Simultaneous equations BY ELIMINATION?
917826
I'm going from memory\n\ntry multiply all terms in top eq by 5\n\ngiving         10x + 20y = 40\n\nthen the bottom by 2\n\ngiving          -10x + 6y = 10\n\nthen add the two gives   26y = 50\n\nwork back  2x + 4(1.923) = 8\nthus               2x + 7.69 = 8\n                                 2x = 8 - 7.69\n                                 2x = 0.31\n                                   x = 0.155\n  \ntherefore y = 1.923  and x =  0.155\n\nyou have to find the lowest common multiple of two of the terms, whether it be x or y.\n\ni hope this helps but it looks wrong to me. did u get the original eq right and what level are they at?\n\ngood luck
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]
833967
This one is easier by subtraction, but whatever.  In order to use addition to eliminate one of the variables, you have to make sure that the coefficient of one of the variables in one of the equations is the opposite of (that is, the negative of) the coeffiecient of the same variable in the other equation.  In this case, multiply the entire second equation by -1 to get:\n-2x + y = -2\n\nAdd this to the first equation to get:\n2x + 5y = -1\n+(-2x + y = -2)\n-------------------\n6y = -3\ny = -1/2\n\nPlug this back into one of the equations (the second one looks easier) to get:\n2x + 1/2 = 2\n2x = 3/2\nx = 3/4
92952
(8x+3)(4x+3)\n32x2+24x+24x+9\n32x2+48x+9
681236
Multiply each term in the first () by each term in the second ():\n\n2x times 2x = 4x^2\n2x times 1 = 2x\n-3 times 2x = -6x\n-3 times 1 = -3\n\nThen combine 2x and -6x\nAnswer: 4x^2 - 4x - 3
191051
First you want to eliminate "a"\nMultiply a-15b=3 by -1 so that it is now -a+15b=-3\nnow line up the two equations so the variables are aligned\n-a+15b=-3    \n  a+3b=21     Cancel out the "a's"\n_____________\n      18b=18\n\nisolate "b" by dividing both sides by 18 so that b=18/18 which means that b=1\n\nNow plug the 1 into either equation (it doesn't matter which..the answer will come out the same)\n\n3b+a=21\n3(1)+a=21      Substitute 1 for "b"\n3+a=21   Multiply\n\n   3+a=21  Subtract -3 from both sides\n  -3     -3  (sorry if the -3's look like they are on the same side)  \n___________\n      a=18\n\nso b=1 and a=18
140559
Using the abc method… \n[a]FexOy+[b]H2SO4 -->[c]Fe2(SO4)3+[d]SO2 +[e]H2O\nFe-->ax=2c------eq.1\nO-->ya+4b=12c+2d+e----eq.2\nH-->2b=2e----------eq.3\nS-->b=3c+d----------eq.4\n\nLet a=1\n\nFrom eq.1\nax=2c\nx/2=2/2c\nx/2=c-----eq.5\n\nForm eq.3\n2b=2e\n2/2b=2/2e\ne=b-----eq.6 \n\nFrom eq.2\nYa+4b=12c+2d+e\n1y+4b=12(x/2) +2d+b (Form eq.5&6)\nY+3b=6x+2d\n2d=y-6x+3b\nd=(y+3b-6x)/2------------eq.7\n\nFrom eq.4\nb=3c+d (but c=x/2)\nb=3(x/2) +d\nReplacing eq.4 by eq.7 we will get\nb=3x/2+(y+3b-6x) ½\nb= (3x+y+3b-6x) ½\n2[b=(y+3b-3x) ½]\n2b=y+3b-3x\n3x-y=b-------eq.8\n\nReplacing eq.6 by eq.8\ne=b\ne=3x-y\n\nReplacing eq.7 by eq.8\nd=(y+3b-6x)/2\nd=[y+3(3x-y)-6x]/2\nd=[y+9x-3y-6x]/2\nd= (3x-2y)/2\n\nFrom the above calculation we know that:\na=1, b=3x-y, c=x/2, d=(3x-2y)/2 & e=3x-y. \n\n[a]FexOy+[b]H2SO4 -->[c]Fe2(SO4)3+[d]SO2 +[e]H2O\nWhen replacing the values we’ll get:\n[1]FexOy+[3x-y]H2SO4 -->[x/2]Fe2(SO4)3+[(3x-2y)/2]SO2 +[3x-y]H2O
810768
Easiest way is to write them as two equations equal to zero.  Then equate them...\n\nStarting with your first example:\nx² + y² - 4 = 0\nx² + (y-6)² - 25 = 0\n\nNow equate these:\nx² + y² - 4 = x² + (y-6)² - 25\n\nExpand out the (y-6)²\nx² + y² - 4 = x² + (y²-12y+36) - 25\n\nNow cancel like terms (like x², y², etc.) which are on both sides:\n-4 = -12y + 36 - 25\n\nNow simplify:\n12y = 36 - 25 + 4\n12y = 15\ny = 15/12\ny = 5/4\n\nNow use the value you found (y=5/4) and solve for the other variable using one of the original equations:\nx² + (5/4)² = 4\nx² + (25/16) = 4\nx² = 64/16 - 25/16\nx² = 39/16\nHere you need to take the sqrt and you'll have two answers:\nx = sqrt(39/16)\nx = +/- sqrt(39)/4\n\nSo your two points of intersection for the first pair of equations are:\n(sqrt(39)/4, 5/4) and (-sqrt(39)/4, 5/4)\n\nThe same method works for the other problems...
693011
You get one variable all by itself on one side of an equation.  Then you take what that variable equals, and replace the same variable in another equation with that answer.  easier to show with an example.  Suppose\n\n2x + 3y = 19\n\nx + y = 7\n\nSubtract x from both sides of the second equation to get y = 7 - x\n\nThen replace the y in the first equation with (7-x):\n\n2x + 3(7-x) = 19\n\nDo this out: 2x + 21 - 3x = 19\n\n-1x + 21 = 19\n\n-1x = -2\n\nx = 2\n\nThen plug in 2 for x and find y: 2 + y = 7\n\ny = 5
926616
2x+3<6x-1\n2x-6x<-1-3 \n-4x<-4\nx>1
863644
take out 4x first\n4x(x2-6x+8)\n4x(x-4)(x-2)
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.
1053464
if you want to simplify it \n\nit is 3x - 4x + 4 = -x + 4
753930
The most intuitive way to answer is to use Gauss-Jordan elimination, as described below.\n\n\nw + x + y + z = 24\n3w + 3x + 3y + 3z = 72\n3w - 2x - 3y + 3z  = -9\n5x + 6y= 81\n\nw + x + y + z = 24\n5w + 5x + 5y + 5z = 120\n5w + 3x - 2y - 2z  = 4\n2x + 7y + 7z = 116\n\nw + x + y + z = 24\n2w + 2x + 2y + 2z = 48\n2w + 4x + 4y - 5z  = 22\n-2x - 2y + 7z = 26\n\n2x + 7y + 7z = 116\n-2x - 2y + 7z = 26\n\n4x + 9y = 90\n5x + 6y= 81\n\n20x + 45y = 450\n20x + 24y = 324\n21y = 126\n\ny = 6\n\n4x + 9y = 90\n4x + 54= 90\n4x  = 36\nx = 9\n\n2x + 7y + 7z = 116\n18 + 42 + 7z = 116\n7z = 56\nz = 8\n\nw + x + y + z = 24\nw + 9 + 6 + 8 = 24\nw = 1\n{w,x,y,z} = {1,9,6,8}\n\nAnother method is to put the coefficiants into a matrix, and put it into reduced-row echelon form.\n\n|1 1   1  1  24|\n|3  -2 -3 3 -9|\n|5  3  -2 -2 4|\n|2  4  4  -5 22|\n\nIn RREF...\n|1 0 0 0 1|\n|0 1 0 0 9|\n|0 0 1 0 6|\n|0 0 0 1 8|\n\nThese are all new valid equations, just in a much nicer form. Specially, 1w+0x+0y+0z=1, etc.\nThis tells us that w=1, x=9, y=6, and z=8.\n\nAlso, you could use Cramer's rule with these equations and get the same answer, but I will not demonstrate that one.
176944
the standard form is: Ax + By = C\n\nin your two equations you dont have any 'y' so you might have typed it wrong but if it is right then 'B' is 0.  \nthen combine like terms. this should leave you with Ax + (0)y = 0\n\n\nto determine 'x' and 'y'...\n\n             PART #1\n1. choose 'x' as zero\n a. this eliminates 'Ax'\n2. divide both sides by coefficient of 'y'\n a. this leaves y = C/B\n3. determine y\n\n             PART #2\n1. choose 'y' as zero\n a. this eliminates 'By'\n2. divide both sides by coefficient of 'x'\n a. this leaves y = C/A\n3. determine x\n\ni hope this helps!
697874
Multiply or divide all components in one question to set the variables in the opposite questions to cancel out if you added the two problems together by combining like terms. For example:\n2x + 3y = -13  and 4x + y = -1\n\nTo keep this simple I will multiply all the components of the 2x + 3y = -13 equation by negative 2, resulting in:\n\n-4x - 6y = 26\n\nThen add the new equation to the other equation 4x + y = -1\n\n-4x -6y = 26\n4x + y = -1\n----------------\n0x -5y = 25  which simplifies to y = -5\n\nInsert the value found for y back into one of the original eqations and solve for x\n\n2x + 3(-5) = -13\n\n2x + -15 = -13\n\n2x = 2\n\nx = 1\n\nYour end result is x =1 and y = -5
664485
nope \n5x+5y\n\nthe 4y and -4y cancel each other out