Math solving for X again?
445066
move all x's to one side and nos. to other side of equals\nNOTE:WHEN MOVING A NUMBER TO ANOTHER SIDE, CHANGE THE SIGNS\n5x-2X=1+8\n3X=9\nX=3  (BCOZ 3 WAS DIVIDED INTO 9)
66565
You cannot "solve" this because it is not an equation.  An equation has two parts that are joined by the symbol =.  However, in early pre-Algebra classes, they might ask you to simplify this, meaning, to write this expression so that the "x" only appears once with its coefficient and the rest of the numerical values are also combined.  For this, you add all of the coefficients of x (3+1 - invisible coefficient before the second appearance of x) to find that you have 4x.  Next, you add the -8 and 4, which you can then express as either adding negative four or as subtracting four (if you need help here, consider the number line, move eight spaces to the left of zero, then move four spaces to the right).  Thus, you can re-write the expression as 4x-4.  Look in your textbook before the problems for a better example of what exactly they are expecting you to do in that section.
939267
Let x = "one number". \nThen, the "another number" is 2x.\n\nWhen the larger number is decreased by 10,\n\n2x - 10\n\nThe result is 2 greater than the smaller number\n\nx + 2\n\nTherefore,\n\n2x - 10 = x + 2, and\nx = 12.\n\nSo the two numbers are x and 2x, or 12 and 24.\n\nLet's put it back into the problem.\n\nOne number is twice another.  24 is twice 12.\nWhen the larger number is decreased by 10 (24 - 10 = 14), the result is 2 greater than the smaller number (2+12 = 14).  Therefore that's the answer!\n\nHappy solving!
859897
***************\nQUESTION 1\n***************\nOn the first question, we first have to find the equation that we will be solving for.  \n\nAssume that x is one integer.\nWe know that x+1 is the other integer because they are consecutive.\n\nSo we now put the wording into an equation\n1/3(x + (x+1)) = x - 5\n\nNow we solve the equation\n1/3(2x + 1) = x - 5\n2/3x + 1/3 = x - 5\n16/3 = 1/3x\n16 = x\n17 = x + 1\n\nOur two integers are 16 and 17.\n\nNow we check our answer.\n1/3 of 33 = 11 & 16 - 5 = 11 We are good! :)\n\n\n***************\nQUESTION 2\n***************\nEvery hour, Amy and Randy together walked 3.6 + 4.2 km apart from each other.\n\n3.6 + 4.2 = 7.8\n\nSo, every hour they will be 7.8 more km apart from each other.  We only want 10 minutes though, and there are 60 minutes in an hour. To get how far apart they will be in 10 minutes, we take:\n7.8 * 10/60 = 1.3\n\nRandy and Amy will be 1.3 km apart after 10 minutes.\n\nWe are done! :)
961595
umm..are you missing an equals sign somewhere or am i really dumb? how do you figure out what x is unless you have both sides of the equation?
904672
let u = x^2 + 1, then you have du = 2xdx, => dx = 1/2x du,\n\nsubstitute back you are on the way  to solve it
859187
I'll do one for u.\nIn the first,\nMultiplying both sides by x*,we get\nx+3x+2x*=7x*\n4x+2x*=7x*\nMoving 4x and 2x* to the R.H.S.\n5x*-4x+0=0\nThus 5x*-4x=0\nAttempt the others in the same manner.
368125
if this is the problem:\n((x+3)/2)*((x^2-x)/(x+3))=21\nthen you solve by\ncancelling out (x+3)\n(1/2)*(x^2-x)=21\nmultiplying by 2\nx^2-x=42\nmaking it quadratic\nx^2-x-42=0\nand using x=(-b(+-)srt(b^2-4ac))/2a\nx=(1(+-)srt(1-4*1*(-42)))/2\nx=(1(+-)srt(169))/2\nx=(1+13)/2  or x=(1-13)/2\nx=7   or x= -6
567493
I'm not sure of your notation in the second one... did you mean x² + x?  I'll assume so.\n\nFirst solve the first equation:\n3x - 1 = 11\n3x = 12 (add 1 to both sides)\nx = 4 (divide both side by 3)\n\nNow plug that into the second equation\nx² + x\n4² + 4\n16 + 4\n\nThe answer (assuming you meant x² + x) is 20!
936579
your question seems to be incomplete.\nso far....\nwidth=x\nlength= 2x-1
443017
Set it up like a long division problem:\n\nx - 1 ) x^5 + x^4 - 2x^3 + 2x^2 - 3x - 7\n\nNow focus on the x terms.  x is on the outside, and x^5 is on the inside.  Just like long division with numbers, how many times does x go into x^5?  x^4 times.  That goes on top, and then you multiply that by x - 1 and subtract, again like normal long division.\n\nx^5 + x^4 - 2x^3 + 2x^2 - 3x - 7\n-(x^5 - x^4)\n--------------------\n2x^4 - 2x^3 + 2x^2 - 3x - 7\n\n\nRepeat process...x goes into 2x^4 2x^3 times:\n\n2x^4 - 2x^3 + 2x^2 - 3x - 7\n-(2x^4 - 2x^3)\n--------------------\n2x^2 - 3x - 7\n\nx goes into 2x^2 2x times:\n2x^2 - 3x - 7\n-(2x^2 - 2x)\n--------------------\n-x - 7\n\nFinally, x goes into -x -1 time:\n-x - 7\n-(-x + 1)\n--------------------\n-8\n\nSo you have x^4 + 2x^3 + 2x - 1 with a -8 remainder, which you can write as -8/(x - 1).
602540
x + 3 >  8 ==> x >  8 - 3  ==> x > 5\nx - 4 < -2 ==> x < -2 + 4 ==> x < 2\n\nAnswer... not possible.
617282
are u sure its x2? shouldn't it be 2x+10x=(a number) usually there is a number at the end of the equation and u have 2 solve for "x". add details to ur question and then i may be able 2 help u.
408554
yes. take out an x \nyou answer should be x(x+3)
781088
Plug in 2 for x to determine this:\n\n3(2) + 5 = 6+5 = 11 ≠ 17\n\nTherefore 2 is not a solution.\n\n~ ♥ ~
40850
-3 <4x\n x < -3/4\n solution set : {- infinity ,.............,-3/4}
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