How do you solve the following by substitution?  1.  y=x+3, x+3y=1?
154468
All you have to do for substitution is substitute one equasion into another so what you do is:\n1. x+3(x+3)=1 because y=x+3\n2. multiply it out to get x2(squared)+6x+9=1\n3. subtract the one so that the equasion equals 0 x2(squared)+6x+8=0\n4. factor the equasion to get (x+2)(x+4)=0\n5. the answer for x could be -2 or -4 because either one of those would make one of the () 0 which would make the equasion equal 0\n6. becasue x=-2 or -4 y=1 or -1\n7. check the answers
826575
Use one equation to substitute into the other,then solve for a one variable problem.  Use your new information (the solved variable) to solve for the second variable.\n\nFor the first set:\n\n-3x - y = 2 rearranged becomes:\ny = -3x -2\n\nUsing this "definition" of y, substitute it into the second equation as such:\n\n-6x + 2(-3x -2) = -2\n-6x - 6x - 4 = -2\n-12x = 2\nx = -1/6\n\nSubstitute -(1/6) for x in either equation to solve for y:\n\n-3(-1/6) - y = 2\ny = -3/2\n\nDo the same for the second pair of equations to make sure you got it (the answers for x and y are x=9/7, y = 1/7).
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
934633
By substitution, the first equation when substituting in y=2+x becomes:\n4x - 4(2+x) = -8 ==> -8 = -8 which is always true.  This means that what you have are two equations with an infinite number of solutions, as the two lines are the same line.  So any pair of the form (x,x+2) such as (0,2) or (1000,1002) will work.
801041
Notice x = -3 + 3y (based on your second equation), so substitute this for x into your first equation to solve for y:\n\n-2x + 7y = 10\n\n-2(-3 + 3y) + 7y = 10\n\n6 - 6y + 7y = 10\n\n6 + y = 10\n\ny = 4\n\nNow plug this into the other equation to solve for x:\n\nx = -3 + 3y\n\nx = -3 + 3(4)\n\nx = -3 + 12\n\nx = 9\n\nANSWER:  x=9, y=4\n\nNRT
269310
Get one variable by itself.  If you choose to get x alone, you would do it like this:\n\n4x-3y=10\n  +3y +3y\n4x=3y+10\n(1/4)(4x)=(1/4)(3y+10)\nx=(3/4)y+(5/2)\n\nThen substitute the above x-value in for x in the second equation.\n\n2x+3y=4\n2[(3/4)y+(5/2)]+3y=4\n(3/2)y+5+3y=4\n      -5    -5\n(9/2)y=-1\n(2/9)(9/2)y=(2/9)(-1)\ny=(-2/9)\n\nSo if y=(-2/9), substitute that in for your original equation to find x.  Hope that aligns with the answer in your book...
769967
removing the brackets\n18+3y=4y-16-y\nsimplifying\n18+3y=3y-16\nadding 16\n34+3y=3y\nadding -3y\n34=0\nwhich is absurd\nso the equation cannot be solved
805693
#1\n4x+1+3=0\n4x+4=0\n4x=-4\nx=-1\n\n#2\n2y+7+4=y\n2y+11=y\ny=-11
1021939
2x+y=3..........eq 1         -2x+5y=-9..............eq 2\nAdd eq. 1 and 2\n6y=-6\ny=-1\nPutting this value in eq 1\nx=2\nS.S={ 2, -1}\n\n3x+5y=0............eq 3           2x-5y=-25.................4\nMultiply eq 3 by 2 and eq 4 by 3\nFrom eq 3: 6x+10y=0................eq 5\nFrom eq 4: 6x-15y=-75..............eq 6\nSubtract eq 6 from 5\ny=3\nPut the value of y in eq 3\nx=-5\nS.S={ -5, 3}
334051
from x + 8y = 2\nwe get x = 2 - 8y\nsubstituting in the other equation, we get\n3( 2 - 8y ) + 3y = 2\n=> 6 - 24y + 3y = 2\ncollecting like terms:\n=> -21y = -4\n=> y = 4/21
413913
x=0,2,3,4 and y=1,-1,0,3
843483
using (x+y)^3 = x^3+3x^2y+3xy^2+y^3\n\nwe get (2a+b)^3 = 8a^3+12a^2b+6ab+b^3\n\n(ab+c)^3 = a^3b^3+3a^2b^2c+3abc^2 + c^3\n\n(6ab-c)^3 = 216a^3b^3-108a^2b^2c+18 abc^2 - c^3
146086
2x+y=6 => y=6-2x \n3x+y=1 => 3x+(6-2x)=1 => x+6=1 => x=-5\ny=6-(2*-5) => y=16\n\nx=-5\ny=16
704941
The second equation can be rewritten as:\n3x + 2y = -1\nMultiply both sides by 2\n6x + 4y = -2\n6x + 5y = -7 (this is the first equation)\nSubtract the first equation from the second one.\ny = -5\nNow, plug the y = -5 into the second equation:\n3x +2(-5) = -1\n3x - 10 = -1\n3x = 9\nx = 3\nSo your solution is x = 3 and y = -5
946761
well to solve the first equation (its a line plot)\n3y+1[-1]=2x+16[-1]\n3y=2x+15\nif x=0, y=5 (y intercept)\nif y=0, x=-7.5 (x intercept)\n\nthe second line graph\n3y+5[+4]=2x\n3y+9=2x\nif x=0, y=-3\nif y=0, x=4.5\n\nNow, these lines are not equal, but do you have to plot both lines to prove? Your question is not clear.
107281
Okay, I see your two equations are 5x + 5y = -30 and \n2x - y = 15.  I'm not certain if these are right so I'll explain how to solve systems of equations in general.\nThe way we solve a system of equations by substitution is to solve one of the equations for one of the variables.  We then take what that variable is equal to and substitute it into the equation we did not yet touch.  We then solve the resulting equation.  \nThe explanation is confusing so I will provide an example.\n4x + y = 9\n-x + 3y = 1\nSo I'll solve the second equation for x like so:\n-x + 3y = 1\n-x = 1 - 3y\nx = 3y -1\nThen, I substitute 3y - 1 into the first equation for x, since this expression is equal to x.\n4x + y = 9\n4(3y-1) + y = 9\n12y - 4 + y = 9\nAdding four to both sides and combining like terms I get:\n13y = 13\nTherefore y = 1.\nWe can then solve for x using one of the original equations, we'll pick 4x + y = 9.\n4x + y = 9, and we know y=1 so:\n4x + 1 = 9\nSubtracting 1 and solving gives us\n4x = 8, and therefore x = 2.\nYou can use this same process to solve your problem.  I suggest starting by solving the second equation for y.\nHope this helps, please remember to come back and choose a best answer!  :-)
727240
Substitute y(0)=1 into y= -1/(-cos x + c)\n\n1 = -1/(-cos0 + c)\n1 = -1/(-1 + c)\n-1 + c = -1\nc = -1 + 1\nc = 0