Modulus Equations PYQ | JEE Main 2025 | Sign Condition Trick

 

❓ Question

If

$$|x+y|+|x-y|=2|x|$$

and

$$|x+1|+|5-x|=6,$$

then find the possible values of $x$.

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✍️ Solution

We use the property

$$|a|+|b|=|a+b| \iff ab\ge0.$$

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Step 1: Using the first equation

Given,

$$|x+y|+|x-y|=2|x|.$$

Since

$$(x+y)+(x-y)=2x,$$

we have

$$|x+y|+|x-y| = |2x|.$$

Therefore,

$$(x+y)(x-y)\ge0.$$
$$x^2-y^2\ge0.$$
$$(x-y)(x+y)\ge0.$$

The critical points are

$$x=-y,\;y.$$

Hence,

$$\boxed{x\le -y\quad\text{or}\quad x\ge y.}$$

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Step 2: Using the second equation

Given,

$$|x+1|+|5-x|=6.$$

Since

$$(x+1)+(5-x)=6,$$

again using

$$|a|+|b|=|a+b| \iff ab\ge0,$$

we get

$$(x+1)(5-x)\ge0.$$

The critical points are

$$x=-1,\;5.$$

Therefore,

$$\boxed{-1\le x\le5.}$$

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Step 3: Taking the intersection

From Step 1,

$$x\in(-\infty,-y]\cup[y,\infty).$$

From Step 2,

$$x\in[-1,5].$$

Since $y\ge0$ (as considered in the solution),

$$(-\infty,-y]\cap[-1,5]=\varnothing,$$

and

$$[y,\infty)\cap[-1,5]=[y,5].$$

Hence,

$$\boxed{x\in[y,5].}$$

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✅ Final Answer

$$\boxed{x\in[y,5]}$$