Solve by elimination 2x+5y=8 5X-2y=20 Show your answer please thanks?
408156
OK, let's first align the equations so the variables and the constants are lined up:\n\n2x + 5y = 8\n5x - 2y = 20\n\nLooking at this, we can either solve for x or y. It really doesn't matter in this case; the difficulty of either is the same. For simplicity, let's solve for x.\nAfter multiplying the top by 2 to get 10y, and the bottom by 5 to get -10y:\n\n4x + 10y = 16\n25x - 10y = 100\n\nCombine the two together:\n\n29x = 116\nx=4\n\nSubstitute x back in either of the equations:\n\n2(4) + 5y = 8\n8 + 5y = 8\n5y = 0\ny = 0\n\nor\n\n5(4) - 2y = 20\n20 - 2y = 20\n-2y = 0\ny = 0\n\nSubstitute x (4) and y (0) back in both equations to check:\n\n2(4) + 5(0) = 8\n8 + 0 = 8\n8 = 8\n\n5(4) - 2(0) = 20\n20 - 0 = 20\n20 = 20\n\nDone. The answer is x=4, y=0.
959550
x-5y=4, therefore x=4+5y and you plug that into the other problem to get 4(4+5y)+8y=-5, 16+20y+8y=-5, 28y=-21, y=-21/28 or -3/4, then plug -3/4 into one of the equations for y and solve for x, although I'm afraid it's been to long for me to remember the answer to dependant, independant, etc.
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.
747439
(5)(5)-2(5)(-4)+(-4)(-4)\n=25-(-40)+(16)\n=25+40+16\n=81\n\nDon't forget that (-)(-)=(+) and (-)(+)=(-).\n\nGood luck!
799916
When you use the elimination method, you pretty much add both of the equations in order to eliminate and x or y.\n\nBefore using the elimination method you may have to multiply every term of one or both of the equations by some number so that equal terms can be eliminated\n\nIn this case multiply the second equation by -2\n\n-2 (2x + 3y) = -2(2)\n\n-4x - 6y = -4\n\nNow the equations can be added to eliminate x.\n\n     4x - 5y = 48\n+ -4x - 6y = -4\n   __________\n          -11y  = 44\n\nNow, solve for y.\n\n-11y / -11 = 44 / -11\n   y = -4\n\nNow plug the y back into one of the equations in order to find x.\n\n2x + 3 (-4) = 2\n2x - 12 = 2\n2x - 12 + 12 = 2 + 12\n2x = 14\n2x / 2 = 14 / 2\nx = 7\n\nSo, \ny = -4\nx = 7\n\nHope it helps!
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
846702
-x + 2y = 18\n x + 8y = -38\n\nadd up these two lines\n\n0x+ 10y = -20\ny= -20/10 = -2\n\nx + 8(-2) = -38\nx - 16 = -38\n\n\nx=-22\ny=-2
250421
Solutions do exist, it’s just possible you got your numbers mixed up (it happens to the best of us)\n\n2x-3y+z=5  --------(1)\n\nx+4y-2z= -3 --------(2)\n\n4x-2y+3z= 6 ---------(3)\n\nConsider (1) . let’s make z the subject of the formula to eliminate it\n\n2x-3y+z= 5\n\nTherefore z= 5+3y-2x    [by rearranging]\n\nYou can substitute z=5+3y-2x in (2) and (3), I’ll put it in (3) first, the order doesn’t matter\n\nTherefore 4x-2y+3(5+3y-2x)= 6\n\n4x-2y+15+9y-6x= 6\n\nSimplifying this you get 2x-7y= 9\n\nPut z=5+3y-2x in (2)\n\nSo  x+4y-2(5+3y-2x)= -3\n\nx+4y-10-6y+4x+-3\n\nsimplifying, you get 5x-2y= 7\n\nso now we have two new equations in terms of x and y, I’ll label them (4) and (5)\n\n2x-7y= 9 ---------(4)\n\n5x-2y= 7 ---------(5)\n\nWe have to solve these simultaneously now, consider (5)\n\n5x-2y= 7\n\nMaking x the subject of the formula we get\n\nx= (7+2y)/5\n\nput this new equation in (4)\n\n2* (7+2y)/5- 7y=9\n\nSimplifying this you get 14-45= 31y therefore y= -1\n\nNow we use ‘backward substitution’ to find x and z, first x\n\nFrom (4) we know 2x-7(-1)=9 therefore x=1\n\nNow put x and y in (1) to find z\n\n2(1)-3(-1)+z= 5\n\nFrom this you find z is 0\n\nTherefore x=1, y= -1 and z=0\n\nP.S: I'm not sure what you mean by the addition method, I used both elimination and substitution methods here, however I hope this helps!
707418
20x^2+60x + 45\n= 5(4x^2+12x + 9) \n= 5((2x)^2+2(2x)3+3^2) = 5(2x+3)^2\n\n3x^2+12x+12 = 0\n=> x^2+4x+4 = 0 (divide by 4)\n=>(x+2) ^2 = 0 or x = -2, -2 (double root)\n\n7x^2-28 =0\n= x^2 = 4\n=> x = 2 or -2\n\narea = xy = 5*2 = 10cm^2
1187525
well start by multiplying things out\n\n-8x+12=-8x+5\n-8x+8x+12=5\n0+12=5\nno solution?\n\n2x+3y=6\n2x=6-3y\nx=6/2-3y/2\nx=3-(3/2)y
913862
3x + 2y - 2z\n\n3(5) + 2(2) - 2(1)\n\n15 + 4 - 2\n\n19 - 2\n\n17\n\nThe answer is 17\n\n- - - - - - -s-
917446
8x^2y\n\nas long as the x's and the y's are the same in both terms, you can just add them up...\n\nboth terms have x^2 and both terms have y and no other variables, so just add them up\n \n(3+5)x^2y = 8x^2y
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
863065
2/3 (5x + 3) + 20 = 4\n\n2 (5x + 3) = 4(3) - 20\n\n10x + 6 = 12 - 20\n\n10x = 12 - 20 - 6\n\nx = -1.4
1061869
5y+2x=-4--------(1)\ny=3x-7-----------(2)\nSubstitute the value of y from (2) into (1), we get\n5(3x-7)+2x=4\n15x-35+2x=4\n17x-35=4\n17x=39\nx=39/17\nnow, y=3(39/17)-7=-2/17
824262
well... imagine 2 functions\n\n2x + 3y = 5\n-2x + 2y = 10\n\nadd the two togather\n\n5y = 15\ny=3\n\nplug y into one of the first equations\n\n2x + 3y = 5\n2x + 3(3) = 5\n2x + 9 = 5\n2x = -4\nx = -2 and y = 3\n\nyou can check it with the other equation too\n\n-2x + 2y = 10\n-2(-2) + 2(3) = 10\n4 + 6 = 10\n10 = 10\n\nand the system is solved... \n\nx = -2 and y = 3
952983
2y = x + 16 - - - - -Equation 1\n5x + 10y = 0- - - - Equation 2\n- - - - - - - - - - - -\n\nRearrange equation 1\n\n2y = x + 16\n\n-x + 2y = x + 16 - x\n\n-x + 2y = 16 - - - -equation 1 rearranged\n- - - - - - - - - - - -\n\nMultiply equation 1 rearranged by 5\n\n5(-x) + 5(2y) = 5(16)\n\n- 5x + 10y = 80- - - - New equation 1\n\n- - - - - - - - - - - -\n\nElimination of x new equation 1 and equation 2\n\n- 5x + 10y = 80\n5x + 10y = 0\n- - - - - - - - - -\n20y = 80\n\n20y/20 = 80/20\n\ny = 4\n\nThe answer is y = 4\n\nInsert the y value into equation 1 \n\n- - - - - - - - - - - - - - - - - - - - - - - -\n\nSolving for x\n\n2y = x + 16\n\n2(4) = x + 16\n\n8 = x + 16\n\n8 - 16 = x + 16 - 16\n\n-8 = x\n\nThe answer is x = - 8\n\nInsert the x value into equation 1\n\n- - - - - - - - - - - - - - - - - - - - - -\n\nCheck for equation 1\n\n2y = x + 16\n\n2(4) = -8 + 16\n\n8 = 8\n\n- - - - - - - - - -\n\nCheck for equation 2\n\n5x + 10y = 0\n\n5(- 8) + 10(4) = 0\n\n-40 + 40 = 0\n\n0 = 0\n\n- - - - - - - - -\n\nThe solution set is { -8, 4 }\n\n- - - - - -s-