My sis doesnt understand System of Equations (dont care if i spell it wrong)?
369017
Set one equation equal to one of the variables, then insert it into the other equation and solve for the second variable.  then insert that variable into the original equation.\n\nx+y=5 or x=5-y\nx+2y=6\n\nput x=5-y into the second equation\n\n5-y+2y=6 \n\nnow solve for y\n\ny=1\n\nput this back into one of the original equations\n\nx+1=5\n\nx=4
754643
im taking Trig. try finding a homework helper web page. or go to the book that ur learning from web page because the majority of the times they have practice help on there.
1020883
i am sure you can find a math tutor that will be happy to help you when you explain what the problem is.\n\nso what IS the problem, anyway?
886265
gotta put the equation into y=mx+b first\ny= 1/3x - 1\n\nperpendicular would be y = -3x (+ or - any #)
428621
I'm not sure what level you are at, so my apologies if this is too simple or too complex.\n\nFirst, simplify the equation so that the dependant variable (usually 'y') is on the left of the equals sign, and the rest (usually 'x' and a bunch of numbers, but sometimes there are more variables such as 'z' and so on) is on the right of the equals sign.\n\nLets use 2y + 6x = 10x + 12 as an example. This simplifies to 2y = 10x - 6x + 12 and then 2y = 4x + 12 and then y = 2x + 6. Don't worry if you can't work out which is the dependant, and which is the independant variables, the graph will still be right if you turn the page through 90 deg.\n\nSecond, set up a small table with x on the first row followed by the range of 'x' that you want to graph. If you're not sure use -2, -1, 0, 1, 2, etc. through to 10.\n\nOn the second row write the first part of the equation. Using the example above this would be '2x'. Now fill in the values of '2x' under the values of 'x' in the first row (e.g. for '2x' these would be  2 times the values above, i.e. -4, -2, 0, 2, 4, etc. through to 20).\n\nOn the third row write the next step in the equation (e.g. 2x + 6) and fill in the values (i.e. 2, 4, 6, 8, 10, etc. through to 26).\n\nThe finished table should look like this:\n\ny = 2x + 6\nx........-2   -1   0   1   2   3   4   5   6   7   8   9   10\n2x......-4   -2   0   2   4   6   8   10 12 14 16 18 20\n2x+6...2     4   6   8   10 12 14 16 18 20 22 24 26\n\nObviously, the 2x + 6 row gives the 'y' values for the graph, because in this example, y = 2x + 6. Also, if your mental arithmetic is good enough, you can miss out the middle row.\n\nNow draw your graph using graph paper or grided paper if possible (if not you'll have to draw very carefully to get a square grid). On the horizontal axis (i.e. the thick black line across the page) put an arrow at the right hand end and mark it 'x', this is the 'x' axis. On the vertical axis (i.e. the thick black line up the page) put an arrow at the top and mark it 'y', this is, you guessed it, the 'y' axis. (If your problem uses 'a' and 'b' or 'p' and 'q' use those letters instead). Mark '0' where the two lines cross.\n\nDepending on the paper that you are using mark -2 and -1 on the x-axis to the left of the '0' and 1,2, 3, up to 10 on the right of the '0'. If you are not using gridded paper be careful to measure equal spaces (5 mm or 1/4 of an inch would be OK).\n\nLook at the table above and you can see that you need to mark off from 1 to 26 on the y-axis (working up the page from the '0'). Use the same space between numbers as on the x-axis.\n\nNow use your ruler to mark a small cross where the vertical line through -2 on the x-axis crosses the horizontal line through 2 on the y-axis (i.e the top and bottom values in the first column of the little table above). Put a little cross where the -1 and 4 lines cross (see the second column in the data table) and so on up to 10 on the x-axis and 26 on the y-axis.\n\nNow use your ruler to draw a line connecting the little crosses, and put little arrows on each end (to indicate that the line is infinite in both directions), and write y = 2x + 6 on the line. If the problem that your teacher has set is a curve then just join the crosses with a freehand curve as best you can (turning the page so that your elbow is where the centre of the circle would be is a good trick)\n\nJob done! Hope the example didn't confuse you and the end result works out OK.
15149
A job and a boyfriend?  Seriously, you sound like her little brother, and none too pleased about it.  Unless you can convince the parental units that having Sis remain there is a danger to your development, you may be stuck.  You could try talking to her to see why she's there.  Maybe she wants to leave but can't figure out how.  The worst she can do is blow you off.  There's always a chance she may ask for help or advice.  Stranger things have happened.
266405
The order of operations gets messed up.  You have to add x+y before you square it.\n\n PEMDAS \n  (You might remember this as "Please excuse my dear Aunt Sally.")\n\nOrder of Operations\n\nParentheses \nExponents \nMultiplication and Division \nAddition and Subtraction
838786
man i suck at math im stupid
817440
Hi sister!!!! I don't think they understood your question :(
107281
Okay, I see your two equations are 5x + 5y = -30 and \n2x - y = 15.  I'm not certain if these are right so I'll explain how to solve systems of equations in general.\nThe way we solve a system of equations by substitution is to solve one of the equations for one of the variables.  We then take what that variable is equal to and substitute it into the equation we did not yet touch.  We then solve the resulting equation.  \nThe explanation is confusing so I will provide an example.\n4x + y = 9\n-x + 3y = 1\nSo I'll solve the second equation for x like so:\n-x + 3y = 1\n-x = 1 - 3y\nx = 3y -1\nThen, I substitute 3y - 1 into the first equation for x, since this expression is equal to x.\n4x + y = 9\n4(3y-1) + y = 9\n12y - 4 + y = 9\nAdding four to both sides and combining like terms I get:\n13y = 13\nTherefore y = 1.\nWe can then solve for x using one of the original equations, we'll pick 4x + y = 9.\n4x + y = 9, and we know y=1 so:\n4x + 1 = 9\nSubtracting 1 and solving gives us\n4x = 8, and therefore x = 2.\nYou can use this same process to solve your problem.  I suggest starting by solving the second equation for y.\nHope this helps, please remember to come back and choose a best answer!  :-)
916972
When all the variables cancel out, then you have an infinite number of solutions, or no solutions.  Here, you have none.\n\nThis problem shows you that you need to check for linear independence.  That is, none of your equations should be linear combinations of any others.  What is a linear combination?  It's when you add equations which may or may not be multiplied by constants.  If you get something like this, then you can't solve your system, as you'll have fewer equations than unknowns.\n\nIn your case, 3[equation1]-[equation3] yields 0=-1.  This is false, so your system has no solution.\n\nIf there was a 3 instead of a 4, you would have an infinite number of solutions, where they satisfied the first and second equations.
961595
umm..are you missing an equals sign somewhere or am i really dumb? how do you figure out what x is unless you have both sides of the equation?
971617
y = sin(3x) + cos(4x)\ny = sin(2x + x) + cos(2x + 2x)\n\nsin(2x + x) = sin(2x)cos(x) + sin(x)cos(2x)\ncos(2x + 2x) = cos(2x)cos(2x) - sin(2x)sin(2x)\n\nsin(2x) = 2sinxcosx\ncos(2x) = cos(x)^2 - sin(x)^2 = 2cos(x)^2 - 1 = 1 - 2sin(x)^2\n\n(sin(2x)cos(x) + sin(x)cos(2x)) + (cos(2x)^2 - sin(2x)^2)\n\n(2sin(x)cos(x)cos(x) + sin(x)(2cos(x)^2 - 1)) + ((2cos(x)^2 - 1)^2 - (2sin(x)cos(x))^2)\n\n(2sin(x)cos(x)^2 + 2sin(x)cos(x)^2 - sin(x)) + (((2cos(x)^2 - 1)(2cos(x)^2 - 1)) - 4sin(x)^2cos(x)^2)\n\n(4sin(x)cos(x)^2 - sin(x)) + ((4cos(x)^4 - 2cos(x)^2 - 2cos(x)^2 + 1) - 4sin(x)^2cos(x)^2)\n\n4sin(x)cos(x)^2 - sin(x) + (4cos(x)^4 - 4cos(x)^2 + 1 - 4sin(x)^2cos(x)^2)\n\n4sin(x)cos(x)^2 - sin(x) + 4cos(x)^4 - 4cos(x)^2 + 1 - 4sin(x)^2cos(x)^2\n\ncos(x)^2 = 1 - sin(x)^2\n\n4sin(x)(1 - sin(x)^2) - sin(x) + 4(1 - sin(x)^2)^2 - 4(1 - sin(x)^2) + 1 - 4sin(x)^2(1 - sin(x)^2)\n\n4sin(x) - 4sin(x)^3 - sin(x) + 4((1 - sin(x)^2)(1 - sin(x)^2)) - 4 + 4sin(x)^2 + 1 - 4sin(x)^2 + 4sin(x)^4\n\n4sin(x) - 4sin(x)^3 - sin(x) + 4(1 - sin(x)^2 - sin(x)^2 + sin(x)^4) - 4 + 4sin(x)^2 + 1 - 4sin(x)^2 + 4sin(x)^4\n\n4sin(x) - 4sin(x)^3 - sin(x) + 4(1 - 2sin(x)^2 + sin(x)^4) - 4 + 4sin(x)^2 + 1 - 4sin(x)^2 + 4sin(x)^4\n\n4sin(x) - 4sin(x)^3 - sin(x) + 4 - 8sin(x)^2 + 4sin(x)^4 - 4 + 4sin(x)^2 + 1 - 4sin(x)^2 + 4sin(x)^4\n\n(4 + 4)sin(x)^4 - 4sin(x)^3 + (-8 + 4 - 4)sin(x)^2 + (4 - 1)sin(x) + (4 - 4 + 1)\n\nsin(3x) + cos(4x) = 8sin(x)^4 - 4sin(x)^3 - 8sin(x)^2 + 3sin(x) + 1\n\nfor a graph, go to http://www.calculator.com/calcs/GCalc.html\n\njust type in sin(3x) + cos(4x), click enter
417188
y=Asin(2x)+Bcos(2x)+(x/4)\n\nNow you get to determine why.
408462
to solve such a system in general, what you do is this: take the first equation, solve it for x treating y as you would a constant. In your case, this is x = 4 + y.\n\nThen, take your expression for x and substitute it everywhere you see x in the second equation; in your case you get 3 *(4+y) + 2y = 7.\n\nSolve this now 1-variable equation for y, once you find y (in this case, -1).\n\nSubstitute the value of y into either equation, and you get the value of x (3 in this case).
195201
i think u typed the question wrong, can u please correct it and if u have 2 unknowns, please give 2 equations, else u cant get a definite solution
795656
The first thing you should remember is the phrase PEMDAS.  This is shorthand for parenthesis, exponents, multiplication/division, and addition/subtraction.  This is the order of operations in math.  You should solve all equations in this order.  So for the first one, you should first distribute (multiply) -7 with (3-3x).  After that combine like terms, i got 25x -30; you can also factor that to be 5(5x-6).  The same steps apply for the rest of the problems.  I hope it helps.