How do I solve systems of equations?
917321
x+y=2\n2x-y=0\n\nx=2-y\n2(2-y)-y=0\n\nX=2-y     \n4-3y=0   \n\nx=2-4/3\ny=4/3
681236
Multiply each term in the first () by each term in the second ():\n\n2x times 2x = 4x^2\n2x times 1 = 2x\n-3 times 2x = -6x\n-3 times 1 = -3\n\nThen combine 2x and -6x\nAnswer: 4x^2 - 4x - 3
819984
restate the first equation in terms of y:\n\ny = 12 - x\n\nNow substitute this in for y in the second equation:\n\n2x² + 3(12 - x)² = 7x(12 - x)\n\nNow, multiply it out and solve for x:\n\n2x² + 3(144 - 24x + x²) = 84x - 7x²\n2x² + 3x² + 7x² - 72x - 84x + 432 = 0\n12x² - 156x + 432 = 0\n12(x² - 13x + 36) = 0\nx² - 13x + 36 = 0\n(x - 9)(x - 4) = 0\n\nx = 9 or 4\n\nSo, {x,y} = {9,3} or {4,8}
91756
Ok if you have like 5x+6=y,and x=2,then obviously use substitution. Substitute the x for the 2 and solve. Y would equal 16. If you have any system that has variable=whatever then that would be easier because in the second equation you would just have to substitute the variable for whatever it equals. Hope this helps.
648696
Define all the known and unknown quantities.\nExpress the relationships between these quantities in equations.\nResolve the equations.
704540
Add the two equations together: (2x+y)+(x-y) = 1+8, so 3x = 9, x = 3.  Then 2*3+y = 1 so y = -5.\n\nSteve
394036
Try to solve the eqs with quad rule:\na*x^2 + b*x + c=0\nx=(-b +- Sqrt(b^2-4*a*c))/(2*a)\n\nx^2+2x+5=0 => Get two Complex answers, \nso no it can't be factorized.\n\nx^2+4x+5=0 => Get two Complex answers, \nso no it can't be factorized.\n\nx^2+4x+3=(x+3)*(x+1)=0; \nSo, yes, this can be factorized. \nSolutions are x=-3 and x= -1\n\nx^2+2x-3=(x+3)*(x-1)=0;\nSo, yes, this can be factorized. \nSols are x= -3 and x=1
1056171
Try www.purplemath.com it shows you how to work out problems step by step
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)]
708644
Use the quadratic formula.\nx1 = 1\nx2 = -1/2
134530
First, generate 3 equations by inserting the values of x and y for each of the coordinates.\n\nEqn 1. For (-1,1): 1=a*-1^2+b*-1+c yields: 1=a-b+c\nEqn 2. For (1,3): 3=a*1^2+b*1+c yields: 3=a+b+c\nEqn 3. For (3,2): 2=a*3^2+b*3+c yields: 2=9a+3b+c\n\nStep 1: solve Eqn 1. for a:  a=1+b-c\nStep 2: insert a from Step 1 into Eqn 2.: 3=(1+b-c)+b+c ==> 2=2b ==> b=1\nStep 3: insert b=1 back into Step 1 and re-solve for a: a=1+1-c ==> a=2-c\nStep 4: insert the results from Step 2 for b and Step 3 for a into Eqn 3 to find c.\n  2=9*(2-c)+3*1+c ==> 2=18-9c+3+c ==> 19=8c ==> c=19/8\nStep 5. Insert the results from Step 4 and Step 2 into Eqn 1 to find a.\n  1=a-1+19/8 ==> a=-3/8\n\nTherefore:\n  a=-3/8\n  b=1\n  c=19/8\n\nStep 6: Verify for -3/8*x^2 + 1*x +19/8\n (-1,1): 1=-3/8*(-1^2)+1*-1+19/8\n (1,3): 3=-3/8*(1^2)+1*1+19/8\n (3,2): 2=-3/8*(3^2)+1*3+19/8
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
239864
Solve for y in the second equation, then plug that into the first equation and solve for z.  After you find the value of z, plug that into the other equation.\nMake sure you always check your answer!  To do that just plug in the values for y and z into each equation to see if you get a true statement.  If you don't then your answer is wrong or else you made a mistake when checking the answer.
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
429488
If you have two unknowns (i.e. x, y) then you must have two equations to solve for x and y. You only have one equation, unless I missed something.
867879
Just plug 0 in the function whenever you see y.\n\n4x + 3(0) + 24 = 0\n4x + 0 + 24 = 0  \n\nThen solve for x.\n\n4x = -24\nx = -6
928663
Are you asking for help or an answer?\n\nTake any 2 equations and multiply all of the terms or one or both by numbers that will make one of the terms in each equation have the same coefficient. Now subtract one from the other and the term will disappear. Now take the resulting equation and do the same with it against the third. Keep doing this until you are down to one term and solve for the variable. Now substitute the variable into the equation and solve for the other variables.