|X+4| = |3X+10|
first conditions:
x+4 ≥0 ---> x ≥ -4
and
3x+10 ≥0 ----> x ≥-10/3
so we know that x ≥ -10/3
case 1:
x+4 = 3x+10
-2x = 6
x = -3
case 2:
-x-4 = 3x+10
-4x = 16
x = -4 , also no good, because of above restriction
So x = -3
check:
if x = -4
LS = |-4+4| = 0
RS = |-12+10| = 2
LS ≠ RS
if x = -3
LS = |1| = 1
RS = |-9+10| = 1
So, x = -3
| x + 4 | = | 3 x + 10 |
x + 4 = 3 x + 10
OR
x + 4 = - ( 3 x + 10 )
1 )
x + 4 = 3 x + 10 Subtract x to both sides
x + 4 - x = 3 x + 10 - x
4 = 2 x + 10 Subtact 4 to both sides
4 - 4 = 2 x + 10 - 4
0 = 2 x + 6 Subtract 2 x to both sides
0 - 2 x = 2 x + 6 - 2 x
- 2 x = 6 Divide both sides by - 2
- 2 x / - 2 = 6 / - 2
x = - 3
2)
x + 4 = - ( 3 x + 10 )
x + 4 = - 3 x - 10
Subtract 4 to both sides
x + 4 - 4 = - 3 x - 10 - 4
x = - 3 x - 14 Add 3 x to both sides
x + 3 x = - 3 x - 14 + 3 x
4 x = - 14 Divide both sides by 4
4 x / 4 = - 14 / 4
x = - 14 / 4 = - 2 * 7 / ( 2 * 2 ) = - 7 / 2
The solutions are :
x = - 7 / 2 and x = - 3
Proof :
For x = - 7 / 2
| x + 4 | = | 3 x + 10 |
| - 7 / 2 + 4 | = | 3 * ( - 7 / 2 ) + 10 |
| - 7 / 2 + 8 / 2 | = | - 21 / 2 + 10 |
| 1 / 2 | = | - 21 / 2 + 20 / 2 |
| 1 / 2 | = | - 1 / 2 |
1 / 2 = 1 / 2
For x = - 3
| x + 4 | = | 3 x + 10 |
| - 3 + 4 | = | 3 * ( - 3 ) + 10 |
| 1 | = | - 9 + 10 |
| 1 | = | 1 |
1 = 1
To determine the values of X that satisfy the equation |X+4| = |3X+10|, we need to consider two cases:
Case 1: (X + 4) = (3X + 10)
In this case, we remove the absolute value symbols and solve for X:
X + 4 = 3X + 10
Subtract X from both sides:
4 = 2X + 10
Subtract 10 from both sides:
-6 = 2X
Divide both sides by 2:
X = -3
Case 2: (X + 4) = -(3X + 10)
In this case, we again remove the absolute value symbols and solve for X:
X + 4 = -3X - 10
Add 3X to both sides:
4X + 4 = -10
Subtract 4 from both sides:
4X = -14
Divide both sides by 4:
X = -3.5
So, there are two solutions to the equation |X+4| = |3X+10|: X = -3 and X = -3.5.