Algebra - how do you solve this?
104055
i solved it for u in the former question!!\n\nhere's how u do it:\nfirst u need to subtract 1/(7x+1) from both sides.\nnow we've got 5/(2x-5) = -1/(7x+1)\nno u need to multiply both sides by (2x-5) and (7x+1) in order to get rid of the denominators.\nafter multiplying we get: 5*(7x+1) = -1*(2x-5)\nwe open all parethesses: 35x+5 = -2x+5\nwe subtract 5 from both sides: 35x = -2x\nwe add 2x: 37x = 0\ndivide by 37: x = 0/37\n--> x = 0\n\n:) good luck
831674
1. replace x with -3, calculate and see if you get 0\n2. divide by (x+3) and solve the remaining quadratic equation
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.
937737
3a2(a-b) -6a(a-b) +21(a-b)\n(a-b)(3a^2-6a+21)\n3(a-b)(a^2-2a+7)
1054627
assuming this is (3a+1)/(a^2-1) - 1/(a+1)\n\nfactorise the (a^2-1) to give (a+1)(a-1)\n\nmultiply the second fraction by (a-1)\n\n-->  (3a+1)/[(a+1)(a-1)] - 1(a-1)/[(a+1)(a-1)]\n\nfactorising this gives [(3a+1)-(a-1)]/[(a+1)(a-1)]\n\nwhich equals (2a+2)/[(a+1)(a-1)]\n\n= [2(a+1)]/[(a+1)(a-1)] \n\nthe (a+1)'s cancel out...\n\ntherefore = 2/(a-1)
908426
Plug in each of the four answers.  The only one that works is (2), because you get\n\n | 3 (-2/3) - 3 | = 5\n | -2 - 3 | = 5\n | -5 | = 5\n 5 = 5
817461
a^4 - b^4\n\n= (a^2 + b^2)(a^2 - b^2)\n\n= (a^2 + b^2)(a + b)(a - b)
873100
Here you go!\n\nhttp://www.math.com/homeworkhelp/Algebra.html
784622
16x^2 - 4x - 2\n= (4x)^2 -(4x)- 2\n= {(4x)^2 - 1} - (4x+1)\n={(4x+1)(4x-1)} - (4x+1)\n=(4x+1)(4x-2)\n= 2*(4x+1)(2x-1)
38366
I think this is the answer\n\na^2-2ab-b^2\n-a^2(b-b^2)\n= -a^2
164954
x^3 – 3x^2 + 4x – 12\n= x^2(x – 3) + 4(x – 3)\n= (x-3)(x^2 +4)
139883
(a+b)(a-b) \n= a^2 - ab + ba - b^2\n= a^2 - b^2 -ab +ab\n= a^2 - b^2
845742
-2(x+1) + 4 = 6(x-2) + 8\n-2x - 2 + 4 = 6x - 12 + 8\n-2x + 2 = 6x - 4\n-2x = 6x -4 -2\n-2x = 6x - 6\n-2x -6x = -6\n-8x = -6\n8x = 6 (The minus signs cancel each other out)\nx = 6/8\nx = 3/4
912865
bring the a's to one side... u then get 9a = 27 ... divide and answer is a = 3
1153940
x - 2 = 3x + 12\nx - 3x = 12 + 2\n-2x = 14\nx = -7
972642
(A^2+2ab+b^2)/(a+b) + (a^2-b^2)/(a-b) + (a-b)/(a+b)/(a^2-b^2)/a+b)\n(a+b)^2/(a+b)+(a+b)(a-b)/(a-b)+ (a-b)/(a+b)*(a+b)/(a^2-b^2)\na+b+a+b+(a-b)/((a+b)(a-b))\n2a+2b+1/(a-b)\n\nif I got all the ( ) in the right places.
993104
take the equation x^2 - 7x + 12 and do the following  (x ) (x) now...see that there is a - and a + in the equation....you need to do this ( x -  ) ( x - ) 'cause you put the minus in the first factor in the second factor you take the - and the + and multiply them....So you get (x- ) (x -) the other 2 terms for each equation comes from the 7x and the 12.  What you do is to find 2 numbers that multiplied gives you 12 and added give you 7....The numbers are 3 and 4 ...Then put them into your factors (x-3)(x-4).\n\nFor the other 2 equations you do the same process\n\n(x^2 - 8x + 7) = (x-  ) (x-  ). In this one, you have to find 2 numbers that multiplied give you 7 and added give you 8. They are 1 and 7. Then (x^2 - 8x + 7) = (x- 1 ) (x- 7 ).\n\nw^2 - 8w + 15 = (w - ) (w- ). As you saw in one and two, three is the same....so the factors are 3 and 5. Then, w^2 - 8w + 15 = (w -3 ) (w-5 )\n\nFor the solve...\n1. x^2 - 6x = 0\n2. x^2 - 2x - 3 = 0\n\nTHe 1 is like this...you have to complete your quadratic equation then you do the following x^2 - 6x = 0 --> x^2 - 6x + 9 =0+ 9...The 9 in the right side was added in order to keep the two sides equal...IN the left side you got a perfect quadratic equation, hence it will be (x-3)^2= 9....Now you pass the 9 of right to the left ...then it will be [(x-3)^2]-9=0\n\n\nThe other equation x^2 - 2x - 3 = 0....This one is quite simple, it is the same as the part 1...The difference is that when you find the factors, they will be 2 numbers that multiplied give you 3 but substracted gives you 2.....The #s are 3 and 1....Here is the thing....when you break the equation into factors, you will have the following (x- ) (x+ )...Since you have a -2 that means that it is the -3 and the +1...Hence your factors will be the following (x-3)(x+1)