Evaluate 3x + 2y -2z when x=5, y =2 and z =1.?
913862
3x + 2y - 2z\n\n3(5) + 2(2) - 2(1)\n\n15 + 4 - 2\n\n19 - 2\n\n17\n\nThe answer is 17\n\n- - - - - - -s-
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
401465
(4x + 5y)(2y^2)\n8y^3 + 10y^3\n\na.)\n(4x + 2y^2)(5y)\n20xy + 10y^3\n\nb.)\n(4x) + (5y)(2y^2)\n4x + 10y^3\n\nc.)\n4x + (5y + 2y^2)\n4x + 5y + 2y^2\n\nd.)\n(2y^2)(4x + 5y)\n8xy^2 + 10y^3\n\nSo the answer is d.) (2y^2)(4x + 5y)\n\nI just did the math out.
720922
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)
699742
-(3x+y)(3x-y)\n-(9x^2-y^2)\ny^2-9x^2
424288
1. y2+6y+8 = (y+4)(y+2)\n2. x2+7x+12 = (x+4)(x+3)\n3. z2+2z-24 = (z-4)(z+6)\n4. x2+7x+10 = (x+2)(x+5)
973910
17.\n5x+7y=-1  (1)\n-2y+3z=9  (2)\n7x-z=27    (3)\n(3)*3\n21x-3z=81\n-2y+3z=9\nadding\n21x-2y=90 *7\n147x-14y=630\n10x+14y=-2\n157x=628\nx=4\nsub in (1)\ny=-3\nsubin (2)\nz=1\n\n\n18.\nr-s+3t=-8    (1)     \n2s-t=15       (2)\n3r+2t=-7      (3)\n\n(2)*2  4s-2t=30\n          3r+2t=-7\nadding\n4s+3r=23\n(2)*3  6s-3t=45\n       r-s-3t=-8\nadding\n5s+r=37\n4s+3r=23\n15s+3r=111\nsubtracting\n11s=88\ns=8\nsub in (2)\n16-t=15\nt=1\nsub in (3)\n3r+2=-7\nr=-3
1144396
you would take the 2x-4y=3 and solve for a variable, such as x=(3-4y)/3   you now have what x equals and you sub that into the other equation so you now have 2[(3-4y)/3]-3y=0 and you should be able to solve that equation
930265
4x + 3y=27\n4x-2y   = 2\nnow subtract the second equation from the first.\n5 y = 25\ny = 25/5  = 5 \n2x-5 = 1\n2x = 6\nx = 3\nTest any of the two equation substituting values for x and y.
516161
x-y+3x-5y+3=0\n\n4x - 6y + 3 = 0\n\n4x + 3 = 6y\n\n2x/3 + 1/2 = y
791030
3x – 2y = 6\n3x = 6 + 2y\nx = 2 + 2/3y
168603
easy if that is the question\n2x+3x=5\nsince they have the same variable\nadd them \n5x=5\nthen divide by 5(no. to reduce the variable with 1 as its coefficient) \nx=1\n\nlonger method\n\n2(x)+3(x)=5\nby distributive property we know that (2+3)x=2(x)+3(x)\nthen: \n(2+3)x=5\n(5)x=5 divide by 5 \nx=1
1065375
5 = 7y - [4 + (3y - 2)]\n5 = 7y - (2 + 3y)\n5 = 4y - 2\n7 = 4y\ny = 7/4
164756
Easy!\n\n2x+2y=2 \n3x-y=1\n\nTake the second equation and isolate the single "y":\n\n3x - y = 1 becomes 3x - 1 = y (without going into how)\n\nSubstitute "3x - 1" in the first equation so it reads:\n\n2x+2(3x - 1)=2\n\nSimplify:\n\n2x + 6x - 2 = 2\n8x - 2 = 2\n8x = 4\nx = 1/2\n\nEarlier, we saw that y = 3x - 1, so:\n\ny = 3(1/2) - 1\ny = 3/2 - 1\ny = 1/2\n\nThere may be another way (less involved) that I've forgotten, but you can check each of those values for x and y in both equations to make sure they work.
958973
y=5x+2\ny=5(2)+2\ny=10+2\ny=12
695176
6x^2-3x-4x+2\n3x(2x-1)-2(2x-1)\n(2x-1)(3x-2)
975557
5x-3y=-12\n-3y=-5x-12\nmultiplying by -1\n3y=5x+12\nm=5/3\nb=12/3=4\n\n2.y=3x+5\n\n3.y+5/5+5=x+3/4+3\ny+5/10=x+3/7\nmultiplying by 70\n7(y+5)=10(x+3)\n7y+35=10x+30\nadding -7y\n35=10x-7y+30\nadding -35\n10x-7y-5=0