Question

|x - 1| + |2x - 3| =|3x - 4|​

Answer 1

Answer:

= 0

Step-by-step explanation:

(x-1) + (2x - 3)=(3x-4)

X - 1 + 2x - 3 = 3x - 4

3x - 4 = 3x - 4

3x - 3x - 4 +4=0

0=0

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Answer 2

$\large\underline{\sf{Solution-}}$

Given equation is

$\rm :\longmapsto\: |x - 1| + |2x - 3| = |3x - 4|$

Let first find the breaking points, by equating 0 to each Modulus term.

$\rm :\longmapsto\:x - 1 = 0\rm :\implies\:x = 1$

$\rm :\longmapsto\:2x - 3= 0\rm :\implies\:x = \dfrac{3}{2}$

$\rm :\longmapsto\:3x - 4= 0\rm :\implies\:x = \dfrac{4}{3}$

So, we get 3 breaking points.

So, 4 cases arises.

Case :- 1

$\rm :\longmapsto\:When \: x \leqslant 1$

We know,

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\: |x| = \begin{cases} &\sf{ - x \: \: if \: x < 0} \\ &\sf{ \: \: \: x \: \: if \: x \geqslant 0} \end{cases}\end{gathered}\end{gathered}$

So, given equation reduces to

$\rm :\longmapsto\: - (x - 1) - (2x - 3) = - (3x - 4)$

$\rm :\longmapsto\: - x + 1 - 2x + 3 = - 3x + 4$

$\rm :\longmapsto\: - 3x + 4 = - 3x + 4$

which is always true.

$\rm :\implies\: \boxed{ \quad \bf \: x \: \in \: ( - \infty , \: 1] \quad}$

Case :- 2

$\bf :\longmapsto\:When \: x \: \in \: (1,\dfrac{4}{3} \bigg]$

The given equation reduces to

$\rm :\longmapsto\: (x - 1) - (2x - 3) = - (3x - 4)$

$\rm :\longmapsto\: x - 1 - 2x + 3= - 3x + 4$

$\rm :\longmapsto\: - x + 2 = - 3x + 4$

$\rm :\longmapsto\: - x + 3x = 4 - 2$

$\rm :\longmapsto\: 2x = 2$

$\rm :\longmapsto\: x = 1$

Hence, there is no solution in this interval.

Case : - 3

$\bf :\longmapsto\:When \: x \: \in \: \bigg(\dfrac{4}{3}, \: \dfrac{3}{2} \bigg)$

The given equation reduces to

$\rm :\longmapsto\: (x - 1) - (2x - 3) = (3x - 4)$

$\rm :\longmapsto\: x - 1 - 2x + 3 = 3x - 4$

$\rm :\longmapsto\: - x + 2 = 3x - 4$

$\rm :\longmapsto\: - x - 3x = - 2 - 4$

$\rm :\longmapsto\: -4 x = - 6$

$\rm :\longmapsto\:x = \dfrac{3}{2}$

Hence, there is no solution in this interval.

Case :- 4

$\bf :\longmapsto\:When \: x \geqslant \dfrac{3}{2}$

The given equation reduces to

$\rm :\longmapsto\: (x - 1) + (2x - 3) = (3x - 4)$

$\rm :\longmapsto\: x - 1 + 2x - 3 = 3x - 4$

$\rm :\longmapsto\: 3x - 4= 3x - 4$

which is always true.

$\rm :\implies\: \boxed{ \quad \bf \: x \: \in \: \bigg[\dfrac{3}{2}, \: \infty )\quad}$

So,

Solution of equation

$\bf :\longmapsto\: |x - 1| + |2x - 3| = |3x - 4|$

is

$\bf:\longmapsto\:x \: \in \: ( - \infty , \: 1] \: \cup \: \bigg[\dfrac{3}{2}, \: \infty )$