solve each system by substitution method?
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.
236017
sorry i don't know mathematica but you can solve it easily using matlab. Enter rthe coefficient matrix as follows for example,\nA=[1 2 3 4; 2 3 3 4;........]\nand the right hand side\nb=[ 2 3 4 4]\nonly for illustration. ";"indicates columns.\nA/b solves this system and gives the solution vector x.
781368
By substitution\ny = x^2\ndy = 2x dx\n\nint x2^(x^2) dx = (1/2) int 2^y dy = (1/(2ln2))2^y \n= (1/(2ln2))2^(x^2)
209958
Replace x, y with k & solve.
886222
Solve for one variable and plug it into the second equation...\n\nSo, let's solve the first equation for y:\n-x + y = -4\ny = x - 4\n\nNow plug this into the second equation:\n4x - 3(x - 4) = 13\n\nDistribute through the parentheses:\n4x - 3x + 12 = 13\n\nSubtract 12 from both sides:\n4x - 3x = 1\n\nSimplify:\nx = 1\n\nNow plug this back into your original substitution:\ny = x - 4\ny = (1) - 4\ny = -3\n\nSo your answer is x = 1 and y = -3.\n\nYou should double-check your answer....\n-x + y = -4,\n-(1) + (-3) =? -4  \n-4 = -4 <-- check\n\n4x - 3y = 13\n4(1) - 3(-3) =? 13\n4 + 9 = 13 <-- check\n\nSo your answer is:\nx = 1\ny = -3
971669
1. 1s2 2s2 2p6 3s1\n2. 1s2 2s2 2p6 \n3. 1s2 2s2 2p4\n4. 1s2 2s2 2p6\n5. 1s2 2s2 2p6 3s2 3p6 3d7 4s2
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
859187
I'll do one for u.\nIn the first,\nMultiplying both sides by x*,we get\nx+3x+2x*=7x*\n4x+2x*=7x*\nMoving 4x and 2x* to the R.H.S.\n5x*-4x+0=0\nThus 5x*-4x=0\nAttempt the others in the same manner.
247965
infinite values for both,\nits caus u cant solve two variables with just one equation
532184
One method is to integrate by parts twice. There is another way where you can use complex methods: \ncos2x = Re e^(2ix)\n[Re means the real part]\n\nIntegral e^(-x)cos2x dx\n= Integral e^(-x) Re e^(2ix) dx\n= Re Integral e^((2i-1)x) dx\n= Re e^((2i-1)x)) / (2i-1) + C\n= Re e^(-x) e^(2ix) / (2i-1) + C\n\nWrite the above in the standard A + iB and then you can find your answer. Good 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!
429670
In matlab, type\n>> syms x\n>> solve('x^3+4*x^2-5=0')\n \nans =\n \n[                1]\n[ -5/2+1/2*5^(1/2)]\n[ -5/2-1/2*5^(1/2)]
896553
4x^2+8x-x-2\n(4x-1)(x+2)
1076564
What do you want to solve for? x? y? or do you want an ordered pair (x,y) that works (integers maybe?)
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
689072
5x + 3y = 16 ------ equation 1\n\n2x + 3y = 1 ------- equation 2\n2x = 1 - 3y\nLet x = (1-3y) / 2 \n\nReplace x with y in equation 1, and solve y\n\n5 [ (1-3y) / 2 ] + 3y = 16\n5 (1-3y) + 6y = 32\n5 - 15y + 6y = 32\n9y = -27\ny = -3\n\nReplace the value of y to solve x\n \ntherefore x = [ 1 - 3(-3) ] / 2\nx = 10/2 = 5\n\nYou can try and solve the other pair of system using the same basis as above.
554172
Um...\n\n2(-X) + 3(-X) + X + 8\n\n-2X - 3X + X + 8\n\n-5X + X + 8\n\n-4X + 8\n\n8 - 4X