find the point of intersection of the two lines. Clearly show an algebraic solution.    5x-2y=-36  4x+3y=-15?
95633
The point of intersection occurs when the two equations are put equal to each other:\n\nNow put the y on one side of the equation for each equation:\n2y=5x+36...1\nand for the 2nd eq.\n3y=-15 -4x\n\nNow make the coordinate of y =1\nThat is:\ny = (5/2)x+ 18....1\n\nand\ny=-5-(4/3)x\n\nNow equalize both equations: Where y (eq 1)=y (eq 2)\nTherefore:\n-5-(4/3)x = (5/2)x+ 18\n\nNow since the equation above has only one variable which is x, solve for x!\n\nThat's all, I bet u can continue here, im already tired!
695994
The eqaution of the line is\ny-y1 =m(x-x1)\ny-0 = -3[x-(-4)]\ny-0 = -3(x+4)\ny= -3x-12
753426
X2   5x - 4y = 1\n10x -8y =2\n-10x + 8y = -3\n-------------------------\n  0=-1                NO SOLUTION
918448
-3x+15y=-30\n\nWe'll divide everything by 3 to make it easier\n\n-x+5y=-10\n\nWe have x - 5y = 10\n\nIf we multiply that equation by -1, we get\n-x+5y=-10!\n\nThis is the same as the simplified -3x+15y=-30, which means that any x and its corresponding y value fits both equations.\n\nIf they are the same equation, then all values of x (and the necessary values of y) will work; the solution is all numbers x.\n\nHope this helps.
472975
Replace x with 4 and y with 5.  If the resulting statement is true, then it is a solution.  If it is not true, then it is not a solution.\n5x +3y = 35\n5(4) + 3(5) = 35\n20 + 15 = 35\n35 = 35\nThis is true, so (4,5) is a solution.\n(There is no need to solve for y.  You can solve for y, but this just adds additional work.)
187542
Answer:\ny=-1.25x + 3\n\nProof:\nWe are given two points to work with:\nPoint 1: (x1=0, y1=3)\nPoint 2: (x2=4, y2=-2)\n\nSo, use the "two point equation" to solve this problem.\ny-y1 = (y2-y1)/(x2-x1) * (x-x1)\n\ny-3 = (-2-3)/(4-0) * (x-0)\ny-3 = -5/4 * x\ny = -5/4 * x + 3\n\ny=-1.25x + 3
943301
Use the standard equation for a line going through two points.\n\nI.e. for two points (x1,y1), (x2,y2) the equation is : \n   (y2-y1)/(x2-x1) = (y-y1)/(x-x1).\n\nIn this case we have (x1,y1)=(0,-5) and (x2,y2)=(8,9), so\n   (9-(-5))/(8-0) = (y-(-5))/(x-0)\nwhich computes to\n   14/8 = (y+5)/x\nor, in standard format\n   y = (7/4)*x - 5\n\nIf you now check (fill in those two points) you will see it is the correct equation for both.\n\nHope this helps :-)
819556
Hey, here's the setup for these types of problems.\nYou want the equation in the form of y = mx + b\nsince we know that x = 2, y = -7, and m = -1/5, we can plug these values into that equation to find the equation of the line that passes through the desired point. SO:\n-7 = (-1/5)(2) + b\n-7 = (-2/5) + b\nb = (-7)+(2/5)\nb = -33/5\nso the equation for the line is:\ny = (-1/5)x - (33/5)\n\nhope that helps!!
1162316
>< you really should read the answers for your other questions and learn... hope you know this afterwards!\n\n\n6x-y=-4\n2x+2y=15\n\nuse second equation to find out what y is\n\nx+y=7.5 (divide everything by 2)\ny=7.5-x\n\nsubstitue 7.5-x as y in the first equation, find x\n\n6x-(7.5-x)=-4\n6x-7.5+x=-4\n7x=3.5\nx=0.5\n\nput in value of x in an equation and find y\n\n6(0.5)-y=-4\n3-y=-4\n-y=-7\ny=7\n\ncheck answers (use equation that was not used to find y):\n2(0.5)+2(7)=15 :)\n\nx=0.5, y=7\n\n\nx+y=1\n2x-y=-2\n\nuse first equation to find y (in relation to x) --- can find x too, but you have to plug in answer into second equation, and y is easier to deal with in 2nd equation...\n\ny=1-x\n\nput 1-x as value of y in the second equation\n2x-(1-x)=-2\n2x-1+x=-2\n3x=-1\nx=-1/3\n\nput in value of x into an equation to find y\n(-1/3)+y=1\ny=4/3\n\ncheck answers:\n2(-1/3)-(4/3)=-2 :)\n\nThe answers are x=-1/3, y=4/3\n\n\n\nthird problem:\n5x-3y=-11\nx-2y=2\n\nuse second equation to find out value of x in relation to y\nx=2+2y (add 2y to both sides to isolate x)\n\nput in that value as x for the first equation:\n5(2+2y)-3y=-11\n10+10y-3y=-11\n7y=-21\ny=-3\n\nsubstitute value of y in x-2y=2 to find x\nx-2(-3)2\nx+6=2\nx=-4\n\ncheck answers\n5(-4)-3(-3)=-11 :) answers match up\n\nhope that helped you learn how to solve this type of algebra questions!
90547
this must be solved graphically\ny >=-3x+2 , draw this line y=-3x+2 and have the area above it\n\ny<2x+2 , draw y=2x+2 and get the area below the line\n\nThe common region is this 2 areas is the answer
913768
y = ax² + bx + c\n\nWhen x = 2 y = 30 so 4a + 2b + c = 30 ...... (1)\nWhen x = 3 y = 15 so 9a + 3b + c = 15 ...... (2)\nWhen x = 4 y = 6 so 16a + 4b + c = 6 ........ (3)\n\n(2) - (1) 5a + b = -15 ....... (4)\n(3) - (2) 7a + b = -9 ......... (5)\n(5) - (4) 2a = 6\na = 3 so b = - 30 and c = 78\n\nie y = 3x² - 30x + 78
1169467
y = x^2 + 5x - 3\n2y=3x - 2\n----------------\n2y = 2x^2 + 10x - 6\n2y = 3x - 2\n-----------------------\n    0 = 2x^2 + 7x - 4\n=>2x^2 + 7x - 4 =0\n=>2x^2 + 8x - x - 4 = 0\n=>2x(x + 4) -1(x+4) = 0\n=> (2x - 1)(x + 4) = 0\ntherefore 2x - 1 = 0 => x = 1/2 or \nx + 4 = 0 or x = -4\nx = - 4 or 1/2  \ngot it?
293210
the graph of |x - 2| + |y - 5| = 5 consists of four lines which are graphs of the following functions:\nx-2+y-5=5 which is  line A: x+y=12\n-x+2-y+5=5 which is line B: x+y=2\n-x+2+y-5=5 which is line C: x-y=-8\nx-2-y+5=5 which is line D: x-y=2\nnext thing to do is graph these four lines. the easiest way is to find the 0s. \nline A: x+y=12\n        x=0 y=12 point A1(0,12)\n        y=12 x=0 point A2(12,0)\nline B:  x+y=2\n         x=0 y=2 point B1(0,2)\n         x=2 y=0  point B2(2,0)\nline C: x-y=-8\n         x=0 y=8 point C1(0,8)\n         x=-8 y=0 point C2(-8,0)\nline D: x-y=2\n        x=0 y=-2  point D1(0,-2)\n        x=2 y=0   point D2(2,0)\nthese four lines enclose a figure that is a rectangle. lets find the length of each side. after you plot the four lines you will notice that line x+y=12 intersects with line x-y=2 lets call the intersection point I1 to find I1 coordinates we rearrange y=12-x and y=x-2 and from y=y we get x-2=12-x or 2x=14 or x=7 if x=7 y=12-7=7-2=5 the point I1 is I1(7,5).\nline x+y=12 also intersects with line x-y=-8. Lets call this intersection point I2. to find coordinates we have y=12-x and y=x+8 y=y or 12-x=x+8 or 2x=4 or x=2 for x=2 y=12-2=2+8=10 so the point is I2(2,10) \nside I1I2 of the rectangle length is side a. to find length of a we do\na^2=(7-2)^2+(10-5)^2=25+25=50\na=sqrt(50)\nwe need the second side, lets call it b.\nto find length of side b we need at least one more point. lets find intersection of x-y=2 and x+y=2 and call it point I3. y=x-2 and y=2-x or x-2=2-x or 2x=4 or x=2. for x=2 y=2-2=2-2=0 point I3(2,0)\nside b:\nb^2=(7-2)^2+(5-0)^2=50\nb=sqrt(50)\nit means a=b or our rectangle is a square!\nso the area of the square is Area=side^2 since our square&#92;s side is a=b=sqrt(50), the area is 50.\nI hope this helps, I really don't have time to explain it in more details.
869796
OK, first you have to find out the point(s) where both curves intersect. To do this, make both equations equal:\n\ny = x*sin(x) = y = cos(x)\n\nx*sin(x) = cos(x)\n\nx = cot(x)\n\nThe above equation gives us the locus of points where the two curves intersect. Say one point of the locus is:\n\nx1 = cot(x1)\n\n\nNow, to prove that both curves intersect at right angles to each other, we have to calculate the product of their slopes at the point of intersection & make sure they equal "-1". For our case:\n\nSlope of curve1 = dy/dx = d{x*sin(x)}/dx\n= x*cos(x) + sin(x)\n\nSlope of curve2 = d{cos(x)}/dx = - sin(x)\n\nProduct of their slopes at point x1:\n\n= {x1*cos(x1) + sin(x1)}*{-sin(x1)}\n= - { x1*cos(x1)*sin(x1) + sin^2(x1) }\n\nBut we already know that x1 = cot(x1), the point of intersection - plugging it into the above equation, we get:\n\n= - { cot(x1)*cos(x1)*sin(x1) + sin^2(x1) }\n= - { cos^2(x1) + sin^2(x1) }\n= -1\n\nThus, we get the product of the slopes equal to "-1" Thus, they intersect at right angles to each other.
1025833
y = -x^2 + 4x + 5\n\nx = (-b/2a)\nx = (-4)/(2(-1))\nx = (-4)/(-2)\nx = 2\n\ny = -(2)^2 + 4(2) + 5\ny = -4 + 8 + 5\ny = 4 + 5\ny = 9\n\nVertex : (2,9)\nAxis of symmetry is x = 2
1152335
Solve,\n\n3x = 2y - 5 ... (1)\n\ny - 5x = -1 ... (2)\n\nRearrange (2), we get,\n\ny = 5x - 1 ... (3)\n\nSubstitute (3) into (1)\n\n3x = 2(5x - 1) - 5\n\n3x = 10x - 2 - 5\n\n3x = 10x - 7\n\n10x - 3x = 7\n\n7x = 7\n\nx = 1 ... (4)\n\nSubstitute (4) into (3)\n\ny = 5(1) - 1\n\ny = 4 \n\nTherefore x = 1 and y = 4.
701107
4x +y = 16   ;2x + 3y = 24      Multiply the 2nd with -2, that gives -4x -6y = -48    Add the two:   0x - 5y = -32  \nDivision by -5 gives  y = 6,4\nInsert this value into the first equation:   4x + 6,4 = 16\n4x = 9,6 ;    x = 2,4\n\nNumber 2 is even easier:\nMultiply the first equation by 2 and add the equations.\n24x + 2y = 50    plus   8x - 2y = 14  gives  32x = 64; x = 2\nYou can now choose:  24 + y = 25  or  16 - 2y = 14\n(That you get when you replace in the original equations \nx by 2)  Anyhow you'll find   y = 1