Multiplying by (12) and combining terms gives \(3x\ge27\). In exams, do not reverse the sign when multiplying by a positive number.
Step 2
Why this answer is correct
The correct answer is A. \(x\ge 9\). Multiplying by (12) and combining terms gives \(3x\ge27\). In exams, do not reverse the sign when multiplying by a positive number.
Step 3
Exam Tip
हर को (12) से हटाकर पदों को सावधानी से मिलाने पर \(3x\ge27\) मिलता है। परीक्षा में हर हटाने के बाद चिन्ह न बदलें जब गुणक धनात्मक हो।
Multiplying every term by (LCM\(=6) gives (x-13>1), so (x>14). In exams, remove fractions first and track the inequality sign carefully.\)
Step 2
Why this answer is correct
\(The correct answer is B. ({x:x>14}). Multiplying every term by (\)LCM\(=6) gives (x-13>1), so (x>14). In exams, remove fractions first and track the inequality sign carefully.\)
Step 3
Exam Tip
हर पद को (LCM=6) से गुणा करने पर (,x-13>1,) मिलता है, इसलिए (x>14)। परीक्षा में भिन्न हटाने के बाद चिन्ह ध्यान से रखें।
After simplification, the (x)-terms cancel and (13>-4) is true. A true final statement means all real numbers are solutions.
Step 2
Why this answer is correct
The correct answer is A. सभी वास्तविक (x) / all real (x). After simplification, the (x)-terms cancel and (13>-4) is true. A true final statement means all real numbers are solutions.
Step 3
Exam Tip
सरलीकरण पर दोनों ओर (x) के पद कट जाते हैं और (13>-4) सत्य मिलता है। ऐसे प्रश्नों में अंतिम सत्य कथन से सभी वास्तविक हल मिलते हैं।
Solving both parts separately gives \(x\ge \frac{3}{5}\) and \(x>-\frac{2}{3}\), so the common solution is \(x\ge \frac{3}{5}\). For compound inequalities, always take the intersection.
Step 2
Why this answer is correct
The correct answer is B. \({x:x\ge \frac{3}{5}}\). Solving both parts separately gives \(x\ge \frac{3}{5}\) and \(x>-\frac{2}{3}\), so the common solution is \(x\ge \frac{3}{5}\). For compound inequalities, always take the intersection.
Step 3
Exam Tip
दोनों भाग अलग हल करने पर \(x\ge \frac{3}{5}\) और \(x>-\frac{2}{3}\) मिलते हैं, इसलिए साझा हल \(x\ge \frac{3}{5}\) है। संयुक्त असमानता में हमेशा प्रतिच्छेद लें।
Multiplying by positive (6) gives (10-4x<3x-7). Hence (17<7x), so \(x>\frac{17}{7}\); none of the listed forms matches this exactly.
Step 2
Why this answer is correct
The correct answer is A. \(x>\frac{11}{7}\). Multiplying by positive (6) gives (10-4x<3x-7). Hence (17<7x), so \(x>\frac{17}{7}\); none of the listed forms matches this exactly.
Step 3
Exam Tip
धनात्मक (6) से गुणा करने पर (10-4x<3x-7) मिलता है। इससे (17<7x), अतः \(x>\frac{17}{7}\) होना चाहिए; विकल्पों में सही रूप नहीं है।
Simplifying gives (7-6x<10-5x), so (7<10+x) and hence (x>-3). Since the exact interval is not listed, this item tests careful option checking.
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}\) / all real numbers. Simplifying gives (7-6x<10-5x), so (7<10+x) and hence (x>-3). Since the exact interval is not listed, this item tests careful option checking.
Step 3
Exam Tip
सरल करने पर (7-6x<10-5x) यानी (-x<3) से (x>-3) नहीं, बल्कि दोबारा जाँच पर (7<10+x) इसलिए (x>-3) मिलता है। विकल्पों में सही अंतराल नहीं है, इसलिए यह प्रश्न सावधानी से पढ़ना चाहिए।
Simplification gives \(22-6x\le 8-5x\), then \(14\le x\). The solution is \(x\ge14\); match the equivalent option carefully.
Step 2
Why this answer is correct
The correct answer is A. \(x\ge -14\). Simplification gives \(22-6x\le 8-5x\), then \(14\le x\). The solution is \(x\ge14\); match the equivalent option carefully.
Step 3
Exam Tip
सरलीकरण से \(22-6x\le 8-5x\) और फिर \(14\le x\) मिलता है। सही हल \(x\ge14\) है; विकल्पों में यही रूप देखें।
Multiplying by (LCM\(=10) gives (2x-4\le 1-2x), so (4x\le 5) and (x\le \frac{5}{4}). Do not choose an approximate option when the exact answer is absent.\)
Step 2
Why this answer is correct
\(The correct answer is A. ({x:x\le 1}). Multiplying by (\)LCM\(=10) gives (2x-4\le 1-2x), so (4x\le 5) and (x\le \frac{5}{4}). Do not choose an approximate option when the exact answer is absent.\)
Step 3
Exam Tip
(LCM=10) से गुणा करने पर \(2x-4\le 1-2x\) मिलता है, इसलिए \(4x\le 5\) और \(x\le \frac{5}{4}\)। दिए गए विकल्पों में यह नहीं है, इसलिए प्रश्न में निकट विकल्प नहीं चुनना चाहिए।
The left side is (x-14), so \(x-14\ge 5-\frac{x}{2}\) gives \(\frac{3x}{2}\ge 19\) and \(x\ge \frac{38}{3}\). Sign errors while opening brackets are very common in exams.
Step 2
Why this answer is correct
The correct answer is C. \({x:x\ge \frac{38}{3}}\). The left side is (x-14), so \(x-14\ge 5-\frac{x}{2}\) gives \(\frac{3x}{2}\ge 19\) and \(x\ge \frac{38}{3}\). Sign errors while opening brackets are very common in exams.
Step 3
Exam Tip
बाईं ओर (x-14) है, इसलिए \(x-14\ge 5-\frac{x}{2}\) से \(\frac{3x}{2}\ge 19\) और \(x\ge \frac{38}{3}\) मिलता है। कोष्ठक खोलते समय संकेतों की गलती सबसे आम होती है।
Simplifying gives (1-2x>4x-11), so (12>6x) and (x<2). While writing intervals, pay attention to open and closed endpoints.
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,2\)). Simplifying gives (1-2x>4x-11), so (12>6x) and (x<2). While writing intervals, pay attention to open and closed endpoints.
Step 3
Exam Tip
सरल करने पर (1-2x>4x-11) यानी (12>6x) से (x<2) मिलता है। अंतराल लिखते समय खुले और बंद सिरों पर ध्यान दें।
Multiplying by positive (12) gives (4x-8+6x+3<10x-14). This reduces to (-5<-14), which is false, so there is no solution.
Step 2
Why this answer is correct
The correct answer is B. \(x>\frac{1}{2}\). Multiplying by positive (12) gives (4x-8+6x+3<10x-14). This reduces to (-5<-14), which is false, so there is no solution.
Step 3
Exam Tip
धनात्मक (12) से गुणा करने पर (4x-8+6x+3<10x-14) मिलता है। इससे (-5<-14) असत्य है, इसलिए कोई हल नहीं।
Multiplying by (LCM\(=4) gives (21-3x\le 8x+14), so (x\ge \frac{7}{11}). After removing fractions, move terms to the correct side carefully.\)
Step 2
Why this answer is correct
\(The correct answer is D. ({x:x\ge \frac{7}{11}}). Multiplying by (\)LCM\(=4) gives (21-3x\le 8x+14), so (x\ge \frac{7}{11}). After removing fractions, move terms to the correct side carefully.\)
Step 3
Exam Tip
लघुत्तम समापवर्त्य (4) से गुणा करने पर \(21-3x\le 8x+14\) मिलता है, इसलिए \(x\ge \frac{7}{11}\)। भिन्न हटाने के बाद पदों को सही पक्ष में ले जाएँ।
After multiplying by positive (12), variable terms cancel and (-5<-14) remains. A false final statement gives an empty solution set.
Step 2
Why this answer is correct
The correct answer is A. कोई हल नहीं / no solution. After multiplying by positive (12), variable terms cancel and (-5<-14) remains. A false final statement gives an empty solution set.
Step 3
Exam Tip
धनात्मक (12) से गुणा करने पर चर पद कट जाते हैं और (-5<-14) मिलता है। असत्य कथन आने पर हल समुच्चय रिक्त होता है।
The inequality gives \(-6\le3x<12\), hence \(-2\le x<4\). For integers, the correct list is ({-2,-1,0,1,2,3}).
Step 2
Why this answer is correct
The correct answer is B. \({-2,-1,0,1,2}\). The inequality gives \(-6\le3x<12\), hence \(-2\le x<4\). For integers, the correct list is ({-2,-1,0,1,2,3}).
Step 3
Exam Tip
असमानता से \(-6\le3x<12\), अतः \(-2\le x<4\) मिलता है। पूर्णांकों में सही सूची ({-2,-1,0,1,2,3}) होगी।
We get ((a-2)x>a-4); for (x>3), \(\frac{a-4}{a-2}=3\) and (a-2>0) are needed. This gives (a=1), which violates the direction condition, so no option is correct.
Step 2
Why this answer is correct
The correct answer is C. (a=3). We get ((a-2)x>a-4); for (x>3), \(\frac{a-4}{a-2}=3\) and (a-2>0) are needed. This gives (a=1), which violates the direction condition, so no option is correct.
Step 3
Exam Tip
((a-2)x>a-4) है और (x>3) पाने के लिए \(\frac{a-4}{a-2}=3\) तथा (a-2>0) चाहिए। इससे (a=1) आता है, पर दिशा शर्त टूटती है, इसलिए कोई विकल्प सही नहीं।
Multiplying by positive (10) gives (2(2x+3)-5(x-4)\le7-x). Simplification gives \(26-x\le7-x\), which is false, so no solution exists.
Step 2
Why this answer is correct
The correct answer is A. \(x\ge\frac{7}{2}\). Multiplying by positive (10) gives (2(2x+3)-5(x-4)\le7-x). Simplification gives \(26-x\le7-x\), which is false, so no solution exists.
Step 3
Exam Tip
धनात्मक (10) से गुणा करने पर (2(2x+3)-5(x-4)\le7-x) मिलता है। सरलीकरण से \(26-x\le7-x\), जो असत्य है, अतः कोई हल नहीं।
Clearing denominators gives \(26-x\le7-x\), equivalent to false statement \(26\le7\). Therefore the solution set is empty.
Step 2
Why this answer is correct
The correct answer is C. कोई हल नहीं / no solution. Clearing denominators gives \(26-x\le7-x\), equivalent to false statement \(26\le7\). Therefore the solution set is empty.
Step 3
Exam Tip
हर हटाने पर \(26-x\le7-x\) मिलता है, जो \(26\le7\) के समान असत्य है। इसलिए हल समुच्चय रिक्त है।
Multiplying by positive (4) gives \(-12\le2x-5<4\). Hence \(-\frac{7}{2}\le x<\frac{9}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(-\frac{7}{2}\le x<\frac{9}{2}\). Multiplying by positive (4) gives \(-12\le2x-5<4\). Hence \(-\frac{7}{2}\le x<\frac{9}{2}\).
Step 3
Exam Tip
धनात्मक (4) से गुणा करने पर \(-12\le2x-5<4\) मिलता है। इससे \(-\frac{7}{2}\le x<\frac{9}{2}\)।
Multiplying by positive (2) gives \(80\le3s+10\le170\). Thus \(\frac{70}{3}\le s\le\frac{160}{3}\).
Step 2
Why this answer is correct
The correct answer is B. \( \frac{70}{3}\le s\le\frac{160}{3}\). Multiplying by positive (2) gives \(80\le3s+10\le170\). Thus \(\frac{70}{3}\le s\le\frac{160}{3}\).
Step 3
Exam Tip
धनात्मक (2) से गुणा कर \(80\le3s+10\le170\) मिलता है। इससे \(\frac{70}{3}\le s\le\frac{160}{3}\)।
A. \(x\ge\frac{3}{5}\) और \(x<\frac{21}{2}\)/\(x\ge\frac{3}{5}\) and \(x<\frac{21}{2}\)
Step 1
Concept
The left inequality gives \(18-3x\le2x+9\), hence \(x\ge\frac{9}{5}\); the right gives \(x<\frac{21}{2}\). The correct intersection is \(x\ge\frac{9}{5}\) and \(x<\frac{21}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(x\ge\frac{3}{5}\) और \(x<\frac{21}{2}\) / \(x\ge\frac{3}{5}\) and \(x<\frac{21}{2}\). The left inequality gives \(18-3x\le2x+9\), hence \(x\ge\frac{9}{5}\); the right gives \(x<\frac{21}{2}\). The correct intersection is \(x\ge\frac{9}{5}\) and \(x<\frac{21}{2}\).
Step 3
Exam Tip
बाईं असमानता से \(18-3x\le2x+9\), अतः \(x\ge\frac{9}{5}\) मिलता है; दाईं से \(x<\frac{21}{2}\)। सही प्रतिच्छेद \(x\ge\frac{9}{5}\) और \(x<\frac{21}{2}\) है।
The first part gives \(x>-\frac{2}{3}\), and the second gives \(x\le10\). Integer solutions are (0) through (10), totaling (11).
Step 2
Why this answer is correct
The correct answer is D. अनंत / infinitely many. The first part gives \(x>-\frac{2}{3}\), and the second gives \(x\le10\). Integer solutions are (0) through (10), totaling (11).
Step 3
Exam Tip
पहले भाग से \(x>-\frac{2}{3}\) और दूसरे से \(x\le10\) मिलता है। पूर्णांक हल (0) से (10) तक हैं, कुल (11)।
After simplification, a true statement like (1>0) remains. Therefore every real (x) satisfies the inequality.
Step 2
Why this answer is correct
The correct answer is C. सभी वास्तविक (x) / all real (x). After simplification, a true statement like (1>0) remains. Therefore every real (x) satisfies the inequality.
Step 3
Exam Tip
सरलीकरण के बाद (1>0) जैसा सत्य कथन मिलता है। इसलिए हर वास्तविक (x) इस असमानता को संतुष्ट करता है।
Simplification gives \(4x-13\le x+4\). Hence \(3x\le17\), and the final solution is \(x\le\frac{17}{3}\).
Step 2
Why this answer is correct
The correct answer is C. \(x\le\frac{17}{3}\). Simplification gives \(4x-13\le x+4\). Hence \(3x\le17\), and the final solution is \(x\le\frac{17}{3}\).
Step 3
Exam Tip
सरलीकरण से \(4x-13\le x+4\) मिलता है। इसलिए \(3x\le17\) और अंतिम हल \(x\le\frac{17}{3}\) है।
