The first, most important step is to be absolutely sure, with a bullet-proof
understanding, of what "solve" means. Sadly, you're not there yet.
To 'solve' an equation that has two variables in it, you need to find a pair of
numbers ... one number for 'x' and another number for 'y' ... that make the
equation true when you write the numbers in place of the letters.
There are an infinite number of these pairs for [ y = 2x + 1 ].
The way to find a pair is to pick any old number for 'x', write it in place of 'x',
and then figure out what 'y' must be.
Here are a few 'solutions':
' X ' . . . ' Y '
-3 . . . . -5
-2 . . . . -3
-1 . . . . -1
0 . .. . . +1
1 . .. . . +3
2 . .. . . +5
There are infinitely many more.
2x + 1 = 9 2x = 9 - 1 2x = 8 2x/2 = 8/2 x=4
X = 0 (zero) Y = -1 (neg 1)
2x + 3y = 9x = 3, y = 1 therefore 6 + 3 = 9
It can be anything. Example: x=1
If we assume that it equals zero and you wish to solve for x, then the answer is:-x2 + 2x - 3 = 0x2 - 2x + 3 = 0x2 -2x + 1 = -2(x - 1)2 = -2x - 1 = ± √(-2)x = 1 ± i√2
2x+32 = 1 2x = 1-32 2x = -31 x = -15.5
2x+1 = 9 2x = 9-1 2x = 8 x = 4
If: x+1+2x = 1x+5 Then: x = 2
2x + 1 = 9 2x = 9 - 1 2x = 8 2x/2 = 8/2 x=4
X = 0 (zero) Y = -1 (neg 1)
2x + 7 = 1 Therefore, 2x = -6 x = -6/2 x = -3
2x + 3y = 9x = 3, y = 1 therefore 6 + 3 = 9
It can be anything. Example: x=1
If we assume that it equals zero and you wish to solve for x, then the answer is:-x2 + 2x - 3 = 0x2 - 2x + 3 = 0x2 -2x + 1 = -2(x - 1)2 = -2x - 1 = ± √(-2)x = 1 ± i√2
2x + 5x - 1 = 9x 7x - 1 = 9x -1 = 9x - 7x -1 = 2x -1/2=x
2x+8 = 6 2x = 6-8 2x = -2 x = -1
2x + 2/3= 5/3 2x= 3/3 2x=1 x=1/2