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Miscellaneous Exercise · Q9

Q.3x−7>2(x−6)3x - 7 > 2(x - 6), 6−x>11−2x6 - x > 11 - 2x

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The solution set is the intersection of two linear inequalities: x>−5x > -5 from the first and x>5x > 5 from the second. The stricter condition x>5x > 5 dominates, so the final answer is x>5x > 5.

Concept and Intuition

When you have two inequalities joined by an implied "and" (as in a system), you're looking for values of xx that satisfy both conditions simultaneously. Think of it like two gates: you can only pass through if both gates are open. The solution is the overlap — the intersection — of the two individual solution sets.

Each inequality is linear, meaning it describes a half-line on the number line. Solving them separately is straightforward: isolate xx by performing the same operations on both sides, being careful with the direction of the inequality when multiplying or dividing by a negative number. Then, find where the two half-lines overlap.

Watch out

A common mistake is to solve each inequality correctly but then take the union instead of the intersection. Remember: "and" means both must hold — you need the region common to both.

Step-by-Step Solution

1. Solve the first inequality: 3x−7>2(x−6)3x - 7 > 2(x - 6)

Start by expanding the right-hand side:

3x−7>2x−123x - 7 > 2x - 12

Now bring the 2x2x term to the left and the −7-7 to the right (or vice versa — just keep it tidy):

3x−2x>−12+73x - 2x > -12 + 7

x>−5x > -5

So the first condition is x>−5x > -5. On the number line, this is all numbers to the right of −5-5, not including −5-5 itself.

2. Solve the second inequality: 6−x>11−2x6 - x > 11 - 2x

Add 2x2x to both sides to get the xx terms together:

6−x+2x>116 - x + 2x > 11

6+x>116 + x > 11

Now subtract 66 from both sides:

x>5x > 5

So the second condition is x>5x > 5. This is all numbers to the right of 55.

3. Find the intersection of the two solution sets

We need xx that satisfies both x>−5x > -5 and x>5x > 5.

Visualise the number line: …

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