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)।
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।
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 \(\frac{70}{3}\le s\le\frac{160}{3}\)।
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 \(-\frac{7}{2}\le x<\frac{9}{2}\)।
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 (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।
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।
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।
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।
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।
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 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।
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।
