How the equations works?? An easy way to explain it.?
955298
The KEY to solving any algebra problem, as I tell my students, is to get your unknown variable all by itself on one side of the equal sign.\n\nYou do this by "undoing" everyingthing that's being done to it right now.  And you "undo" the easy stuff first, like addition and subtraction, and then the more complicated stuff like multiplication and division.\n\nSo if you're adding 3 to X, then you would subtract 3.  And remember, what you do to one side of the equal sign, you MUST do to the other, so you would need to subtract 3 from both sides in my example.
114500
the way u phrased that doesnt necessarily make sense. x= 2n+1 is 3   ....something along those lines. your wording confused me...my name is my messenegr msg me there and i can try and help
647266
(Ok, by the way, this is not a linear equation. A linear equation is when the equation includes "x" and "y")\n\nBefore you can get the answer, you have to obviously solve it. \nSo for this equation, you want to get the variable alone; to one side. Now looking at the equation, everything isn't simplified. The left side is, you can not do anything with "x" and "-15"...Though, on the right side, there are like terms (expressions/terms that can be simplified)..."x" and "2x" become 3x. See 1x (1x is same as x) plus 2x equals 3x. So now your equation is x-15=3x+1 Now that everything is simplifed, you can solve it. As I said, get the variable to one side. Usually, the variable is on the left side of the equation, but it doesn't really matter which side its on...So to get x alone from the left side, you have to do what to get rid of NEGATIVE 15?...Do the opposite, POSTIVE 15, which then cancel each other out and x will be left alone. However when you add 15 to the left side, you have to add 15 to the right side. So adding 15 to the right side, 3x+1+15= 3x+16...Now the equation should be x=3x+16. So now, you have to get the 3x from the right side to the left side. How do you get sixteen alone? You minus 3x from both sides. The right side will result with having just 16 as the other side will have -2x. The equation is ow -2x=16. Again, get x alone. And how to do that is divide -2 to both sides. So -2x divided by -2 is 1x or just x and 16 divided by -2 is -8. At the end you should have x=-8\n\nHere's the work:\nx-15=x+2x+1\nx-15=3x+1\nx-15+15=3x+1+15\nx=3x+16\nx=3x+16-3x\nx-3x=16\n-2x=16\n-2x/-2=16/-2\nx=-8\n\nAnd the "check" works too(when you plug in -8 as x)\n\nx-15=x+2x+1\n-8-15=-8+2(-8)+1\n-23=-8-16+1\n-23=-23
58450
use the distributive property (poor u, u seem to be having a lot of trouble with your math homework tonight!! hope this helps:)\n\nthe property states that a(b+c)=ab+ac, basically you take the "a" and multiply it by the other 2, or distribute it.  say a=2, b=3, and c=1.  your equation would be 2(3+1) and, using the distributive property, make it become 6+2 which equals 8.  so its the same when your using x.  3(-4x+7)= -12x+21!!! yay!! hope this helped!
180174
Confirm, if you insert the values for x and y the equations work out with the solutions.
576570
algebra...usually (well i do) set up a equation for everything, and solve for the variable. If u need help u can always get help in www.algebra.com .....
380004
I don't understand.  Can you explain the problem a little better?
589722
Step 1: Solve one equation for one variable in terms of the other.\nStep 2: Plug that into the other equation and solve for the second variable.\nStep 3: Plug the second variable into the first equation and solve for the first variable.
663777
SOrry, I can't help you. If I had the formulas I could do it, but there is no way I remember that off the top of my head! Good luck
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
891393
are they all related? or are these just different equations??????\nnot clear at all..................................................\n \n\n3x-q=p: 3x=p+q, x=(p+q)/3\nx-2=r: x=2+r\ny-a=7: y= 7+a\nx-c=d: x= c+d\n4x-5c=3c: 4x= 8c, x= 2c\nbx=9b^2: x=9b
620528
There are two equations there, not one.\n\nFor the first equation, plug the numbers on your X-axis into "x", calculate "y", then draw the points on your graph.  I'll get you started:\n\nx=0, y = -3\nx=1, y = -1\nx=2, y = 1\netc.\n\nThe second equation simply means that x = 1.5 and y = 3.  This is just a single dot on a graph.
639859
its simple..see\n\n2x + 3y = 1 -------- 1st expression\n5x + 3y = 16 ------- 2nd expression\n\nnow..first take 1st expression\n\nStep 1: 2x + 3y = 1\n        2x = 1 - 3y\n        x = (1 - 3y) / 2  -------- this is value of x\n\nnow..take 2nd expression\n\nStep 2: 5x + 3y = 16\n        5(1 - 3y) / 2 + 3y = 16  ---here put the value of x\n        2.5(1 - 3y) + 3y = 16\n        2.5 - 7.5y + 3y = 16\n        2.5 - 16 = 7.5y - 3y\n        -13.5 = 4.5y\n        y = -13.5 / 4.5\n        y = -3 ------ this is value of y\n\nNow again put the value of y in the 1st expression\n\nStep 3: 2x + 3y = 1\n        2x + 3(-3) = 1\n        2x - 9 = 1\n        2x = 1 + 9\n         2x = 10\n         x = 10 / 2\n         x = 5\n\nSo, final value of x = 5 and value of y = -3 \n\nHope ya understand ;-)
833452
INSTEAD of giving you the answer directly -- here are the steps to solve and simplify.\n\nWRITE down the equations on separate lines -- leave about 4 lines between each problem.\n\nNOW -- draw connecting lines between the LIKE TERMS (for example -- in the first problem (4x+6-2x+6) the like terms are those that have the 'x' value with them and the other set of like terms are the constants with NO variables with them) -- I would use two different color pencils to do that)\n\nso add the 4x and 2x together (Answer is _x) -- then add the constants (those numbers without variables together (or +6+6)) together (Answer is __).\n\nNow put these two results together on the 3rd line.\n\n\nFinally, factor out any common terms in the results.\n\nTHAT is your final answer and the simplified expression.\n\n\n\n\nFOLLOW these steps once again for the 2nd problem -- but now instead of the variable being 'x' it is now 't'.\n\nNOW
786553
Manipulate your equation so y is on one side of the equal sign, and x variables and numbers are on the other side.  A proper answer would look like this:\n         y= 3x+4\ny should have a coefficient of one, for example 2y=....would be wrong.  To move something from one side of the equation to the other remember to add the opposite to both sides!  So if you originally had y-3x -4=0 you would do this:\n+3x +y -3x -4=+3x  the x's cancel out on the left side leaving you with y-4=3x....now you have to get rid of the -4 on the left side..so add +4 to each side of the equation....leaving you with y=3x+4, which is y in terms of x!!!
859187
I'll do one for u.\nIn the first,\nMultiplying both sides by x*,we get\nx+3x+2x*=7x*\n4x+2x*=7x*\nMoving 4x and 2x* to the R.H.S.\n5x*-4x+0=0\nThus 5x*-4x=0\nAttempt the others in the same manner.
498659
1) y=x+4\n   3x+y=16 (I guess you made a mistake by writing +16)\n--------------\nsubstitute x+4 for y in second equation and you'll get\n\n   3x+x+4=16\n    4x+4=16\n    4x=12\n     x=3\n\nnow y=x+4=3+4=7\n\nsolution (x,y)=(3,7)\n\n2) 4x+5y=48\n    y=3x+2\n--------------\nsubstitute 3x+2 for y in first eq.\n\n4x+5(3x+2)=48\n\n4x+15x+10=48\n\n19x=38\n\nx=2\n\nnow y=3x+2=3*2+2=6+2=8\n\nsolution (x,y)=\n(2,8)\n\n3)3x-y=0\ny=6\nin first equation instead of y write 6\n\n3x-6=0\n3x=6\nx=2\n\nso, the solution is (2,6)\n\n4) x-2y=-10\n3x-y=0\n\nfrom the first equation you have\nx=2y-10\nnow substitute 2y-10 for x in second one\n\n3(2y-10)-y=0\n6y-30-y=0\n\n5y=30\ny=6 \nnow x=2y-10=2*6-10=12-10=2\nso x=2 and y=6