Solve the system of equations 3r - 4s = 0 and 2r + 5s = 23. \n\nR = ?\nS = ?\n\nWhat does R = ? ... and what does S = ? ...\n\nThanks.
120490
3r - 4s = 0\n2r + 5s = 23\n\nMultiply the first equation by 5 and the second equation by 4:\n\n5 * (3r - 4s = 0)  --> 15r - 20s = 0\n4 * (2r + 5s = 23) -->  8r + 20s = 92\n\n15r + 8r = 23r\n-20s + 20s = 0\n0 + 92 = 92\n\n23r = 92\n\nDivide each side by 23:\n\n23r / 23 = 92 / 23\n\n23r / 23 = r\n92 / 23 = 4\n\nTherefore, r = 4.\n\nSo solve for s, let's use the first equation 3r - 4s = 0.\n\n3(4) - 4s = 0\n\nAdd 4s to both sides of the equation:\n\n3(4) - 4s + 4s = 0 + 4s\n\n3 * 4 = 12\n-4s + 4s = 0\n\n12 = 4s\n\nDivide both sides by 4:\n\n12 / 4 = 4s / 4\n\n12 / 4 = 3\n4s / 4 = s\n\nTherefore, s = 3.
123098
35 Student tickets\n15 Adult tickets\n\n5a+3s=$180\na+s=50\na=50-s  solve for a\n\n5(50-s)+3s=180  substitute the value of a into the equation\n250-2s=180\n2s=75  solve for s\n\ns=35\na=50-s=50-35=15
357953
1st one:\n\n5x-2y=-5; flip 2y to other side\n5x=2y-5; divide by 5\nx=2/5y-1; put this into 2nd equation for x\ny-5(2/5y-1)=3\ny-2y+5=3\n-y=3-5\n-y=-2\ny=2\n\nput this into 2nd equation for y\n2-5x=3\n-5x=1\nx=-1/5\n\n2nd one\n8x-4y=16; divide by 4\n2x-y=4; flip 2x to other side\n-y=4-2x; multiply by -1\ny=2x-4; lo and behold this is exactly the same as the 2nd equation - can't be solved with substitution, equations 1 and 2 are the same\n\n3rd one\n4x-12y=5; divide by 4\nx-3y=5/4; flip 3y to other side\nx=5/4+3y; substitute this into x in 2nd equation\n-(5/4+3y)+3y=-1; simplify\n-5/4-3y+3y=-1; cancel out the -3y and + 3y\n-5/4=-1; well, how can you have -5/4 ever equal to -1, equations are not the solvable (unless either of the numbers equals infinity)
429833
2 + 1 = 3\n4 + 1 = 5\nTake the numerals out of the two equations above which are in the next equation 2+2=5.\nWe will get 2 & 5.Now take out the nos. which are similar in both the equations.now we will have 1,1,2,5.As we have 3 numerals in the final equation we will add up the smallest digits 1+1.Then we will have 2,2,5which also means 2+2=5.\n2+2=5 is equal to 2+1+3and 4+1+5\nA little confusing.\nI hope u understand.
79344
This is not an equation unless it is equal to something.  It is an expression.\n\nFirst, you can simplify the middle: a-5a = -4a\n\nThen, multiply the first 2 together: 3a(-4a)=-12a^2\n\nNow multiply this by the last part: -12a^2(a-35)= -12a^3 + 420a^2.\n\nIf you were supposed to set this equal to 0 and solve for a, you would not need to multiply it out.  You would just set each part = 0 and solve\n\n3a = 0\n-4a = 0\na-35 = 0
119380
Use the identity cos 2x = 2 cos^2 x - 1\n\nThen you have 3(2cos^2 x -1) - 5cosx - 1 = 0\n\n6 cos^2 x - 3 - 5cosx - 1 = 0\n\n6 cos^2 x - 5 cosx - 4 = 0\n\n(3 cosx - 4)(2 cosx + 1) = 0\n\ncosx = 4/3 (impossible) or cosx = -1/2\n\nIn degrees, x = 120, 240, or if you need, 120+360n, 240+360n\nIn radians, x = 2pi/3 or 4pi/3 etc.\n\nHope that helps you see what to do.
477228
No, I don't think you are right.  I assume this is a system of simultaneous equations and you're trying to find a solution set.\n\n4x - 12y = 5\n-x + 3y = -1\n\nFirst, multiply the second equation by 4 on both sides, giving\n\n4x - 12y = 5\n-4x + 12y = -4\n\nNext, add the two equations, giving\n\n4x-4x +12y -12y = 5 -4\n0x + 0y = 1\n\nThere are no solutions (x,y) for which this is true, so the solution set is the empty set.\n\nTaking the solution set you've determined, (5,0), putting that into the first equation gives\n4*5 -12 * 0 = 5\n20 -0 = 5 or 20 = 5, which is not true, so (5,0) cannot possibly be a solution.\n\nYou've made several errors, as follows:\n\n4x-12y=5  OK\n-x+3y= -1 OK\n\n-x=3y+-1 NO!, should be -x = -3y + -1 or x = 3y +1\n4(3y+-1)-12y=5 OK\n12y+-4-12=5 NO!, should be 12y -4 -12y = 5 or -4 = 5 \n0y=1 I have no idea how you got this\ny=0 NO!, if 0y =1, y is not 0, there is no y such that 0y =1\n-x=12y+5  I have no idea how you got this\n-x=12(0)+5 I understand how you got this, but y does not equal 0\n-x=0+5 OK\n-x=-5 previously you had -x = 0 + 5 , which was wrong but if it was right, you should get x = -5, not -x = -5\n\nOMG
81336
Ah, systems.\n\nthe first problem- there are many ways you can choose to do this, but I think we should multiply both sides of the first equation by two so that they positive and negative y's will cancel out.\n\nSo now we have:\n2(2x + y)=2(4)\n3x - 2y = -1\n\nSimplified:\n4x + 2y = 8\n3x - 2y = -1\n\nnow you add down, so you get:\n7x +0y = 7\n7x =7\nx=1\n\nsubstitute 1 for x in any equation:\n2(1)+y = 4   \n2 + y = 4\ny = 2\n\n\n\nfor the second one- lets cancel out the y's again.\n\n5x - 3y = 14\n-3(2x - y) = (6)-3\n\n5x - 3y = 14\n-6x +3y = -18\n\nAdd them together:\n-x = -4\nx = 4\n\nThen solve for y:\n5x - 3y = 14\n5(4) - 3y = 14\n20 -3y = 14\n-3y = -6\ny = 2
447762
let - 5+ 5 sqrt(3) i = r cos t + ir sin t\n\nr = sqrt(5^2+(5sqrt(3))^2) = sqrt(25+75) = 10\n\ncos t = -1/2\n\nt = 120 degrees as cos -ve and sin +ve\n\nt = 2pi/3\n\nso C is the ans
394573
-4(3a+4b)= -12\n3(4a-5b)= -16\n\n-12a-16b= 48\n12a-15b= -48\n\n-b=0\nb=0\n\n3a+4(0)= -12\n3a= -12\na= -4\n\ncheck: 3(-4)+4(0)= -12
534029
in order to solve the system of equations you can do one of two things.  The easiest thing to do is to set the two equations to be equal. i.e. \n\nsince y=2x and y = -3x+5, then 2x=-3x+5.  In solving for x you will find that 5x=5 or x=1.  Then you can plug your resulting x back into the equation to get y=2(1) =2.  The main thing you are trying to do is get one of your variables to be equal to some combination of your other variables.  For example, if you are trying to solve 3y+2x-1=5x+3 and 6y=2x+7.  The first thing you want to do is to get a single variable by itself on one side of an equation.\nLet's start with the first equation, 3y+2x-1=5x+3. You would try to group the y terms by themselves on one side of the equation, an group the x terms and constants on the other side of the equation.  In this example, in order to group the terms correctly, you would have 3y=3x+4.  Then solve for y by itself by dividing both sides by 3.  The result is y=x+4/3.  You take this result and substitute for y in the second equation, 6y=2x+7.   You will get 6(x+4/3)=2x+7.  Then you solve for x by itself.  6x+8=2x+7\n4x=-1;  x=-1/4.  Then you can substitute this result back into either of your original equations to produce your solution for this system.  6y=2(-1/4)+7 => 6y=13/2 =>y=13/12.  In order to check your answer, you can graph the two linear equations.  Where the two lines intersect is the solution to your system.\n\n\nIf you are looking for a more advanced way to solve a system of equations, then you may want to consider using an augmented matrix.  Rearrange your equations so that you have only variables on one side of your equations, and only constants on the other side of your equations, then sole the system of Ax=b.  The matrix A is the coefficients of your variables, and b is a vector of your constants.  x is a vector of your variables.  Augment the matrix of your coefficients with your constant vector and do Gaussian Elimination until the matrix is in reduced row echelon form.  This will provide the exact solution to the system of equations.
45358
-3+5=5-3y+4y       \n   2=5+y\n  2-5=y \n   -3=y\n because...          2=5-3(-3)+4(-3)\n                     2=5+9-12 \n                     2= 14-12         \n                     2=2
366078
(1) 5x-2y+z=11\n(2) 3x+4y-2z=-9\n(3) -2x+2y-z=-4\nadd (1) 5x-2y+z=11\nand (3) -2x+2y-z=-4\n3x=7\nx=7/3\n\nnow the equations become\n(4) 5*7/3+2y-z=11\nor 2y-z=11-35/3=-2/3\n\n(5) 3*7/3+4y-2z=-9\nor 4y-2z=-9-7=-16\n\n(6) -2*7/3+2y-z=-4\nor 2y-z=-4+14/3=2/3\n\n4 & 6 are the same, just multiplied by -1\n5 is parallel to them. There is no solution
289906
where is the rest of the equation?\n\n2x+5-x+3x+2-2x+5x-3x+8-4x+2+1x\n\nCollect like terms:\n\n2x-x+3x-2x+5x-3x-4x+1x+5+2+8+2\n\nRelocate the terms:\n\n1x+2x+3x+5x-x-2x-3x-4x+5+2+8+2 = 37\n\nAdd like terms:\n\n11x - 10x + 17 = 37\n\nNow, subtract all terms and solve from there.\n\nx + 17 = 37\nx = 20
348694
Is this your equation?\nx - 2 - 2 - 3x = 0\n\n-2x - 4 = 0\n-2x = 4\nx = 4/-2\nx = -2\n\nThe answer is:\n x can either equal *-2* or *you made a mistake*
339421
1.4x^2-3x-1=0\nx= (3+- sqrt(9-4(4)(-1) ) )/2(4)\nx=(3+- sqrt(25) )/8\nx=(3+- 5)/8\nx= (3+5)/8 =1 or x=(3- 5)/8 = -1/4\nthen \n4x^2-3x-1 = (x-1) (4x+1) =0\n\n2.(x^2-14)/(x-2)(x+2)+3/(x-2) =  4 / (x+2)\n [ (x^2-14)+3(x+2) ]/(x-2)(x+2)= 4 / (x+2)\nmultiply the equation by (x-2)(x+2) then we get:\n(x^2-14)+3(x+2) = 4(x-2)\nx^2 +3x -4x -14 +6 +8 =0\nx^2 -x =0\nx(x-1) =0\nso x =0 or x=1\n.
368598
4b + 3m = 3\n3b + 11m = 13\n\nMULTIPLY 1ST EQUATION BY 3\n3(4b + 3m = 3)\n12B + 9 M = 9\n\nMULTIPLY 2ND EQUATION BY -4\n-4(3b + 11m = 13)\n-12B -44M = -52\n12B +9M   = 9\n\nTHE B'S CANCEL\n-35M = 43 \n\nM= -35/43 0R -0.81395348837209302325581395348837\n\nNOW REPLACE THAT IN THE 1ST OR 2ND EQUATION TO GET B\n\n4b + 3(-35/43) = 3\n4B - 2.4418604651162790697674418604651 = 3\nADD 2.4418604651162790697674418604651 TO BOTH SIDES\n4B = 5.4418604651162790697674418604651\nDIVIDE 4 ON BOTH SIDES\nB = 0.73504273504273504273504273504275