What would be the answer to this system by using substitution?
152532
y = 4x - 3 and y = 5x - 1, so\n4x - 3 = 5x - 1 (I substituted the value of y in the first eq'n to the second eq'n)\n-x = 2\nx = -2\ny = 4(-2) - 3\ny = -8 - 3\ny = -11\n;-)
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
1171615
I think this is also the 'trial and error method'\nYou will have an equation where you do not know the value of x.\nKeep on substituting different values of x into the equation until you come up with the right answer.\nIf you are lucky, you may have been told that x lies somewhere between 2 values.\nBest way to do it is to choose a number, get the answer, then choose another number a fair bit larger than the first. If the answer you want is between these 2 values, choose a number about half way between the 2 and so on.
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.
81081
g(x)=(x-a)(x+a)/(x-a)\n= x+a\nSince g(x) =4\ntherefore, x+a = 4 \nbut, x=a so substituting the value of x\na+a=4\n2a=4\na=2\nThe constant a = 2.
137569
(3/4P_ + 1/4pp) x (3/4S_ + 1/4ss)---maybe
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]
118612
Kerplode is right - but that's kind of complicated.  Basically you solve both in exactly the same way.  The only difference is that instead of an = you have an > or <.  \n\nIt just looks different, there really is no difference in how you solve them.
866437
Let one side be 'x', second will be 'x + 5 ' third will be  ( x+5) - 3\nx  + x + 5 + ( x + 5 ) - 3 = 16
71723
You just plugin one function into another function where there's an X.\n\ng(f) means use function G and plugin function F to replace the X\n\n1.  g(f)=3-2x f(g)=2(3-x)=6-2x\nI just replaced 2x with x for g(f) and 3-x for x for f(g)\n\nUse this method to figure out the rest.\n\nNotice that it says "x is not equal to whatever".\nThis just states that it will be undefined if it is that number and for now, should be ignored as all problems will have real answers.
255958
Substitute 2 for x and solve for c.\n2^2-3(2)+c=0\nRemember do powers, then multiplication, then subtraction.
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).
387063
X=[-(30y-433)]/1232\nY=[-(1232x-433)]/30
22859
Subract the two terms on the left giving:\n-2t < 16\n\nAdd 2t to both sides giving\n0 < 16 + 2t\n\nSubtract 16 from both sides\n\n-16 < 2t\n\nDivide both sides by 2\n\n-8 < t
160067
When you substitute with y=x^(1/6), you're assuming :\nI) x>0 (which should be put as a starting condition)\nII) y>0 . This is to be considered after solving the 2nd degree equation, as you get a fake solution y= -3 that you must reject as non-acceptable.\n\nThe only acceptable solution is x= 1/64
565039
For this you are asked to take the the equation y = 2n + 3 and substitute the rest of the values in for n and solve the equation. The implied operation between 2 and n is multiplication.\n\nSo you take the n out of the equation and put in the values that you have listed.\n\nFirst Value:\ny = 2(2) + 3  Do the multiplication first then add the 3\ny = 4 + 3\ny = 7\n\nSecond Value:\ny = 2(1/2) + 3\ny = 1 + 3\ny = 4\n\nThird Value:\ny = 2(0) + 3\ny = 3
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