use substitution to solve the system of equations: x=2, 3x+y=5=?
594458
The steps in substituting "2" for "x" are:\n\n3(2)+ y = 5\nwhich results in:\n6 + y = 5\nthen you isolate "y" by subtracting "6" from each side of the equation:\n6 - 6 + y = 5 - 6\nwhich results in the answer:\ny = -1\n\nI hope that helps.
769640
Solve one equation for x (or y), then substitute that value into the other equation.\n\nExample:  x-2y=5, so x=2y+5\n\nNow substitute this for x in "5x-2y=5"\n\n5(2y+5) - 2y = 5\n\n10y + 25 - 2y = 5\n\n8y = -20\n\ny = -20/8 = -5/2\n\nPut this value for y back into your first equation to solve for x:\n\nx = 2y + 5\nx = 2(-5/2) + 5\nx = -5+5\nx = 0\n\nANSWER: x=0, y=-5/2
169903
3x-5=5x-3\n3x=5x-3+5   (gets the numbers together)\n3x-5x=5x-5x+2   (gets the variables together)\n-2x=2\nx/-2=2/-2\nx=-1
238302
3x-y=3\ny=3x-3\n\n2x+5y=19\n2x+5(3x-3)=19\n2x+15x-15=19\n17x=34\nx=2\n\ny=3x-3\ny=3(2)-3\ny=3
472975
Replace x with 4 and y with 5.  If the resulting statement is true, then it is a solution.  If it is not true, then it is not a solution.\n5x +3y = 35\n5(4) + 3(5) = 35\n20 + 15 = 35\n35 = 35\nThis is true, so (4,5) is a solution.\n(There is no need to solve for y.  You can solve for y, but this just adds additional work.)
1152335
Solve,\n\n3x = 2y - 5 ... (1)\n\ny - 5x = -1 ... (2)\n\nRearrange (2), we get,\n\ny = 5x - 1 ... (3)\n\nSubstitute (3) into (1)\n\n3x = 2(5x - 1) - 5\n\n3x = 10x - 2 - 5\n\n3x = 10x - 7\n\n10x - 3x = 7\n\n7x = 7\n\nx = 1 ... (4)\n\nSubstitute (4) into (3)\n\ny = 5(1) - 1\n\ny = 4 \n\nTherefore x = 1 and y = 4.
1174662
2(x-4y+1)+3(-2+y-2)\n2x-8y+2-6+3y-6\n2x-8y+3y-6-6\n2x-5y-12\n\n2x-5y-12= answer\n\nHope this helps you.
56056
x= -1 ; y = 3\n5x - y = 5(-1 )- 3 = -5-3 = -8\n\nall you need to do is to subsitute (replace) the values of x and y with the numbers in the polynomial expression. then its plain maths
431687
you have to set up your equations first. y=x-9 and x=5-y. Now you have to substitute. y=(5-y)-9 and x=5-(x-9). You have to have variables on one side, so you add the -y to the 1st equation, you get 2y=5-9. Same for the 2nd equation, you get 2x=5+9. Divide by 2 for each equation, you get y=-2 and x=7.
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.
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
771836
Simple substitution.\n\n5(3) - 2(2) + 4(4)\n15 - 4 + 16\n27
408462
to solve such a system in general, what you do is this: take the first equation, solve it for x treating y as you would a constant. In your case, this is x = 4 + y.\n\nThen, take your expression for x and substitute it everywhere you see x in the second equation; in your case you get 3 *(4+y) + 2y = 7.\n\nSolve this now 1-variable equation for y, once you find y (in this case, -1).\n\nSubstitute the value of y into either equation, and you get the value of x (3 in this case).
733063
5x^2-35x+3x-21\n5x^2-32x-21
394124
y=5\nx=-2/3\n\nSame method as the other one.
763084
or solve equation for x and y components separately. \nfor x    0 +3+?=0 \nfor y    3+0 +?=0
870259
You can do it by hand or with a calculator. Either way, it would look like this:\n\n2x + y = 3 \n(move the 2x to the right side of the equation, so only y is on the left side)\ny = 3 - 2x\n\nNow put 1, 2, 3, etc. in for x, and see what y equals:\n\nwhen x=1, y=1\nwhen x=2, y=(-1)\nwhen x=3, y=(-3)\nand so on.....\n\nNow put these values into a table:\n\nx --- y\n1 --- 1\n2 --- (-1)\n3 --- (-3)