I need help solving these simultaneous linear equations, not just the answers but show me how to work them?
919291
I will show one example on how to do it and then you take that example and apply it to the other two problems:\n\nOn the first problem, start with these two equations:\n2x-y+z=15\nx-y+2z=12\n\nTake the second equation and multiply it by -2:\n2x-y+z=15\n-2x+2y-4z=-24\ny-3z=-9\n\nLet's focus on the other part:\nx-y+2z=12\nx+2y-z=-3\n\nMultiply the second equation by -1:\nx-y+2z=12\n-x-2y+z=3\n-3y+3z=15\n\nLet's put those two equations together:\ny-3z=-9\n-3y+3z=15\n-2y=6\ny=-3\n\nPlug it into a y-value:\n-3-3z=-9\n-3z=-6\nz=2\n\nPut the y and z value into this equation:\nx-(-3)+2(2)=12\nx+3+4=12\nx+7=12\nx=5\n\n(5,-3,2)\n\nCheck:\n5-(-3)+2(2)=12\n5+3+4=12\n8+4=12\n12=12\n\nI hope that helps!\n\nP.S. I checked with the original equations and they equaled out also. You can check the answers with a graphing calculator. Whoever is giving the people for thumbs down for their answers, whether it is right or wrong, is being very unkind and rude for doing so. I am sure that I am not only person who does not appreciate the thumbs down given too.
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
744692
Consider the following equations:\n1)   2x + y = 4 \n  \n2)   x − y = −1 \n\nTo solve them simultaneously, first, you need to solve one equation with respect to one variable. Let's solve the first one:\n\n2x + y = 4  Rearranging, you get:\ny = 4 - 2x\nNow you know the value of y, so plug it into the second equation, like this:\n\nx − y = −1 Rearrange using the value for y that you determined above:\n\nx - (4 - 2x) = -1\nNow solve this equation, since it now has only one variable:\n\nx - 4 + 2x = -1\n3x = 4 - 1 = 3\nx = 1 Now that you have the real value for x, go back to the first equation and solve for the real value of y:\n\n2x + y = 4  \ny = 4 - 2x  we know x = 1 so, y = 4 - 2\ny = 2\n\nYou can prove these are the right answers by using them in the equations.
999333
There are several simple rules for re-arranging and solving equations.  It would be in your best interest to try and learn them, they aren't very hard.
729005
a.xa+xb\nb.21x^3+24x^2y\nc.3x(8x+5)+2(8x+5)\n=24x^2+15x+16x+10\n=24x^2+31x+10\nd.4x(2x+1)-6(2x+1)\n=8x^2+4x-12x-6\n=8x^2-8x-6\ne.8x(5x^2+7x+3)+2(5x^2+7x+3)\n=40x^3+56x^2+24x+10x^2+14x+6\n=40x^3+66x^2+38x+6\nf.x^2-1\ng.x(x^2+2x+4)-2(x^2+2x+4)\n=x^3+2x^2+4x-2x^2-4x-8\n=x^3-8
642018
Wonderful.\nTHis is some sort of simultaneous equation.\nAccording to the principle of this equation, we try to rationalize one term of the equation so as to get one unknown term. We can either try finding the x term or the y term. For your question as to find the y term.\n\nFirstly, we rationalize the x terms by multipling them by the inverts. then it will result to \n                                    12x + 20y = 32\n                                    12x + 18y = 39\nThe we apply the subtraction rule. ie. we multiply all the terms beneath by the negative (-) sign. This will result to \n                                    12x + 20y = 32\n                                   -12x - 18 = -39\nWe later subtract so as to attain\n                                2y = -7\nWe apply the law of linear equation by dividing both sides by the coefficient which in this case is 2.\n                              2y/2 = -7/2\nThis will eventually cancel the 2 in the first term therby leavin y alone. \n The answer will be in either of the following cases through simplification.  -7/2, -3 1/2 or -3.5
893823
Use the quadratic equation \n\nt = -b + or - sqr(b^2 - 4ac)/2a\n\nyour equation is already in the form:\n\nat^2 + bt + c = 0\n\njust plug in the variables and solve for t twice
892382
To solve the first one, divide each part of the first equation by two, so 2x-3y=-13; since -2x+3y = 13, adding the two equations gives 0x+0y=0.  No help.  Try substitution.\nFrom the first equation, 4x = 6y-26; x = 1.5y - 6.5\nsubstitute into the second equation -2(1.5y-6.5) + 3y = 13; -3y +13 + 3y = 13; 13=13.  Still no help.\nBoth equations are equivalent; the line that represents them is the solution.\n\nProblem two:\nSubtract equation 2 from equation 1.\n3x-3x+2y-3 = 6 - y\n2y-3 = 6-y\n3y = 9\ny = 3\n3x+6=6\nx=0\nchecking\n3x+3=y=3\n3x=0\nx=0
919864
1.  ax + xb\n2.  21x^3 + 24x*2y\n3.  24x^2 + 31x + 10\n4.  24x^2 - 8x - 6\n5.  40x^3 + 66x^2 + 38x + 6\n6.  x^2 - 1\n7.  x^3 - 8
835073
okay well\n\n1) combine two to figure out the first variable\nx-2y+z=4\n-x+2y+3z=-4\nso 4z=0; z=0\nthen figure out the next variable with a different set of equations\nlike and multiply so the ys will cancel [take Zs out cause zero]\nx-2y+z=4\n2(2x+y)=6\n\ntherefore\nx-2y=4\n4x+2y=6\n5x=10\nx=2\n\nthen substitute for y\n\n-x+2y=-4\n-2+2y=-4\n2y=-2\ny=-1\n\nso y=-1, x=2, z=0\n\n\ncontinue the process with the rest.
589722
Step 1: Solve one equation for one variable in terms of the other.\nStep 2: Plug that into the other equation and solve for the second variable.\nStep 3: Plug the second variable into the first equation and solve for the first variable.
282293
The techniques used to solve systems of linear equations (substitution and elimination) tend to do a pretty good job of solving quadratic systems.\n\nThis page does a pretty good job of demonstrating how:
967876
1) 2a^3 - 13a^2b + 11ab^2 + 14b^3\n\n2) (2a-4b) (4a^2+8ab+16b^2)\n\n3)  (b-8)(b-1)=0\n      b=8   or    b=1\n\n4) 4m^3 - 26m^2 - 14m=0\n2m(2m^2 - 13m - 7)=0\n2m=0  or  2m^2 - 13m -7=0\nm=0  or    (2m+1)(m-7)=0\n                  m=-0.5   or  m=7\n\n so the final answer is m=0 and m=-0.5 and m=7
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
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
239058
You have to solve for one and substitue into the other...\n\nLets solve for x...\n\n2x + 3y = 40 \n2x = 40 - 3y\nx = (40 - 3y)/2\n\nThen Substitute in the other.\n-2x + 2y = 20\n-2(40 - 3y)/2 + 2y = 20  (the 2's in the first set cancel becaue 2/2 = 1 so...\n-(40-3y) + 2y = 20  (then distribute the -)\n-40 + 3y + 2y = 20 (combine like terms)\n5y = 60\ny =12\n\nThen sub y into the equation to find x\n\n 2x + 3y = 40 \n2x + 36 = 40\n2x = 4\nx=2\n\nDouble Check in the other equation...\n-2x + 2y = 20\n-2(2) + 2(12) = 20\n-4 + 24 = 20\n20 = 20 These equal so x=2 and y=12!
932620
solve the equations of any two and make sure that the solution satisfies the equation of the third perpendicular bisector