हिंदी

Solve the inequation given below. Write the solution set and represent it on the number line: $$3x - 16 < \frac{2x}{5} - 3 \leq -\frac{3}{5} + 2x ; x \in \mathrm{R}$$

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प्रश्न

Solve the inequation given below. Write the solution set and represent it on the number line:

$$3x - 16 < \frac{2x}{5} - 3 \leq -\frac{3}{5} + 2x ; x \in \mathrm{R}$$

योग
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उत्तर

Step 1: Split the given double inequation into two separate linear inequations $$3x - 16 < \frac{2x}{5} - 3$$ and $$\frac{2x}{5} - 3 \leq -\frac{3}{5} + 2x$$

Step 2: Solve the first inequation $$3x - 16 < \frac{2x}{5} - 3$$

Multiply both sides by $$5$$: $$15x - 80 < 2x - 15$$

$$15x - 2x < 80 - 15$$

$$13x < 65$$

$$x < \frac{65}{13}$$

$$x < 5$$

Step 3: Solve the second inequation $$\frac{2x}{5} - 3 \leq -\frac{3}{5} + 2x$$

Multiply both sides by $$5$$: $$2x - 15 \leq -3 + 10x$$

$$-15 + 3 \leq 10x - 2x$$

$$-12 \leq 8x$$

$$8x \geq -12$$

$$x \geq -\frac{12}{8}$$

$$x \geq -\frac{3}{2}$$ (or $$x \geq -1.5$$)

Step 4: Combine both parts $$-\frac{3}{2} \leq x < 5$$ (or $$-1.5 \leq x < 5$$)

Step 5: Solution Set Since $$x \in \mathrm{R}$$, the solution set in set-builder notation is: $$\text{Solution set} = {x \in \mathrm{R} : -1.5 \leq x < 5}$$

Step 6: Representation on the Number Line

Mark a solid (darkened) circle at $$-1.5$$ to indicate that $$-1.5$$ is included in the solution.

Mark a hollow (open) circle at $$5$$ to indicate that $$5$$ is not included in the solution.

Darken the continuous line segment connecting $$-1.5$$ and $$5$$.

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अध्याय 4: Linear Inequations - EXERCISE 4 [पृष्ठ ४५]

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आर.एस. अग्रवाल Mathematics [English] Class 10 ICSE
अध्याय 4 Linear Inequations
EXERCISE 4 | Q 22. | पृष्ठ ४५
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