Compare equations with one variable to equations with two variables.How are they alike?How are they different?
491343
Equations with one variable can be solved with just that equation for example 2x=4 can be simplified to x=2.\n\nAn equation with two variables cannot be solved purely by itself. For example 4x = 2y can only be simplied to 2x = y. You cannot find the values of x and y without a second equation. If you were given the following two equations 4x = 2y and x + y = 3 then you could use them to calculate x = 1 and y = 2.
851603
study + no study = no failed + failed\nstudy [ 1 + no ] = failed [ no + 1 ]\n\ndivide [ 1 + no ] from each side :\n\nstudy = failed\n\neyeonthescreen you have no sense of humor. However, if "study", "failed", and "no" are variables, then my equations do in fact work (assuming no≠-1)!  Also, it is "no failed" not "not failed"; therefore not Boolean either!
323124
== equal, != not equal, >=, greater than or equal, <= less than or equal, < less than, > greate than. Used for comparing to variables.\n\nexample:\nif (a == b)\n //they are equal\n\nif (a != b)\n //they are not equal\n\netc...
550918
Type "=B1-C1+D1*A1" without the quotes inside another cell such as E1.  This is assuming that your equation is the correct algebraic form and solved for one variable.  Your text is a bit jumbled so I didn't parse all of it.
869887
If you have 2 equations, just solve for one variable for one of the equations. Take that equation and plug it into the same variable for the other equation. You will solve for that variable, and plug in that number to any of the 2 equations to solve for the other variable.
580733
First equation:\n(x+a+b)^2 = x^2 + (x+a)^2 - (x+b)^2\nor x^2 + 2x(a+b) + (a+b)^2 = x^2 + x^2 + 2ax + a^2 - x^2 - 2bx - b^2\nor x^2 + 2ax + 2bx + a^2 + 2ab + b^2 - [x^2 + x^2 + 2ax + a^2 - x^2 - 2bx - b^2] = 0\nor 4bx + 2ab + 2b^2 = 0\nor 2b(2x+a+b) = 0\nor  2x + (a + b) = 0\nor 2x = -(a+b)\nTherefore x = -(a+b)/2\n\nSecond equation:\n(x+a)^2=(x+b)^2\nor (x+a)^2 - (x+b)^2 = 0\nor (x+a+x+b)(x+a-x-b) = 0\nor (2x+a+b)(a-b) = 0\nIf a = b then the equation becomes identity. So a is not equal to b.\nHence 2x + a + b = 0 \nor x = -(a+b)/2\n\nThus from both equations we have x = -(a+b)/2
895454
There is no ordered pair {a,b} which solves both equations\n\nif a+b = a-b , then 2b = 0 or b = 0 and a is anything\n\nBut the second equation is not defined for b=0, no matter what a is (specifically, a/b is not defined for b=0)
830984
You have to first eliminate one variable from two of the equations which would leave two equations with two variables.\nThen eliminate one of those two variables to get one value.  Plug this in to get a second value; plug these in to get the third.\n\nFor example in your problem b, if you multiply the second row by -1 and add it to the first row, that would eliminate z\n\n(2x - 1y = 1)\n\nThen divide the third row by 2 and add it to the (original) first row, which also eliminates z\n\n(20x - 4y = 16)\n\nNow take those 2 rows, eliminate either x or y, and start plugging in
724411
Sure...\n\nfirst we want to combine like terms on each side of the equals sign.  Think of the equals like the balance point on a scale... you can do anything on one side as long as you match it (keep it balanced) on the other!\n\nso we want to combine the variables and the constants...\n\n8a - 3a + 4 = -a -4 + 2\n5a + 4 = -a - 2\n\nnow we want all the variables on one side, and constants on the other:\n\n5a -(-a) + 4 - 4 = -a-(-a) - 2 - 4\n5a + a = -2 - 4\n6a = -6\n\nnow we just divide by 6 to get a alone:\n\n6a/6 = -6/6\na = -1\n\nHope this helps! :)
41587
If something is equal to other thing, then you can substitute that something with the other thing.\n\nExample:\n\nIf x = 2y\n\nthen in: x+ 3y = 5\n\nyou can substitute x by 2y\n\n2y + 3y = 5
886222
Solve for one variable and plug it into the second equation...\n\nSo, let's solve the first equation for y:\n-x + y = -4\ny = x - 4\n\nNow plug this into the second equation:\n4x - 3(x - 4) = 13\n\nDistribute through the parentheses:\n4x - 3x + 12 = 13\n\nSubtract 12 from both sides:\n4x - 3x = 1\n\nSimplify:\nx = 1\n\nNow plug this back into your original substitution:\ny = x - 4\ny = (1) - 4\ny = -3\n\nSo your answer is x = 1 and y = -3.\n\nYou should double-check your answer....\n-x + y = -4,\n-(1) + (-3) =? -4  \n-4 = -4 <-- check\n\n4x - 3y = 13\n4(1) - 3(-3) =? 13\n4 + 9 = 13 <-- check\n\nSo your answer is:\nx = 1\ny = -3
790758
Quit thinking x and y and start thinking independent variable and dependent variable, or independent variable and function result.\n\n\nAll you care about is that the independent variable is along the abcissa, and the dependent variable is along the ordinate.\n\nThat said, just pretend that the y in your equation is x for the TI and the x in your equation is the y for the TI.
408980
when playing with < and >\nif you multiply all the equation by -1 the sign < changes to >, and the sign > changes to <\nbut if you just switch the number on the other side the sign doesnt change\nin ur case :\n2x-y < 4 \nequals\n-y < -2x + 4\nif you multiply it with -1 it will become\ny > 2x - 4\n\nlike that. hope it helps :)
214090
I count 4 methods, although the last 2 are pretty similar.\n\n1. Graph both equations and see where they intersect.  This can be tricky if it's not on a clear point, though.\n\n2. Substitution - isolate one variable in one of the equations (i.e. solve for y in terms of x, or x in terms of y) and then substitue the answer everywhere it appears in the other equation.\n\nEx: 2y = 6 + 8x; 4y = 5 + x\nIt's easiest to isolate x in the second equation because the coefficient is already 1, so x = 4y - 5.  Now subsitute 4y - 5 for x in the first equation. \n\n2y = 6 + 8(4y - 5)\n2y = 6 + 32y - 40\n2y = 32y - 34\n-30y = -34\ny = 34/30, or 17/15\n\nThen substitute 17/15 back into the other equation, 4y = 5 + x\n\n4(17)/15 = 5 + x\nx = 68/15 - 5 or -7/15\n\nSo your solution is x = -7/15, y = 17/15\n\n3. Another choice is to manipulate one of the equations so that one of the variables cancels out.  There's 2 ways to do this\n2y = 6 + 8x\n4y = 5 + x\n\nMultiply the top equation by 2 -- 4y = 12 + 16x\nAnd since you have 4y = (something) on both equations, then you can set the somethings equal to eachother.\n5 + x = 12 + 16x\n-7 = 15x\nx = -7/15. Now substitute for y in one of the equations\n\n4. Alternately, you could have multiplied the first equation by -2\n\n-4y = -12 - 16x\n4y = 5 + x\n\nAnd added them together\n\n0 = -7 - 15x\n15x = -7\nx = -7/15\n\nAgain, substitute back in to find y.  The last method is particularly useful if you have more then just 2 variables/equations, and is done using matrices, but you probably haven't got to that point yet :)
167128
There's an infinite number of answers.\n\nIf you do it all out, they both equal the same line...\nI multiplied the second equation by a -4 and got 0=0.\n\nIf you just multiply it by a 4 you'd see that it's the same exact line.
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.
835073
okay well\n\n1) combine two to figure out the first variable\nx-2y+z=4\n-x+2y+3z=-4\nso 4z=0; z=0\nthen figure out the next variable with a different set of equations\nlike and multiply so the ys will cancel [take Zs out cause zero]\nx-2y+z=4\n2(2x+y)=6\n\ntherefore\nx-2y=4\n4x+2y=6\n5x=10\nx=2\n\nthen substitute for y\n\n-x+2y=-4\n-2+2y=-4\n2y=-2\ny=-1\n\nso y=-1, x=2, z=0\n\n\ncontinue the process with the rest.