What situation could i use to model the equation y=10 -  x?
98833
ryanray didn't pay attn: it's more like this: John can only carry 10 bottles of beer or wine at a time. if he carries x bottles of wine, how many bottles of beer can he carry (10-x)?
391343
i think 2.\n\nRearrange the eqution to give x= (10-2y)/5\n\nX intercept occurs when y=0\n\nput 0 in place of y and solve the equation.\n\n(10-0)/5\n\ntherfore x = 2
229858
if you had an inverse variation, then x*y would always be the same number.\n\nif you had a direct variation, then x/y (or y/x) would always be the same number.\n\nonce you know which variation you have, then use the equations y=kx (for direct) and y=k/x (for inverse) to find the relation
785524
y = 3x-5\n\nx^2 +(3x-5)^2 = 25\nx^2 + 9x^2 -30x + 25 = 25\n10x^2 -30x = 0\n10x(x-3) = 0\n\nso x = 0 or x = 3\n\nWhen x = 0 y = -5\n\nWhen x = 3  y = 4
277095
For x=10 and y=5:\n\n10^2-5y^2 ?= (10+5)(10-5)\n100-25 ?= (10+5)(10-5)\n75 ?= (10+5)(10-5)\n75 ?= (15)(5)\n75 = 75\n\nFor x=10 only:\n\n10^2-y^2 ?= (10+y)(10-y)\n100 - y^2 ?= (10+y)(10-y)\n100 - y^2 ?= 100 + 10y - 10y - y^2\n100 - y^2 = 100 - y^2\n\nFor y=5 only:\n\nx^2-5^2 ?= (x+5)(x-5)\nx^2 - 25 ?= (x+5)(x-5)\nx^2 - 25 ?= x^2 + 5x - 5x - 25\nx^2 - 25 = x^2 - 25\n\nThe idea with such things is to make both sides look the same to prove (or if not, disprove) equality.
806897
First equation:\n4x + 2y = 10\n\nRearrange to get the y-term alone:\n2y = 10 - 4x\n\nDivide both sides by 2 to get it in slope-intercept form (y = mx + b):\ny = -2x + 5\n\nNow you can graph it by drawing a line through (0,5) as the y-intercept and have it go down at a slope of -2 through points (1,3), (2,1), (3,-1), etc.\n\nFor the second equation, draw a horizontal line for y = 4\n\nYou'll see they intersect at the point (½, 4).  I've attached a graph of the two lines...\n\nSo your solution is:\nx = 0.5 (or ½)\ny = 4
207006
you can generalize further --> y = -2x
242388
Example 1: If y varies inversely as x, and y = 6 when x = , write an equation describing this inverse variation.\n\nk = (6) = 8\nxy = 8 or y =  \nExample 2: If y varies inversely as x, and the constant of variation is k = , what is y when x = 10?\n\nxy = \n10y = \ny = × = × =  \nk is constant. Thus, given any two points (x1, y1) and (x2, y2) which satisfy the inverse variation, x1y1 = k and x2y2 = k. Consequently, x1y1 = x2y2 for any two points that satisfy the inverse variation. \nExample 3: If y varies inversely as x, and y = 10 when x = 6, then what is y when x = 15?\n\nx1y1 = x2y2\n6(10) = 15y\n60 = 15y\ny = 4\nThus, when x = 6, y = 4. \n\n\ndo u get it now
1009010
y= 40-5x\ny=40-5(6)\ny=10\n10
855878
you have to set up two equations....x+y=24 and .10x+.25y=3.00\n\nnow you can do to things...one way is to solve the first equation for x or y then    x=24-y.... sub this into x in the second equation    \n.10(24-y)+.25y=3.00\n2.4-.1y+.25y=3.00  combine terms\n2.4+.15y=3.00\nsubtract 2.4 from both sides\n.15y=.6 divide both sides by .15\ny=4  x=20
144204
-10\n\nand to prove it -10 x -10 x -10 = -1000
1082168
y=x^2+10x+8\ny=(x+5)^2 - 17\n\nVertex form: (5,-17)
801726
You solve one of the equations for one of the variables, and put the solution into the other equation to solve the other variable:\n\n3x+y=4\ny=4-3x  then go to the other equation, and put this in for y:\n\n-2x-10(4-3x)=-12\n-2x-40+30x=12\n28x=28\nx=1     Now drop this answer into one of the original equations:\n\n3(1)+y=4\ny=4-3=1     So both x & y =1
133024
y = 0.\n\nETA: That's if we're to assume that the value of x is constant between the equations.
896619
slope is 2,  y intercept is  - 10
756495
log(base x) y=[log(base 10)y]/[log(base10)x]. I hope you can take up from here!
807209
I can tell you just by looking at this that it won't work, but here's the process:\nMultiply the first equation by 2 so the x coefficient in the first equation is the opposite (negative) of the x coefficient in the second equation:\n2x + 10y = 20\nMove the y and the constant in the second equation to make it match the formet of the first\n-2x -10y = -20\nAdd the two equations to get:\n2x + 10y = 20\n+(-2x -10y = -20)\n----------------------\n0 = 0\n\nThere is no single solution, because the two equations are actually the same equation written differently.