Complete the ordered pairs for the equation  2x + y = 10    (5, ?), (?, 10), (? ,-2), (0, ?)?
604216
In an ordered pair, the first number stands for the x value and the second number stands for the y value.  \n\nTaking the first ordered pair you have (5,?), you know that the x value is 5.  So you plug 5 into the equation where it says x.  \n\n2(5)+y=10.  Now solve as a regular algebraic equation\n10+y=10  \ny=0\n\nNext one: (?,10).  This time you know the y value, so you plug the 10 into the y value and you have \n\n2x+10=10   Solve algebraically. \n2x=0, \nx=0.\n\nI will leave the last two for you to figure out, although I'm sure others will be posting it.\n\nI hope this helped you to understand HOW to do it...\n\nGood Luck.
778763
1) (a - b)(x - y)\n2) (x^9 + y^3)(x^9 - y^3)\n3) (6x + 5)(x + 8)\n4) (x - 9)(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.
380604
yes, 2(3)+5(-5)=-19 .... 6+(-25) = -19 .... -19 = -19\n\nthe ordered pair 3 is the x variable and 5 is the y variable so just substitute to check the answer \n\nthe ordered pair works for the second equation as well
243986
2+2[24+(-4)+(10)]\n2+2[30]\n2+60\n62!
1000586
# 1    \n(x+1)*(x+6)=x2 +x+6x+6=x2 +7x +6\n#2\n(y-6)*(y-5)= y2-5y-6y+30= y2-11y+30\n#3\n(a-2) * (a+10)=a2 +10a-2a -20= a2  +8a -20\n#4\n(x-1) *(x+1)= x2 -x+x -1= x2 - 1
737993
equation of line through (4,3) and (2,4) is 2y=-x+10
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.
860625
(10x+8) - (5x+4)\n= 10x - 5x +8 -4\n= 5x + 4
298404
4x+y=10\n\n6x-y=20 \n+y   +y\n_______\n6x = 20+y\n-20  -20\n6x-20 = y\n\nreplace 6x-20 in for y with the other equation\n\n4x+(6x-20)=10\n10x-20=10\n+20     +20\n10x=30\nx=3\n\nnow plug in 3 in the equation as x\n\n4x+y=10\n4(3)+y=10\n12+y=10\ny=-2\n\nso x=3 and y=-2\nI think\n_____________________________\nAnother way to do it is get y's alone and set them equal\n6x-20=10-4x\n10x-20=10\n10x=30\nx=3\nand then just plug in like what I did b4
908374
Okay, the goal for solving simultaneous equations is to isolate one variable, so rearrange one of the equations (it doesn't matter which one) until you get either x or y alone on one side.\n\n5x - 2y = 10\n5x - 2y + 2y = 10 + 2y (if you do something to one side, do it to the other)\n5x = 10 + 2y\n5x/5 = (10 + 2y) / 5\nx = (10+2y)/5\n\nThen, plug this new value for x (the right side of the equation) into the other equation, hence the "substitution."\n3x + 2y = 6\n3 ((10+2y)/5) + 2y = 6\n(30 + 60y)/5 + 2y = 6\n6 + 12y + 2y = 6\n14y = 0\ny = 0\n\nNow, solve for x by substituting this y value back into one of the equations.\n\n5x - 2y = 10\n5x - 2(0) = 10\n5x = 10\nx=2\n\nSo, x=2 and y=0
797737
Ok, take your equation \nf(x) = 2x^2 - 10\n\nreplace f(x) with y\n\ny = 2x^2 - 10\n\nswap y and x\n\nx = 2y^2 - 10\n\nsolve for y\nto do this first add 10 to both sides\n\nx + 10 = 2y^2\n\ndivide both sides by 2\n\n(1/2)x + 5 = y^2\n\ntake the sqaure root of both sides\n\nsqrt((1/2)x + 5) = y\n\nchange y to f(x)^-1 [the inverse function notation]\n\nf(x)^-1 = sqrt((1/2)x + 5)\n\nhope this helps
676814
First equation: sqrt[x] + y = a\nor sqrt[x] = a - y\nSquaring both sides\nx = (a - y)^2__________(1)\n\nSecond equation: x + sqrt[y] = b\nor sqrt[y] = b - x\nSquaring both sides\ny = (b - x)^2___________(2)\n\nSubstituting 'x' from eqn. (1) in eqn. (2)\ny = [b - (a - y)^2]^2\n  = [b - a^2 + 2*a*y - y^2]^2\n\nSo it is clear that you have to solve a 4-th order equation.\nIt is really tough.
1169467
y = x^2 + 5x - 3\n2y=3x - 2\n----------------\n2y = 2x^2 + 10x - 6\n2y = 3x - 2\n-----------------------\n    0 = 2x^2 + 7x - 4\n=>2x^2 + 7x - 4 =0\n=>2x^2 + 8x - x - 4 = 0\n=>2x(x + 4) -1(x+4) = 0\n=> (2x - 1)(x + 4) = 0\ntherefore 2x - 1 = 0 => x = 1/2 or \nx + 4 = 0 or x = -4\nx = - 4 or 1/2  \ngot it?
498659
1) y=x+4\n   3x+y=16 (I guess you made a mistake by writing +16)\n--------------\nsubstitute x+4 for y in second equation and you'll get\n\n   3x+x+4=16\n    4x+4=16\n    4x=12\n     x=3\n\nnow y=x+4=3+4=7\n\nsolution (x,y)=(3,7)\n\n2) 4x+5y=48\n    y=3x+2\n--------------\nsubstitute 3x+2 for y in first eq.\n\n4x+5(3x+2)=48\n\n4x+15x+10=48\n\n19x=38\n\nx=2\n\nnow y=3x+2=3*2+2=6+2=8\n\nsolution (x,y)=\n(2,8)\n\n3)3x-y=0\ny=6\nin first equation instead of y write 6\n\n3x-6=0\n3x=6\nx=2\n\nso, the solution is (2,6)\n\n4) x-2y=-10\n3x-y=0\n\nfrom the first equation you have\nx=2y-10\nnow substitute 2y-10 for x in second one\n\n3(2y-10)-y=0\n6y-30-y=0\n\n5y=30\ny=6 \nnow x=2y-10=2*6-10=12-10=2\nso x=2 and y=6
119395
Just solve the two eqautions simultaneously:\n1.Use M, and arrange it, to give y=10-2x. \n2.Substitute this value of y to give 5x-3(10-2x)=14. (equation 3)\n3.Solve the eqaution, this will give 5x-30+6x=14 which will lead to, 11x=44, x=4\n4. Substitue the value of x in either eqaution y=10-2(4), y=10-8,y=2\nThe intersection point is (4,2)
850518
5 x _ - 28 = 12 + 3 x _\n\nLet the first blank be Y and th second blank be X\n\nThen 5Y - 28 = 12 + 3X\nie 5Y = 3X + 40\nie Y = 3/5 X + 8\n\nSo ANY ordered pair on this line will obey the rule\n\nEg when X = 5 Y = 11\n\nWhen X = 20 Y = 20\n\nWhen X = - 15 Y = -1\n\nThe solutions are infinite in number but all obey the set rule that the first blank = 3/5 of second blank + 8