Solve by elimination 2x+5y=8 5X-2y=20 Show your answer please thanks?
479219
2x + 5y = 8\n5x - 2y = 20\n\nSince none of the coefficients are multiples of one another, you must choose either and find a common factor.  One easy way to do it is to multiply the top equation by 5 and the bottom by -2\n\nThen they become\n\n10x + 25y = 40\n-10x + 4y = -40\n\nAdd the two equations and you get\n29y = 0 this gives y = 0 as one solution\n\nIf y = 0 then substitute the value into either equation:\n\nFor example, 2x + 5(0) = 8\n\nThen x = 4\n\nThe final solution is (4, 0) You can check the answer by substituting the answer into both equations to see if it makes them both true.
801397
equation (2)*2\n6x-4y+8z=12\n9x+4y-z=0\nadding\n15x+7z=12    (4)\n(1)*2\n18x+8y-2z=0\n6x-8y-3z=3\nadding\n24x-5z=3        (5)\n(4)*5        75x+35z=60\n(5)*7      168x+35z=21\nadding     243x=81\nx=1/3\nsubstituting\n8-5z=3\n-5z=-5\nz=1\nsub\n3+4y-1=0\n4y=-2\ny=-1/2\nsolution set{1/3,-1/2,1}
605260
Multiply the first equation by x, multiply the second equation by -y, then add the two equations together.\n1 + xy = 5x\n-xy - 1 = -8y\n0 = 5x - 8y\nAdd 8y to each side\n8y = 5x\nDivide both sides by 5y\n8/5 = x/y\nAnswer: 8/5
659444
25x + 5y = 25........(1)\ny = 5 – 5x...........(2)\n\n25x+5y=25............(1)\n5x+y=5...............(2)\n\nOn rearranging u find both equations are the same.\ni.e equation1=5 * equation(2)\nso solution is x=0,y=0;\n\nif u r looking for values\nu can use trial and error methods\n\none solution would be x=-1 and y=10\ni.e 5 * -1+y=5\ny=10\n\nlike wise u can put values for x and find the corresponding value for y.
278998
You mean the two equations:\n4x - 5y = 7 and x + 5y = 17, right?\nThe equations are already set up for elimination because if you add the coefficients of the y's you get zero.\n  4x - 5y = 7\n+  x + 5y = 17\n5x = 24   Divide both sides by 5 to get\nx = 24/5\nNow plug in 24/5 for x in either equation and solve for y.\nIn this case the second equation is easier.\n24/5 + 5y = 17 Subtract 24/5 from each side and get a common denominator of 5\n5y = 85/5 - 24/5\n5y = (85-24)/5\n5y = 61/5  Divide both sides by 5 (or multiply by 1/5)\ny = 61/25\nThus, the answer is the ordered pair (24/5,61/25)\nYou can check your answer by substituting these values into each equation to see if you get a true equation.
854561
from 2nd equ\nx=2y+5\n\n5x-6y=21\n5(2y+5)-6y=21\n4y=-4\ny=-1\n\nx=2(-1)+5\nx=3
1068646
hmm see you can do it in simple way,\n\n(8x-3x) = (5x+ 3x)\ni.e.x ( 8-3) = x (5+3)\ni.e 5x = 8x\ni.e 5 = 8\n\nCheers
599677
i know how 2 do this.\n4x+2y=2\n x-2y=8\n---------\n5x   =10\n5x=10\nx=10/5\nx=2.\nsubstitute x=2 in any1 of the equations.\nlet us take x-2y=8.\n  2-2y=8\n  -2y=8-2\n  -2y=6\n    y=6/-2\n    y=-3.\n therefore ur x=2 nd y=-3.
239058
You have to solve for one and substitue into the other...\n\nLets solve for x...\n\n2x + 3y = 40 \n2x = 40 - 3y\nx = (40 - 3y)/2\n\nThen Substitute in the other.\n-2x + 2y = 20\n-2(40 - 3y)/2 + 2y = 20  (the 2's in the first set cancel becaue 2/2 = 1 so...\n-(40-3y) + 2y = 20  (then distribute the -)\n-40 + 3y + 2y = 20 (combine like terms)\n5y = 60\ny =12\n\nThen sub y into the equation to find x\n\n 2x + 3y = 40 \n2x + 36 = 40\n2x = 4\nx=2\n\nDouble Check in the other equation...\n-2x + 2y = 20\n-2(2) + 2(12) = 20\n-4 + 24 = 20\n20 = 20 These equal so x=2 and y=12!
18182
2x^2 + 5x + 20x + 50\n2x^2 + 25x + 50
1031270
x^2 + 5 + 5x + x y +5y\n\nx^2 + xy + 5x + 5y + 5\n\nx(x+y) + 5(x+y) + 5\n\nx(x+y) + 5(x+y+1)       OR     (x+5)(x+y)+5\n\nCynthia
904822
Hello Dear Blue \n\n2x^3+8x^2 - 5x + 5 ÷ x-2\n\n▪ Step1\n( 2x^2 ) ( x-2 ) = 2x^3 - 4x^2 ; { let Q1 = 2x^2 }\n2x^3+8x^2 - 5x + 5 - ( 2x^3 - 4x^2 ) = \n2x^3+8x^2 - 5x + 5 - 2x^3 + 4x^2 = +12x^2 - 5x + 5 {remainder} \n\n▪ Step 2\n+12x^2 - 5x + 5 ÷ x - 2 \n(x - 2) (12x) = 12x^2 - 24x ; { let Q2 = +12x }\n+12x^2 - 5x + 5 - ( 12x^2 - 24x  ) =\n+12x^2 - 5x + 5 - 12x^2 + 24x = +19x +5 {remainder}\n\n▪ Step 3 \n19x +5 ÷ x -2\n(x -2) (+19) = +19x - 38; { let Q3 = +19}\n19x +5  - (+19x - 38) =\n19x +5  -19x + 38 = +43 {remainder}\n\n▪ Step 4\nSum Q1, Q2, Q3\nQ1 + Q2 + Q3\n2x^2 +12x +19\n\nSo;\n(2x^3+8x^2-5x+5) ÷ (x - 2)    = >\n\nQ(x) OR quotient = 2x^2 +12x +19 \nR(x) OR Remainder is + 43 \n\nGood Luck dear.
95804
You know that\n              1)3x+4y=12y\n              2)-2x+10=12y\nSo you can substitute 3x+4y for 12y in equation 2. Hence:\n              3)3x+4y=-2x+10\n                   5x=10-4y\n                    x=2-4y/5\nSubstitute this value for x into equation 1 and it becomes:\n              1)8y/5+6=12y\n                     y=15/26\nPlug this back into 3 and you get\n              3)x=2-(4/5)*(15/26)\n                x=20/13\nTherefore the answer to the system is (x,y)=(20/13,15/26)
958973
y=5x+2\ny=5(2)+2\ny=10+2\ny=12
190156
2x+5y=6------(1)\nx+y=7--------(2)\n\n(2) x 2; 2x+2y=14----(3)\n\n(1) - (3);\n3y=-8\ny=-8/3\n\nReplace y=-8/3 into (2)\nx+(-8/3)=7\nx-8/3=7\nx=7+8/3\nx=29/3\n\nConclusion: x=29/3 and y=-8/3
975557
5x-3y=-12\n-3y=-5x-12\nmultiplying by -1\n3y=5x+12\nm=5/3\nb=12/3=4\n\n2.y=3x+5\n\n3.y+5/5+5=x+3/4+3\ny+5/10=x+3/7\nmultiplying by 70\n7(y+5)=10(x+3)\n7y+35=10x+30\nadding -7y\n35=10x-7y+30\nadding -35\n10x-7y-5=0
714030
8y=-23-x\n\n-5x+8y=19\n-5x+(-23-x)=19\n-6x-23=19\n-6x=42\nx=-7\n\n-7+8y=-23\n8y=-16\ny=-2\n\n\nThe solution is (-7,-2).  I've checked it, so it works.  However, it is a good idea to always check the solutions.  Since you end up using only one equation to find the second variable, make sure the pair also works in the second equation.\n\nHope this helps!