Solution equation: absolute value 3x + 1 = absolute value 2x-5
Absolute value 3x + 1 = absolute value 2x-5
Ψ 3x + 1 = 2x-5 or 3x + 1 = 5-2x
3x-2x = - 5-1 or 3x + 2x = 5-1
X = - 6 or x = 0.8
RELATED INFORMATIONS
- 1. Solving equation 3x + 2 / 2x + 1 = [3x + 2 / 2x + 1] (absolute value) process
- 2. The solution equation | 3x-1 | = 1 | 2x - | 3x-1 | | - x + 3 | - | - X-2 | = 3 "|" is an absolute value
- 3. Solve the equation [3x + 1] = 2x-0.5, where [] is the integer sign
- 4. The formula method is used to solve the quadratic equation AX + BX + C = 0 (a is not equal to 0)
- 5. If the square of the quadratic equation 2x minus 4x plus 3 = 0 is x1, X2, then x2 divided by X1 plus X1 divided by x2 =? .
- 6. Let X1 and X2 be the two roots of the equation 2x + 4x-3, then (x1 + 1) (x2 + 1) and X1 + x2
- 7. Weida theorem | x1-x2| X1 and X2 are two steps of the quadratic equation 2x ^ + 5x-3 = 0 with one variable. The process of finding the value of | x1-x2 | is written by WIDA's theorem
- 8. If (K-4) X2 (the square of x) + (K-2) x + 3Y = 1 is a quadratic equation with respect to x, y, then k =?
- 9. Given the equation x & sup2; + 2 (2-m) x + 3-6M + 0 about X, if two real roots X1 and X2 of the equation satisfy the condition X1 = 3x2, find the value of real number M
- 10. When the fractional equation 2XX − 1 − 5x − 5x + 3 = 0 is solved by the substitution method, if XX − 1 = y, then the original equation can be reduced to () A. 2y2+3y-5=0B. 2y2-5y+3=0C. y2+3y-5=0D. y2-5y+3=0
- 11. The absolute value of solving equation - 12 + 3x = 18 + 2x
- 12. If the equation with the sign of four absolute values is |||||||||||x-1 |- 2 |- 3 |- 4 | = 0, then its root is
- 13. Sign the absolute value of | X-1 | - | Y-X | x
- 14. The sign | - 2 |, denotes that the absolute value of - 2 is 2, and | + 2 |, denotes that the absolute value of + 2 is 2. If | - x | = 2, then x = 2 or x = - 2. If the equation | - X-1 | = 2 is solved, then X-1 = 2 or X-1 = - 2 can be obtained, and x = 3 or x = - 1 can be obtained by solving the equation respectively. Using the above knowledge, the equation can be solved as follows: | - 2x-1-7 = 0
- 15. The absolute value of X + 2 = 3 11 of x-9-2 of X + 2 = (x-1) - 2 of X-2 to solve the equation And 113x - 1 / 12 = 105x plus 1 / 6 4x-3 (20-x) = 6x-7 (9-x)
- 16. Absolute value of solving equation x + 4 (X-2) = x + 1
- 17. Solving equation (x + X + one third x) times 3 = 3080.3x + 0.2 times (160-x) = 36 x + 10 = 40% times (36-x + 0) PS: This is what I listed on the application questions. It doesn't matter if you can't understand and solve it! Thank you all the same!
- 18. 46 times 1 / 3-6 / 5x = 10 how to solve the equation
- 19. A student forgot to multiply 1 by the least common multiple 12 of each denominator when solving the equation 10-x of 4 / 4 = x-m of 1-3. The result is x = - 1. Please help him find the correct solution of the equation
- 20. When solving the equation 10-4-x = 1-3-x-m to remove the denominator, a student forgot to multiply 1 by 12, and the result = - 1 Find the right solution