HELP!!!! on the graph of the equation 5x-y=8, the point whose y-coordinate is 7 has an x-coordinates of???
404835
plug in y\n\n5x-7=8\n5x=15\nx=3
963370
In the cartesian coordinate system:\n1) the "x" coordianate is horizontal with positive numbers being on the right and negative numbers being to the left of the origin (0,0)\n2) the "y" coordinate is vertical with positive numbers being above and negative numbers being below the origin.\n\nExample:  the point (3,5) is 3 units left of zero and 5 units above zero.\n\nSince these are linear equation, you can just try some sample values for x or y (whichever is easier) and calculate the other.  With two or three points you should be able to graph the line for the equation given.\n\nFor the first equation  2y=5x-1 I got the following points:\n(1,2) (2,4.5) and ( 3,4)\n\nFor the second:  x=y+2 I got:\n(3,1) (4,2) and (5,3)\n\nSince these are linear equations you can plot the points and draw a line to connect them.  If you're supposed to find where they intersect, plot them on the same graph and extend the lines until they intersect. (You can calculate more points to help you do this.)  Then determine the point of intersection.  There are other ways to do this but it says to solve them graphically.\n\nHope this helps.
749030
y = m (x-x1) + y1\nm=slope (so m=2)\nx1 and y1 are the coordinates. Just plug your numbers in and expand the m(x-x1) part.
838763
if y=0\n\n2x+4(0)=-8\n2x+0=-8\nx=-4\n\n\nif x=0\n\n2(0)+4y=-8\n0+4y=-8\ny=-2\n\nIf your doing co-ordinate geometery, two points on the line are (0,-2) and (-4,0)
570575
In a diagram where x denotes the horizontal axis and y the vertical axis, linear equations with only one variable (x or y) is either a horizontal or vertical line.\n\nexample 1:\n\ny = 3 is a horizontal line that connects all points in the diagram which have the y-value 3.\n\nexample 2:\n\nx = 5 is a vertical line that connects all points in the diagram which have the x-value 5.\n\nI hope this is clear?
375327
Where x = 0, the line intercepts the y-axis:\n\n1. 5x-2y =10 \n2y = 5x - 10\ny = (5/2)x - 5 \ny-int = (0, -5)\n\n2. 2y+7x =14\n2y = -7x + 14\ny = (-7/2)x + 7\ny-int = (0, 7)\n\n3. x=3\nHas no y-intercept\n\n4. 1/2x-2y=8\n2y = (1/2)x - 8\ny = (1/4)x - 4\ny-int = (0, -4)
106013
y = -2x + 8\n\nsolve it yourself
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!
901793
x/(x-8) + 7 = 8/(x-8)\nMultiply everything by (x-8)\nx + 7(x-8) = 8\nDistribute the 7\nx + 7x - 56 = 8\nCombine x and 7x\n8x - 56 = 8\nAdd 56 to each side\n8x = 64\nDivide both sides by 8.\nx = 8\n\nNote that x-8 cannot equal zero, but we just found x to be 8.\nThat means x-8 = 8-8 = 0\nThus, we have no solution.
973040
Ok, since you didnt give any equation, i would assume that the graphs are linear in nature\n\nThen for find the points where the graph intersects the Axes,\n\n  Substitute x=0 in the equation to find the point where the point where the graph intersects the Y axis\n\n    Substitute y=0 in the equation to find the point where the point where the graph intersects the X axis\n\nEg.   2x + y = 4\n\nSubstituting y=0,\n\n                          2x=4\n                           x=2\n\nHence the the graph meets the x axis at (x,y) that is (2,0)\n\nSimilarly, substituting for x=0, we get..\n\n                         y=4\n\nTherefore the point where hte graph meets the x axis is (x,y), in this case (0.4)
918264
IF you mean like an ordered pair it is a point on a graph.  An example is:  (x,y)  so you can have (3,5)  or (-7,-88)  Otherwise I'm not sure what you're talking about.
318213
eh? you using simultaneous equations with a straight line equation?\n\ny = mx + c is a straight line equation, where m is the gradient & c is where the line cuts the y axis.\n\nThere for lets say y = -2x + 8:\ny = mx + c\nCuts x-axis Put y = 0\n-2x + 8 = 0\n-2x = -8\nx = -8/-2\nx = 4\n\nHope that was correct :| lol
117178
Well, your third question is incomplete..\n\nBut I can answer the first two:\n\nIf you have an equation, assuming we're on cartesian coordinates, the simplest thing to do is to solve the Y variable to a couple or more X variables.\nLike:\nY=3x + 1\n\nx=0, Y= 3(0) + 1 = 0 + 1 = 1    (0,1)\nx=5, Y= 3(5) + 1 = 15 + 1 = 16  (5,16)\nand so on...\n\nThen you plot the coordinates on the graph and unite the resulting dots with a curve. In my example it should yeild a straight line.\n\nGraphs help describe trends on phenomenons that are tied to the equations.
939024
Draw yourself a cartesian graph, vertical line is the y axis, horizontal is the x axis.\n\npick points on the x axis like 1, 2, 3, etc.  plug thos values into the equation and solve for y. then pair the numbers, place points in the graph by taking the x value going straight up (or down if y is a negative number) and place the dot accross from the y value.  do this for each set of points then connect the dots with a line or a curve. for instance in your equationstart with x=1\n\ny= 7(1-5)(1-2)^4(1+.5)(1+2.5)^3\ny= 7(-4)(-1)^4(1.5)(3.5)^3\ny=7(-4)(1)(1.5)(42.875)\ny=-1800.75\n\nso in XY format the first point would go at (1, -1800.75)
735905
a). If it touches the x-axis, then y is zero.  \nSo x^2 + 0 + 0 = 7\n\nx^2 = 7\n\nx = sqrt 7 or - sqrt 7  \n\nThe points are (sqrt 7, 0) and (-sqrt 7, 0)\n\nb).  Differentiate implicitly like this:\n\nd/dx x^2 + d/dx xy + d/dx y^2 = d/dx 7\n\n2x + (x dy/dx + y) + 2y dy/dx = 0\n\nx dy/dx + 2y dy/dx = -2x - y\n\ndy/dx (x+2y) = -2x - y\n\ndy/dx = (-2x-y)/(x+2y)\n\nc,d).  Just plug in the coordinates from (1)
502159
?\n\n(x-coordinate, y-coordinate)\n(0,0)\n(1,3)\n(2,6)\n(3,9)\n\nCos y-coordinate is 3 times of x-coordinate?
859504
(x-7)(x-5)\nso x= +7, or +5