SOLUTION: How can i solve these type of equations??
{{{ abs (x-4) - abs (x+2) =6}}}
would the same rules apply to these?
{{{ abs (2x-1) + abs (4-2x) = 10 }}}
Algebra ->
Absolute-value
-> SOLUTION: How can i solve these type of equations??
{{{ abs (x-4) - abs (x+2) =6}}}
would the same rules apply to these?
{{{ abs (2x-1) + abs (4-2x) = 10 }}}
Log On
You can put this solution on YOUR website! I assume you want to solve for x over the integers
***************************************************************
1) abs(x-4)-abs(x+2) = 6
:
: *** we want an abs value on each side of the =
: Add abs(x+2) to both sides
:
abs(x-4) = 6+abs(x+2)
:
Split the equation into two possible cases:
x-4 = 6+abs(x+2) or x-4 = -6-abs(x+2)
:
Subtract 6+abs(x+2) from both sides:
-10+x-abs(x+2) = 0 or x-4 = -6-abs(x+2)
:
Subtract x-10 from both sides:
-abs(x+2) = 10-x or x-4 = -6-abs(x+2)
:
Multiply both sides by -1:
abs(x+2) = x-10 or x-4 = -6-abs(x+2)
:
Split the equation into two possible cases:
x+2 = x-10 or x+2 = 10-x or x-4 = -6-abs(x+2)
:
Subtract x+2 from both sides:
0 = -12 or x+2 = 10-x or x-4 = -6-abs(x+2)
:
0 = -12 is trivially false:
x+2 = 10-x or x-4 = -6-abs(x+2)
:
Subtract 2-x from both sides:
2 x = 8 or x-4 = -6-abs(x+2)
:
Divide both sides by 2:
x = 4 or x-4 = -6-abs(x+2)
:
Add 6+abs(x+2) to both sides:
x = 4 or 2+x+abs(x+2) = 0
:
: ****** now work on second equation
:
Subtract x+2 from both sides:
x = 4 or abs(x+2) = -2-x
:
Split the equation into two possible cases:
x = 4 or x+2 = -2-x or x+2 = x+2
:
Subtract 2-x from both sides:
x = 4 or 2 x = -4 or x+2 = x+2
:
Divide both sides by 2:
x = 4 or x = -2 or x+2 = x+2
:
x+2 = x+2 is trivially true:
:
*******************************
The integer solution is x = -2
*******************************
2) |2x - 1| + |4 - 2x| = 10
yes - same technigue
