Complete the ordered pairs so that each is a solution for the given equation:\n\n3x + 4y = 12 (0,  ),(  ,3/4),(  ,0),(8/3,  ) \n\n...and...\n\nWhen x number of units are sold, the price of each unit (in dollars) is given by p = -x/2 + 75. Find the unit price when the following quantities are sold: 2, 7, 9, 11.\n\nThanks!
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3\n3\n4\n1\n\n74\n71.5\n70.5\n69.5
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I'm not sure what you mean by ***, so I'll answer the second part of your question.\n\n-4x/-4x + 12/-4x = 1 - 3/x
447775
We need to factor the equation down. This involves finding the roots of the equation. This will look like:\n\n(ax+b)(cx+d) = 12x^2-31x+20 = 0\n\nBecause the final term is positive and the middle term is negative, my first hunch is to have b = -4 and d = -5. By doing so, the equation looks like:\n\n(ax-4)(cx-5) = 12x^2-31x+20 = 0\n\nThe factors needed for 12x^2 could be 2 and 6 or 3 and 4. We should test each combination to see that they work out. The factoring ends up looking like this:\n\n(3x-4)(4x-5) = 0\n\nNow all that remains is to solve for x in each set of parentheses. If the first term is zero, then\n\n3x-4 = 0\nx = 4/3\n\nIf the second is zero, then\n\n4x-5 = 0\nx = 5/4\n\nThus, x = 5/4, 4/3
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
352402
3 variables, 2 equations, a number of solutions are possible,\n\nSuppose you have bought x no. of sheeps, y no. of pigs & z no. of chickens.\nHence, x + y + z = 100\ncost of x sheeps = $10x\ncost of y pigs = $2y\ncost of z chickens = $0.5z\nHence, 10x+2y+0.5z = 100\nor, 20x + 4y + z= 200\nor, 20x + 4y + z - x - y - z = 200-100\nor, 19x + 3y = 100\nif x = 1 , 3y = 100 - 19 = 81\nor, y = 27\nhence, z = 100 - 27 -1 = 72,\nif x = 2, 3y = 100 - 38 = 62\nhence, in that case y is not an integerwhich is impossible (you can't buy a half of a pig)\nfor similar reason x is not equal to 3, 5\nif x=4, 3y = 100 - 76 = 24\nor, y = 8\n hence z = 100 - 8 - 4= 88\nx can't exceed 5, as in that case y becomes negative.\n\nso the two possible solutions are,\nYou have to buy 1 sheep, 27 pigs & 72 chickens\nor,\nYou have to buy 4 sheeps, 8 pigs & 88 chickens\n\nIn both the cases you have to spent $100 & you buy 100 animals.
452445
To figure this out, you need to know that an ordered pair is in the form: (x, y). Just take each ordered pair and stick it into the equation:\na) 3(1) = 2(-1) -1 ... 3 = -3 (Not correct.)\nb) 3(-1) = 2(-1) - 1 ... -3 = -3 (Correct.)\nc) 3(-2) = 2(-5/2) - 1 ... -6 = -6 (Correct.)\nd) 3(0) = 2(-1/2) - 1 ... 0 = -2 (Not correct.)\n\nSo, the correct answers are (b) and (c).
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extract value of x from eq 1 \n\nx=(8+4y)/3 ------ (3)\n\nSubstitute value of x in eq 2\n\n6((8+4y)/3)-2y=10\n2(8+4y)-2y=10\n16+8y-2y=10\n6y=10-16\n6y=-6\n\ny=-1\n\nput y=-1 in eq 3\n\nx= (8+4(-1))/3\nx= (8-4)/3\n\nx= 4/3\n\nso the solution set is\n\nsolution set= {4/3, -1}
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ok let's make it like this\n\n4x+2y=5 ...(i)\n13x-3y=2... (ii)\n\nnow we are multiplying (i) with 3 and (ii) with 2 and we get\n12x+6y=15 .. (i)\n26x-6y=4 .. (ii)\n\nnow adding (i) and (ii) we get\n\n38x = 19\nso x = 0.5\n\nnow putting this value of 'x; in (i) we get\n2 + 2y= 5\nor 2y = 3\nor y =3/2 = 1.5\n\nthus we get two values\n\nx=0.5\ny=1.5 \n\nand that is the answer
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WHAT is the QUESTION ?\n\nThis is NOT an equation.\n\nYou can simplify the expression to:  4x+3\n\n(divide each CONSTANT in the BINOMIAL by 2)
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There've already been a lot of good answers, but I thought I'd see if I could explain it a little more clearly.\n\n1) 3x - 7 = 2x + 8.  Start by adding 7 and subtracting 2x from each side:  3x -7 +7 -2x = 2x +8 +7 -2x.  Combine the terms to get x = 15.\n\n2) 4(7x + 10) = 3(9x - 5) + 20.  Start by multiplying 4 by everything in the first set of parentheses and 3 by everything in the second set:  28x + 40 = 27x - 15 + 20.  Now, subtract 27x and 40 from each side:  28x - 27x +40 -40 = 27x - 27x -15 +20 - 40.  Combine the terms to get x= -35.\n\n3) Let Ricardo's age = R and his sister's = S.  We are told that R = 2S - 4 and R + S = 20.  Since R and 2S - 4 are the same, we can write the second equation as 2S - 4 + S = 20.  Add 4 to each side and combine: 3S = 24.  Then divide by 3:  S = 8.  So his sister is 8 years old.  Remember that their ages added together are 20, so he must be (20 - 8) = 12 years old.\n\n4) First, find the actual amount of the discount.  The original price was $350, and the sale price is $308, so the discount is ($350 -$308) = $42.  Now, you want to find the rate of discount, or what percent of the price was taken off.  To find that, you divide the discount ($42) by the original price ($350) and mutiply by 100%.  It comes out to 12%.\n\n5) -3(x + 1) = 2(x - 8) + 3.  Start by multiplying out the parentheses:  -3x -3 = 2x - 16 + 3.  Then subtract 2x and add 3 to each side and combine the terms:  -5x = -10.  Divide by -5:  x = 2.
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Divide this equation by 4 on both sides.\n\nYou get \n\n3y^2-x^2+18y+4x+44 = 0\n\nso, (3y^2+18y+27)-(x^2+4x+4)-12 = 0\n\nso, 3(y^2+6y+9)- (x^2+4x+4)-12 = 0\n\nor, 3(y+3)^2 - (x-2)^2 = 12\n\nThis is clearly a hyperbola.\nNow, what you have to do is take Y = y+3 and X = x-2 (note the difference between caps and lower case letters) and divide the whole equation by 12\n\nSo you get Y^2/(2)^2 - X^2/(2sqrt3)^2 = 1\n\nhence a = 2, a^2= 4, b = 2sqrt3, b^2=12, c^2=a^2+b^2=16 and c = 4.\n\nVertices = (0,2),(0,-2) in the X,Y system so it is (2,-1) and (2,-5) in the x,y system. Remember x= X+2 and y = Y-3\n\nFoci = (0,4) and  (0,-4). So, they should be (2,1) and (2,-7) in the x,y system.\n\nEquation of asymptote lines = \n\nY = (X)(1/sqrt3) and Y = -X(1/sqrt3)\n\nso they will be (y+3) = (x-2)(1/sqrt3) \n\nor (y+3) = (2-x)(1/sqrt3)\n\nThe other things can easily be calculated.
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Alright, don't listen to the above post...\n\nGrade 12 here, answering questions all day for mere fun.\n\n12)\ny - 3 = -1/3 (x + 4)\n\nEXPLAINED\n\nadd 3 to both sides\ny = -1/3 (x + 4) + 3\n\nexpand 1/3(x + 4) \n= x * 1/3 + 4 * -1/3\n= x/3 + 4/3\ny = -1/3x - 4/3 + 3\n\nTo add -4/3 and 3, use 3 = 9/3\ny = -1/3x - 4/3 + 9/3\ny = -1/3x + 5/3\n\nTrue, my bad. I originally added 4/3 instread of -4/3\n\n\n16)\n(-2, -1) (4, -4)\n\nm = Dy / Dx\nm = ((-1) - (-4)) / ((-2) - (4))\nm = 3 / -6\nm = -1/2\n\ny = mx + b\n(-1) = -1/2(-2) + b\n-1 = 1 + b\n-1 - 1 = b\n-2 = b\n\ny = -1/2x - 2\n\nCHECK\ny = -1/2(4) - 2\ny = -2 - 2\ny = -4\nTherefore it works.\n\n18)\n(0, 0) (2, 3)\n\nm = Dy / Dx\nm = ((3) - (0)) / ((2) - (0))\nm = 3/2\n\nb = 0\n\ny = mx + b\ny = (3/2)x + 0\ny = (3/2)x\n\nNote, the lower post copied my answers
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1. x-y=9---------------------(1)\nx+y=7-----------------------(2)\nadding\n2x=16\ndividing by 2\nx=8\nsub in 1,\ny=-1\n\n2. 5x-4y=12--------(1)\n3x-4y=4-------------(2)\n(1)-(2)\n2x=8\ndividing by 2\nx=4\nsub in 1 \ny=2\n\n3. 3x+3y=6-----------(1)\n2x-y=1-----------------(2)*3\n6x-3y=3---------------(3)\n(1)+(3)\n9x=9\ndividing by 9\nx=1\nsub in (1)\ny=1\n\n4. 3x-5y=13-----------(1)*3\n4x+3y=17--------------(2)*5\n9x-15y=39\n20x+15y=85\nadding\n29x=124\ndividing by 29\nx=124/29\nsub and find y\n\n5. x-y=1---------(1)\n3x-y=10---------(2)\n(2)-(1)\n2x=9\ndividing by 2\nx=9/2\nsuby=7/2\n\n6. 4x=2y-7\nthe other equation is missing\nx=(2y-7)/4\n\n7. y=-2x-----------(1)\n5x+3y=-3---------(2)\nsub y=-2x in (2)\n5x+3(-2x)=-3\n5x-6x=-3\n-x=-3\nx=3\nsub\ny=-6\n\n8. x+y=6---------(1)\n-2x+y=-3---------(2)\nsubtracting\n3x=9\ndividing by 3\nx=3\nsub\ny=3\n\n9. 2x+y=45-----------(1)\n3x-y=5-----------------(2)\nadding\n5x=50\ndividing by 5\nx=10\nsub\ny=25
330224
in order to answer this question, just plug in a number for x.  Forexample, let's use 3.  \nfirst, we substitute 3 for x\n(8*3)-4y=16\ny=(2*3)-4\nlets follow that though\n24-4y=16\ny=6-4------->  y=2\nnow, let's substitue for y with the answer that we just got\nand check it to see if its right.\n(8*3) - (4*2) = 16-------> 24 - 8 = 16\n2 = (2*3) - 4     ------->  2 = 6 - 4\n\ny=2, x=3
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1)x+y-z=0.........(eqn. 1)\n2x-y+z=3...........(eqn. 2)\n4x+2y-2z=-2       (eqn.3)\nMultiplying eqn.1 by by 2 and subtracting it from eqn.2,we get\n-3y+3z=3...........(eqn. 4)\nMultiplying eqn.2 by 2 and subtracting it from eqn.3,we get\n4y-3z=-4...........(eqn 5)\nAdding eqn, 4 and 5 we get,y=7\nPutting the value of y in eqn 4 .we get\n-21+3z=3 or,3z=24  or z=8\nputting the value of y and z in eqn.1,we get\nx+7-8=0 \n=>x-1=0 or x=1\nTherefore,x=1.y=7 and z=8  ans\n2)x+2y=9....(eqn 1)\n 2x-3y=4     (eqn.2)\nmutiplying eqn.1 by 2 and subtracting it from eqn 2,we get\n-7y=-14 or y=2\nPutting the value of y in eqn 1,we get.\nx+4=9 or x=5\nTherefore x=5 and y=2 ans.
606389
1) (3xy-12x)-2y+8\ncuz 3x times y = 3xy, 3x times -4 = -12x\n-2 times y = -2y, -2 times -4 = 8 (-times- = +)\n2) second u leave as it is.....\n3) ax-3x + cx-3c\ncuz atimes x = ax, a times -3 = -3x \ncuz c times x = cx and c times -3 = -3c\nu should do ur hw on ur own next time though
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One could buy eight horses and chariot sets and then sell them as suggested in the multiple choice to see how it works or one could set up two sets of equations and solve:\n\n-10%*(Horse Price) + 20%*(Chariot Price) = 0 (no profit)\n 10%*(Horse Price) + -5%*(Chariot Price) = 15 (profit)\n\nCombine the equations to get:\n                     -15%*(Chariot Price) = 15\n\nSolve for Chariot Price... 100\n\nPlug back into either equation to get Horse Price... 200\n\nThe answer is c