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Similar expressions
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(|x-3|)+2*(|x+1|)=4 equation
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The solution
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|x - 3| + 2*|x + 1| = 4
$$\left|{x - 3}\right| + 2 \left|{x + 1}\right| = 4$$
Detail solution
For every modulo expressions in the equation
allow cases, when this expressions ">=0" or "<0",
solve the resulting equation.
1.
$$x - 3 \geq 0$$
$$x + 1 \geq 0$$
or
$$3 \leq x \wedge x < \infty$$
we get the equation
$$\left(x - 3\right) + 2 \left(x + 1\right) - 4 = 0$$
after simplifying we get
$$3 x - 5 = 0$$
the solution in this interval:
$$x_{1} = \frac{5}{3}$$
but x1 not in the inequality interval
2.
$$x - 3 \geq 0$$
$$x + 1 < 0$$
The inequality system has no solutions, see the next condition
3.
$$x - 3 < 0$$
$$x + 1 \geq 0$$
or
$$-1 \leq x \wedge x < 3$$
we get the equation
$$\left(3 - x\right) + 2 \left(x + 1\right) - 4 = 0$$
after simplifying we get
$$x + 1 = 0$$
the solution in this interval:
$$x_{2} = -1$$
4.
$$x - 3 < 0$$
$$x + 1 < 0$$
or
$$-\infty < x \wedge x < -1$$
we get the equation
$$\left(3 - x\right) + 2 \left(- x - 1\right) - 4 = 0$$
after simplifying we get
$$- 3 x - 3 = 0$$
the solution in this interval:
$$x_{3} = -1$$
but x3 not in the inequality interval
The final answer:
$$x_{1} = -1$$
allow cases, when this expressions ">=0" or "<0",
solve the resulting equation.
1.
$$x - 3 \geq 0$$
$$x + 1 \geq 0$$
or
$$3 \leq x \wedge x < \infty$$
we get the equation
$$\left(x - 3\right) + 2 \left(x + 1\right) - 4 = 0$$
after simplifying we get
$$3 x - 5 = 0$$
the solution in this interval:
$$x_{1} = \frac{5}{3}$$
but x1 not in the inequality interval
2.
$$x - 3 \geq 0$$
$$x + 1 < 0$$
The inequality system has no solutions, see the next condition
3.
$$x - 3 < 0$$
$$x + 1 \geq 0$$
or
$$-1 \leq x \wedge x < 3$$
we get the equation
$$\left(3 - x\right) + 2 \left(x + 1\right) - 4 = 0$$
after simplifying we get
$$x + 1 = 0$$
the solution in this interval:
$$x_{2} = -1$$
4.
$$x - 3 < 0$$
$$x + 1 < 0$$
or
$$-\infty < x \wedge x < -1$$
we get the equation
$$\left(3 - x\right) + 2 \left(- x - 1\right) - 4 = 0$$
after simplifying we get
$$- 3 x - 3 = 0$$
the solution in this interval:
$$x_{3} = -1$$
but x3 not in the inequality interval
The final answer:
$$x_{1} = -1$$
The graph