Solve by the addition method.\n4x+5y+z= 6\n2x-y+2z= 11\nx+2y+2z= 6
200336
I don't condone getting Homework answers here but I can help you. first you can eliminate 1 of the variables for example x by multiplying the 3rd eq. by -2 and adding eq. 2 and  eq.3 \n\n2x-y+ 2z = 11 + (-2x - 4y - 4z = -12)\nthis will leave you with 0x -5y - 2z = -1\n\nthen you can do the same thing with eq. 1 and eq. 2\nnamely (4x + 5y + z = 6) +(-2)(2x-y+2z = 11)\nthis simplifies to \n\n0x + 7y -3z = -16\n\nyou have to repeat the process and eventually you get a solution to 1 of the variable which you can then put back into one of the equations to find the other answers
440299
Using the third equation, you can see that z = x + y + 1.\n\nSubstituting into the first two equations:\n\n4x + 2y - (x + y + 1) = 4\n2x - 3y + 2(x + y + 1) = 4\n\nSimplifying:\n\n3x + y = 5\n4x - y = 2\n\nAdding the two equations:\n\n7x = 7\nx = 1\n\nSubstituting x into the first of the two equations that I have containing only x's and y's:\n\n3(1) + y = 5\ny = 2\n\nSubstituting x and y into the original equation that I derived:\n\nz = (1) + (2) + 1\nz = 4\n\n(x, y, z) = (1, 2, 4)\n\nYou can verify that this is a solution by substituting values into the original 3 x 3 system.
454306
9x/5 + 5y/4 = 47/10\n2x/9 +3y/8 = 5/36\n\n36x+25y=94\n16x+27y=10\n\ny=(94-36x)/25\n400x+2538-972x=250\n\ny=-2\nx=4\n\nfor another one use the same method!
411003
6z+11=5z+3\n=>6z-5z=3-11\n=>z=--8\n/Therefore, (C) is corret
532001
There are 2 ways.\n1st - Substitution\n In the first equation solve for x.\n x=4y+6\nIn the second equation substitute 4y+6 for x.\n5(4y+6)+3y = -16\n23y=-46\ny=-2 No use -2 for y in either equation and ckeck the other\nwith x and y. Using the first -x+4(-2)=-6\nso x=-2\nChecking  5(-2)+3(-2)=-10-6=-16   OK\n\n2nd  by adding (or subtracting) simultanious equations\n\nMultiply the first thru by 5\n-5x + 20y=-20  and add this equation to the second.\nThis gives 23y=-46 so y=-2 and you can do the rest.\n\nTry the last 2 yourself befor reading on so you get the\nmental exercise of doing. Reading the answer is not\nthe same as doing it\n 2nd pair of equations - Subtract the second equation from the \nfirst to get rid of the y.  -5x+x=23-3 ==>  -4x=20 \nx=-5.   -(-5)+y=3  ==> y=-2\nChecking:  -5(-5)+(-2)=25-2=23  OK\n\n3rd pair\n  Add equations to eliminate x\n-2y+5y=36 ==>  3y=-12 ==> y=-4\nx-2(-4)=12 ==> x=4\nChecking: -4 +5(-4)=-24  OK
288419
Multiply the first equation by x, multiply the second equation by -y, then add the two equations together.\n1 + xy = 5x\n-xy - 1 = -8y\n0 = 5x - 8y\nAdd 8y to each side\n8y = 5x\nDivide both sides by 5y\n8/5 = x/y\nAnswer: 8/5
677040
well start by multiplying things out\n\n-8x+12=-8x+5\n-8x+8x+12=5\n0+12=5\nno solution?\n\n2x+3y=6\n2x=6-3y\nx=6/2-3y/2\nx=3-(3/2)y
530101
5x+3y+z+5w=-2\n3x+2y+2z+4w=2\n2x+4y+3z-3w=-11\n4x-3y-2z+2w=-3\n\n=>\n5     3     1     5   I  -2\n3     2     2     4   I   2\n2     4     3    -3   I  -11 \n4    -3    -2     2   I  -3\n\nR4 - (4/5)*R1 , R3 - (2/5)*R1 and R2 - (3/5)*R1 will give\n\n5.0000    3.0000    1.0000    5.0000   -2.0000\n         0    0.2000    1.4000    1.0000    3.2000\n         0    2.8000    2.6000   -5.0000  -10.2000\n         0   -5.4000   -2.8000   -2.0000   -1.4000\n\nR4 - (-5.4/0.2)*R2 , R3 - (-5.2/0.2)*R2 will give\n\n5.0000    3.0000    1.0000    5.0000   -2.0000\n         0    0.2000    1.4000    1.0000    3.2000\n         0    0.0000  -17.0000  -19.0000  -55.0000\n         0    0.0000   35.0000   25.0000   85.0000\n\nR4 - (-37/17)R3 will give\n\n5.0000   3.0000    1.0000    5.0000   -2.0000\n         0   0.2000    1.4000    1.0000    3.2000\n         0   0.0000  -17.0000  -19.0000  -55.0000\n         0   0.0000    0.0000  -14.1176  -28.2353\n\nR4 / -14.1176 will give\n\n5.0000    3.0000    1.0000    5.0000   -2.0000\n         0    0.2000    1.4000    1.0000    3.2000\n         0    0.0000  -17.0000  -19.0000  -55.0000\n         0    0.0000    0.0000    1.0000    2.0000\n\nR3 - (-19)R4 will give\n\n    5.0000    3.0000    1.0000    5.0000   -2.0000\n         0    0.2000    1.4000    1.0000    3.2000\n         0   -0.0000  -17.0000         0  -17.0000\n         0    0.0000   -0.0000    1.0000    2.0000\n\nR3 /(-17) will give\n\n5.0000    3.0000    1.0000    5.0000   -2.0000\n         0    0.2000    1.4000    1.0000    3.2000\n         0    0.0000    1.0000    0.0000    1.0000\n         0    0.0000    0.0000    1.0000    2.0000\n\nR2- R4 will give\n\n5.0000    3.0000    1.0000    5.0000   -2.0000\n         0    0.2000    1.4000             0    1.2000\n         0    0.0000    1.0000             0    1.0000\n         0    0.0000    0.0000    1.0000    2.0000\n\nR2- (1.4)R3 will give\n\n5.0000    3.0000    1.0000    5.0000   -2.0000\n         0    0.2000    0.0000             0   -0.2000\n         0    0.0000    1.0000             0    1.0000\n         0    0.0000    0.0000    1.0000    2.0000\n\nR3 /(0.2) will give \n\n5.0000    3.0000    1.0000    5.0000   -2.0000\n         0    1.0000    0.0000             0   -1.0000\n         0    0.0000    1.0000             0    1.0000\n         0    0.0000    0.0000    1.0000    2.0000\n\nR1- 5R4 will give\n\n5.0000    3.0000    1.0000             0  -12.0000\n         0    1.0000    0.0000             0   -1.0000\n         0    0.0000    1.0000             0    1.0000\n         0    0.0000   -0.0000    1.0000    2.0000\n\nR1-R3 will give \n\n5.0000    3.0000             0             0  -13.0000\n         0    1.0000    0.0000             0   -1.0000\n         0    0.0000    1.0000             0    1.0000\n         0    0.0000   -0.0000    1.0000    2.0000\n\nR1-3R2 will give \n\n5.0000             0   -0.0000             0  -10.0000\n         0    1.0000    0.0000             0   -1.0000\n         0    0.0000    1.0000             0    1.0000\n         0    0.0000   -0.0000    1.0000    2.0000\n\nR1/5 will give \n1.0000             0   -0.0000             0   -2.0000\n         0    1.0000    0.0000             0   -1.0000\n         0    0.0000    1.0000             0    1.0000\n         0    0.0000   -0.0000    1.0000    2.0000\n\nAnd thats the final answer.\n\nSo, your answer is x=-2, y=-1,z=1, w=2\n\nVerification\na=\n     5     3     1     5\n     3     2     2     4\n     2     4     3    -3\n     4    -3    -2     2\nb= \n    -2\n     2\n   -11\n    -3\ninverse of a is\n\n    0.0250   -0.0250    0.1250    0.1750\n    0.4000   -0.4000    0.0000   -0.2000\n   -0.5125    0.6792    0.1042    0.0792\n    0.0375    0.1292   -0.1458   -0.0708\n\nThus,\ninv(a)*b = [-2 -1 1 2]'\n\nI hope i answered your question.\n\nAll the best
354183
2x+y=15, 5x-6y+-22...\n\nwould like to clarify...\n\ngiven: \n2x+y=15\n5x-6y-22=0 --> change to 5x-6y=22\n\n(are the givens correct???)\n\ngiven1: 2x+y=15\ngiven2: 5x-6y=22\n\n12x+6y=90 (multiply by 6 to cancel the y terms)\n5x-6y=22\n-------------\n7x=112\n\nx=112/7\nx=6\n\nif x=6, get teh value of y, using\n2x+y=15\n2(6)+y=15\ny=15-12 (transfer to the other side, change the sign)\ny=3\n\nto check if x is correct, substitue to one of the given equations\n2x+y=15\n2(6) + 3 =15\n\nusing the other equation:\n5x-6y=22\n5(6)-6(3)=22\n30-18=22\n\nanswer: x=6, y=3
477309
2x+y=15  *6\n12x+6y=90\n5x-6y=-22\nadding\n17x=68\nx=4\nsub\n8+y=15\nadding -8\ny=7\nso x=4,y=7
3688
X+Y+Z=24\nY=X+2\nZ=2*X-2\n\nX+(X+2)+(2*X-2)=24\nCollect terms: 4*X=24; X=6\nY=6+2; Y=8\nZ=2*6-2; Z=10\n\n6 cm, 8 cm, 10 cm
160631
Set one equation equal to one of the variables, then insert it into the other equation and solve for the second variable.  then insert that variable into the original equation.\n\nx+y=5 or x=5-y\nx+2y=6\n\nput x=5-y into the second equation\n\n5-y+2y=6 \n\nnow solve for y\n\ny=1\n\nput this back into one of the original equations\n\nx+1=5\n\nx=4
431136
Multiply equation 1 by 2 so that when adding both equation, the x variable is cancelled,\n\n6x + 12y = 16\n-6x + 3y = 2\n0 + 15y = 18\ny = 6/5\n\nSubstituting y in equation1\n3x + 36/5 = 8\n3x = 4/5\nx = 4/15\n\nTherefore  x = 4/15 , y = 6/5
414392
I'll get you started.\n\n1) x + y = 6, so x = 6 - y\n\nx + z = 8, so since x = 6 - y, (6 - y) + z = 8,  y = 6 - 8 + z\n\ny + z = 2, so (6 - 8 + z) + z = 2, and z + z = 2 - 6 + 8, so \n\n2z = 4, and then z = 2\n\nnow put that number back into the other formulas, \n\ny + z = 2, so y + 2 = 2, so y = 0\n\nthen x + y = 6, so x + 0 = 6, so x = 6\n\n\n2) 2x + 4y = 9 & -2x + y = 5\n\nso 2x = 9 - 4y, so x = (9 - 4y) / 2\n\nsubstitute x, so -2((9 - 4y) / 2) + y = 5, so -9 + 4y + y = 5\n\nso 5y = 5 + 9, 5y = 14, y = 14/5 = 2.8\n\n2x + 4(2.8) = 9,   so 2x + 11.2 = 9,   so 2x = 9 - 11.2\n\nthen x = (9 - 11.2) / 2,  so x = -1.1, \n\ncheck 2x + 4y = 9, 2(-1.1) + 4(2.8) = 9\n\n-2.2 + 11.2 = 9,   9 = 9
662703
r = 5z\n\n15z = 3y // Solve for z\nz = (3y)/15  \n\nr = 5 * (3y)/15  // Substitute z\nr = 15y/15  // Simplify\nr = y
339221
(y+4)(y+4)-2y -11=0\ny^2 +8y + 16 -2y -11 =0\ny^2 +6y +5=0\n(y+5)(y+1)=0\ny=-5, x=-1 , and   y=-1, x=3\n\nCheck with y=-5\n(-1)^2 -2(-5) = 1+10=11  OK\n\nCheck with y=-1\n3^2 -2(-1) = 9+2=11  OK
102831
Solutions do exist, it’s just possible you got your numbers mixed up (it happens to the best of us)\n\n2x-3y+z=5  --------(1)\n\nx+4y-2z= -3 --------(2)\n\n4x-2y+3z= 6 ---------(3)\n\nConsider (1) . let’s make z the subject of the formula to eliminate it\n\n2x-3y+z= 5\n\nTherefore z= 5+3y-2x    [by rearranging]\n\nYou can substitute z=5+3y-2x in (2) and (3), I’ll put it in (3) first, the order doesn’t matter\n\nTherefore 4x-2y+3(5+3y-2x)= 6\n\n4x-2y+15+9y-6x= 6\n\nSimplifying this you get 2x-7y= 9\n\nPut z=5+3y-2x in (2)\n\nSo  x+4y-2(5+3y-2x)= -3\n\nx+4y-10-6y+4x+-3\n\nsimplifying, you get 5x-2y= 7\n\nso now we have two new equations in terms of x and y, I’ll label them (4) and (5)\n\n2x-7y= 9 ---------(4)\n\n5x-2y= 7 ---------(5)\n\nWe have to solve these simultaneously now, consider (5)\n\n5x-2y= 7\n\nMaking x the subject of the formula we get\n\nx= (7+2y)/5\n\nput this new equation in (4)\n\n2* (7+2y)/5- 7y=9\n\nSimplifying this you get 14-45= 31y therefore y= -1\n\nNow we use ‘backward substitution’ to find x and z, first x\n\nFrom (4) we know 2x-7(-1)=9 therefore x=1\n\nNow put x and y in (1) to find z\n\n2(1)-3(-1)+z= 5\n\nFrom this you find z is 0\n\nTherefore x=1, y= -1 and z=0\n\nP.S: I'm not sure what you mean by the addition method, I used both elimination and substitution methods here, however I hope this helps!