Question

2 * |x-3| - |2x -2| = 4

Answer 1

Ответ:

(2x-3)^2-4x^2=0

(4х^2-12x+9)-4x^2=0

12x=9

x=0.75

Пошаговое объяснение:

Answer 2

To solve the equation $2 \cdot |x-3| - |2x -2| = 4$, we can break it down into cases based on the absolute value expressions. Remember that the absolute value of a number $a$ is defined as $|a| = a$ if $a$ is positive or zero, and $|a| = -a$ if $a$ is negative.

Case 1: $x - 3$ and $2x - 2$ are both nonnegative (or zero): In this case, the equation becomes: $2 \cdot (x - 3) - (2x - 2) = 4$ Simplify and solve for $x$: $2x - 6 - 2x + 2 = 4$ $-4 = 4$ This is not true, so there are no solutions in this case.

Case 2: $x - 3$ is nonnegative (or zero), but $2x - 2$ is negative: In this case, the equation becomes: $2 \cdot (x - 3) + (2x - 2) = 4$ Simplify and solve for $x$: $2x - 6 + 2x - 2 = 4$ $4x - 8 = 4$ $4x = 12$ $x = 3$

Case 3: $x - 3$ is negative, but $2x - 2$ is nonnegative (or zero): In this case, the equation becomes: $2 \cdot (3 - x) - (2x - 2) = 4$ Simplify and solve for $x$: $6 - 2x - 2x + 2 = 4$ $-4x + 8 = 4$ $-4x = -4$ $x = 1$

Case 4: $x - 3$ and $2x - 2$ are both negative: In this case, the equation becomes: $2 \cdot (3 - x) + (2x - 2) = 4$ Simplify and solve for $x$: $6 - 2x + 2x - 2 = 4$ $4 = 4$ This is true for all values of $x$, so any value of $x$ that makes $x - 3$ and $2x - 2$ negative satisfies this case.

Putting it all together, the solutions are $x = 1$ and $x = 3$.