8 which is always true.  This means that what you have are two equations with an infinite number of solutions, as the two lines are the same line.  So any pair of the form (x,x+2) such as (0,2) or (1000,1002) will work.
191567
Please use parentheses to explain the problem more effectively. Assuming the problem as,\n\n(x-1)/2x + x/(x+3)\n\nThe solution can be,\n\n(3x^2+2x-3)/(2x^2+6)
825795
OK,as the first one is already done,I am doing the rest for you\n4(2x-6)=x+2 =>8x-24=x+2  =>8x-x=2+24  =>7x=26  =>x=26/3=8 2/3  (3) 2(x+5)+1=3(x-1)+4   =>2x+10+1=3x-3+4 =>2x-3x=-3+4-10-1  => -x=-4  =>x=1  (5)x+6=4(4x-1)   =>x+6=16x-4  =>x-16x=-4-6  => -15x=-10  =>x=-10/-15=2/3 [after reducing]
780025
x ≤ 5\n\n-- - - - - - -s-
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
11448
Substitution Method\nSolve this system of equations:\n3y - 2x = 11\ny + 2x = 9\n \n\n1.  Solve one of the equations for either "x =" or "y =".\nThis example solves the second equation for "y ="\n \n\n3y - 2x = 11\ny = 9 - 2x\n \n\n2.  Replace the "y" value in the first equation by what "y" now equals.  Grab the "y" value and plug it into the other equation. \n\n \n\n3(9 - 2x) - 2x = 11\n \n3.  Solve this new equation for "x".\n \n\n(27 - 6x) - 2x = 11\n27 - 6x - 2x = 11\n27 - 8x = 11\n-8x = -16\nx = 2\n4.  Place this new "x" value into either of the ORIGINAL equations in order to solve for "y".  Pick the easier one to work with!  \n\n \n \n\ny = 9 - 2(2)\ny = 9 - 4\ny = 5\n \n\n5.  Check:  substitute x = 2 and y = 5 into BOTH ORIGINAL equations.  If these answers are correct, BOTH equations will be TRUE!\n   \n\n3y - 2x = 11\n3(5) - 2(2) = 11\n15 - 4 = 11\n11 = 11\n\ny + 2x = 9\n5 + 2(2) = 9\n5 + 4 = 9\n9 = 9 \n\nElimination:\n x - 2y = 14\nx + 3y = 9\n \n\na.  First, be sure that the variables are "lined up" under one another.  In this problem, they are already "lined up".\n \n\nx - 2y = 14\nx + 3y = 9\n \nb.  Decide which variable ("x" or "y") will be easier to eliminate.  In order to eliminate a variable, the numbers in front of them (the coefficients) must be the same or negatives of one another.  Looks like "x" is the easier variable to eliminate in this problem.\n  \n\nx - 2y = 14\nx + 3y = 9\n \n\nc.  Now, subtract to eliminate the "x" variable.  \n(Remember:  when you subtract signed numbers, you change the signs and follow the rules for adding signed numbers.)\n \n x - 2y = 14\n-x - 3y = - 9    \n\n    - 5y = 5  \nd.  Solve this simple equation.   \n\n-5y = 5\n   y = -1\n \n\ne.  Plug "y = -1" into either of the ORIGINAL equations to get the value for "x".\n   \n\nx - 2y = 14\nx - 2(-1) = 14\nx + 2 = 14\nx = 12\n \n\nf.  Check:  substitute x = 12 and y = -1 into BOTH ORIGINAL equations.  If these answers are correct, BOTH equations will be TRUE!\n   \n\nx - 2y = 14\n12 - 2(-1) = 14\n12 + 2 = 14\n14 = 14  \n\nx + 3y = 9\n12 + 3(-1) = 9\n12 - 3 = 9\n9 = 9
553253
http://en.wikipedia.org/wiki/Liskov_substitution_principle click the link to know in detail.
929743
∫t²cos(1 - t³) dt\n\nI note that d(1 - t)/dt = -3t² \nThis is set for substitution of 1 - t³\n\nLet u = 1 - t Thus du = -3t²dt ie  -1/3 . du = t²dt\n\nSo ∫t²cos(1 - t³) dt\n= ∫cos(1 - t³) t²dt\n= ∫cosu (-1/3) . du\n= -1/3 sinu + c\n= -1/3 sin (1 - t³) + c