solve sytms of equatns by graphing see if both yield line. x+y=8 & x-y=-4?
921939
Good Evening\n\nI think we've been avoiding this question because we can't graph here... however to get you started, create an x|y chart for each line, plug in a couple of values, and connect the dots  (for a line you only need two points, but some books recommend finding three, so you can verify you didn't make a mistake somewhere)\n\nx+y = 8\nx|y\n----\n4|4\n3|5\n2|6\n\nx - y = -4\nx|y\n0|4\n1|5\n2|6\n\nGraph the two lines, and they should intersect at (2,6) since it appears in both of the charts...\n\n~i~
163088
x = y + 7 y= 8 + 2x\ny=8+2(y+7)\ny=8+2y+14\n-y=22\ny=-22\nx=-22+7=-15
679397
y = 0.5x-4 \n\nor similarly\n\nx= 2y +8
411757
since y=+-(x)^(3/2) Range [0,oo[\nnP=(n,n*n^1/2) and x_n/y_n= [(n^1/2^)/n]=n^-1/2
60538
if [ -7x + x^2 =0 ]\nthe you will get y = 7/x
939650
x+y=-3\nx=-y-3\nx -3  -2  -1  0  1  2  3 \ny  0  -1  -2 -3-4-5 -6\n\nx^2+2y^2-12y-18=0\nwhen y=-1\nx^2+2+12-18=0\nx^2-4=0\nx^2=4\nx=+/-2\nso (-2,-1),(2,-1)\nwhen y=3\nx^2+18-36-18=0\nx^2-36=0\nx=+/-6\nso(-6,3),(6,3)\n\npoints of intersection\n(-2,-1) and (-6,3)
735754
just plug the numbers in.  if y=8, and x=2, \nits 8=a(2^2)\nso 8=a(4)\nso a=2.\nhope that helps!
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 :-)
770253
You did everything correct.  If you eliminate the x, you also eliminate the y.  The resulting equation is not ?=-9, it is 0=-9.  Since this is an untrue statement, there is no answer to the problem.  Both equations represent lines that are parallel to each other and will never touch.
417188
y=Asin(2x)+Bcos(2x)+(x/4)\n\nNow you get to determine why.
767931
4x+2y=8\n\nSubtract the 4x to the other side to get\n\n2y= -4x + 8\n\nDivide the 2 on all sides to get:\n\n2y(/2)= -4x(/2) + 8 ( /2)\n\nYour final answer will be:\n\ny= -2x + 4\n\n4= y intercept and -2 equals the slope.
1055471
You need to graph this function. \nIt is a line.\ny = (x^2)^ 1/2 - 16 \ny = x - 16\nx = 0, y = -16, \nx = 16, y = 0
676814
First equation: sqrt[x] + y = a\nor sqrt[x] = a - y\nSquaring both sides\nx = (a - y)^2__________(1)\n\nSecond equation: x + sqrt[y] = b\nor sqrt[y] = b - x\nSquaring both sides\ny = (b - x)^2___________(2)\n\nSubstituting 'x' from eqn. (1) in eqn. (2)\ny = [b - (a - y)^2]^2\n  = [b - a^2 + 2*a*y - y^2]^2\n\nSo it is clear that you have to solve a 4-th order equation.\nIt is really tough.
725212
Yeah, make them graphable on your calc.\n\ny = 4-x\ny = 2x-2\n\nIf you want to solve, just set 4-x and 2x-2 equal to each other (since y must also be equal in both equations)\n\n4-x=2x-2\n3x=6\nx=2\n\n\n2+y=4\ny=2\n\nTheir intersection is (2,2)\n\nThe y-intercept of the first is 4, and the second is 2.\nThe slope for the first is -1, and the second is 2.\nThe x-intercept of the first is 4, and the second is 1.
704540
Add the two equations together: (2x+y)+(x-y) = 1+8, so 3x = 9, x = 3.  Then 2*3+y = 1 so y = -5.\n\nSteve
810768
Easiest way is to write them as two equations equal to zero.  Then equate them...\n\nStarting with your first example:\nx² + y² - 4 = 0\nx² + (y-6)² - 25 = 0\n\nNow equate these:\nx² + y² - 4 = x² + (y-6)² - 25\n\nExpand out the (y-6)²\nx² + y² - 4 = x² + (y²-12y+36) - 25\n\nNow cancel like terms (like x², y², etc.) which are on both sides:\n-4 = -12y + 36 - 25\n\nNow simplify:\n12y = 36 - 25 + 4\n12y = 15\ny = 15/12\ny = 5/4\n\nNow use the value you found (y=5/4) and solve for the other variable using one of the original equations:\nx² + (5/4)² = 4\nx² + (25/16) = 4\nx² = 64/16 - 25/16\nx² = 39/16\nHere you need to take the sqrt and you'll have two answers:\nx = sqrt(39/16)\nx = +/- sqrt(39)/4\n\nSo your two points of intersection for the first pair of equations are:\n(sqrt(39)/4, 5/4) and (-sqrt(39)/4, 5/4)\n\nThe same method works for the other problems...
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