A. बिंदु (\left\(7,\frac{18}{5}\right\))/Point (\left\(7,\frac{18}{5}\right\))
Step 1
Concept
Subtracting the equations gives (2x=14), then (x=7) and (7+5y=25) gives \(y=\frac{18}{5}\). This is the graphical intersection.
Step 2
Why this answer is correct
The correct answer is A. बिंदु (\left\(7,\frac{18}{5}\right\)) / Point (\left\(7,\frac{18}{5}\right\)). Subtracting the equations gives (2x=14), then (x=7) and (7+5y=25) gives \(y=\frac{18}{5}\). This is the graphical intersection.
Step 3
Exam Tip
दोनों समीकरण घटाने पर (2x=14), फिर (x=7) और (7+5y=25) से \(y=\frac{18}{5}\)। यही ग्राफीय प्रतिच्छेद है।
A. बिंदु (\left\(5,3\right\))/Point (\left\(5,3\right\))
Step 1
Concept
Subtracting the equations gives (3x=15), so (x=5) and (y=3). In a real situation this is the meeting point.
Step 2
Why this answer is correct
The correct answer is A. बिंदु (\left\(5,3\right\)) / Point (\left\(5,3\right\)). Subtracting the equations gives (3x=15), so (x=5) and (y=3). In a real situation this is the meeting point.
Step 3
Exam Tip
दोनों समीकरण घटाने पर (3x=15), इसलिए (x=5) और (y=3)। वास्तविक स्थिति में यही मिलन बिंदु है।
A. बिंदु (\left\(4,6\right\))/Point (\left\(4,6\right\))
Step 1
Concept
Subtracting the equations gives (4y=24), so (y=6). Then (4x-6=10) gives (x=4).
Step 2
Why this answer is correct
The correct answer is A. बिंदु (\left\(4,6\right\)) / Point (\left\(4,6\right\)). Subtracting the equations gives (4y=24), so (y=6). Then (4x-6=10) gives (x=4).
Step 3
Exam Tip
दोनों समीकरण घटाने पर (4y=24), इसलिए (y=6)। फिर (4x-6=10) से (x=4)।
A. बिंदु (\left\(5,4\right\))/Point (\left\(5,4\right\))
Step 1
Concept
Subtracting the second from the first gives (5y=20), so (y=4). Then (3x-4=11) gives (x=5).
Step 2
Why this answer is correct
The correct answer is A. बिंदु (\left\(5,4\right\)) / Point (\left\(5,4\right\)). Subtracting the second from the first gives (5y=20), so (y=4). Then (3x-4=11) gives (x=5).
Step 3
Exam Tip
पहले से दूसरे को घटाने पर (5y=20), इसलिए (y=4)। फिर (3x-4=11) से (x=5)।
A. बिंदु (\left\(\frac{79}{19},\frac{113}{19}\right\))/Point (\left\(\frac{79}{19},\frac{113}{19}\right\))
Step 1
Concept
Using (y=6x-19) from the first equation gives \(x=\frac{79}{19}\). Then \(y=\frac{113}{19}\).
Step 2
Why this answer is correct
The correct answer is A. बिंदु (\left\(\frac{79}{19},\frac{113}{19}\right\)) / Point (\left\(\frac{79}{19},\frac{113}{19}\right\)). Using (y=6x-19) from the first equation gives \(x=\frac{79}{19}\). Then \(y=\frac{113}{19}\).
Step 3
Exam Tip
पहले समीकरण से (y=6x-19) रखकर \(x=\frac{79}{19}\) मिलता है। फिर \(y=\frac{113}{19}\) है।
A. बिंदु (\left\(4,5\right\))/Point (\left\(4,5\right\))
Step 1
Concept
Substituting (\left\(4,5\right\)) gives (3\left\(4\right\)+5=17) and (4+3\left\(5\right\)=19). This is the common point of both lines.
Step 2
Why this answer is correct
The correct answer is A. बिंदु (\left\(4,5\right\)) / Point (\left\(4,5\right\)). Substituting (\left\(4,5\right\)) gives (3\left\(4\right\)+5=17) and (4+3\left\(5\right\)=19). This is the common point of both lines.
Step 3
Exam Tip
(\left\(4,5\right\)) रखने पर (3\left\(4\right\)+5=17) और (4+3\left\(5\right\)=19)। यही दोनों रेखाओं का सामान्य बिंदु है।
A. बिंदु (\left\(\frac{25}{7},\frac{23}{7}\right\))/Point (\left\(\frac{25}{7},\frac{23}{7}\right\))
Step 1
Concept
(\left\(\frac{25}{7},\frac{23}{7}\right\)) satisfies both equations. Read fraction coordinates carefully using the graph scale.
Step 2
Why this answer is correct
The correct answer is A. बिंदु (\left\(\frac{25}{7},\frac{23}{7}\right\)) / Point (\left\(\frac{25}{7},\frac{23}{7}\right\)). (\left\(\frac{25}{7},\frac{23}{7}\right\)) satisfies both equations. Read fraction coordinates carefully using the graph scale.
Step 3
Exam Tip
(\left\(\frac{25}{7},\frac{23}{7}\right\)) रखने पर दोनों समीकरण संतुष्ट होते हैं। भिन्न निर्देशांक को ग्राफ के पैमाने से सावधानीपूर्वक पढ़ें।
A. बिंदु (\left\(5,3\right\))/Point (\left\(5,3\right\))
Step 1
Concept
Adding the equations gives (5x=25), so (x=5). Then (x+3y=14) gives (y=3).
Step 2
Why this answer is correct
The correct answer is A. बिंदु (\left\(5,3\right\)) / Point (\left\(5,3\right\)). Adding the equations gives (5x=25), so (x=5). Then (x+3y=14) gives (y=3).
Step 3
Exam Tip
दोनों समीकरण जोड़ने पर (5x=25), इसलिए (x=5)। फिर (x+3y=14) से (y=3)।
A. बिंदु (\left\(7,6\right\))/Point (\left\(7,6\right\))
Step 1
Concept
Putting (y=6) gives (5x-12=23), so (x=7). In a horizontal line, the value of (y) is fixed.
Step 2
Why this answer is correct
The correct answer is A. बिंदु (\left\(7,6\right\)) / Point (\left\(7,6\right\)). Putting (y=6) gives (5x-12=23), so (x=7). In a horizontal line, the value of (y) is fixed.
Step 3
Exam Tip
(y=6) रखने पर (5x-12=23), इसलिए (x=7)। क्षैतिज रेखा में (y) का मान तय रहता है।
A. बिंदु (\left\(-5,-9\right\))/Point (\left\(-5,-9\right\))
Step 1
Concept
Putting (x=-5) gives (4\left\(-5\right\)-3y=7), so (y=-9). In a vertical line, (x) is already fixed.
Step 2
Why this answer is correct
The correct answer is A. बिंदु (\left\(-5,-9\right\)) / Point (\left\(-5,-9\right\)). Putting (x=-5) gives (4\left\(-5\right\)-3y=7), so (y=-9). In a vertical line, (x) is already fixed.
Step 3
Exam Tip
(x=-5) रखने पर (4\left\(-5\right\)-3y=7), इसलिए (y=-9)। ऊर्ध्वाधर रेखा में (x) पहले से निश्चित होता है।
A. बिंदु (\left\(5,2\right\))/Point (\left\(5,2\right\))
Step 1
Concept
Substituting (\left\(5,2\right\)) gives (5\left\(5\right\)+3\left\(2\right\)=31) and (5+2=7). If both equations are true, that is the intersection.
Step 2
Why this answer is correct
The correct answer is A. बिंदु (\left\(5,2\right\)) / Point (\left\(5,2\right\)). Substituting (\left\(5,2\right\)) gives (5\left\(5\right\)+3\left\(2\right\)=31) and (5+2=7). If both equations are true, that is the intersection.
Step 3
Exam Tip
(\left\(5,2\right\)) रखने पर (5\left\(5\right\)+3\left\(2\right\)=31) और (5+2=7)। दोनों समीकरण सत्य हों तो वही प्रतिच्छेद है।
A. बिंदु (\left\(\frac{21}{5},\frac{16}{5}\right\))/Point (\left\(\frac{21}{5},\frac{16}{5}\right\))
Step 1
Concept
Using (x=y+1) from (x-y=1) gives \(y=\frac{16}{5}\) and \(x=\frac{21}{5}\). On the graph this is the intersection point.
Step 2
Why this answer is correct
The correct answer is A. बिंदु (\left\(\frac{21}{5},\frac{16}{5}\right\)) / Point (\left\(\frac{21}{5},\frac{16}{5}\right\)). Using (x=y+1) from (x-y=1) gives \(y=\frac{16}{5}\) and \(x=\frac{21}{5}\). On the graph this is the intersection point.
Step 3
Exam Tip
(x-y=1) से (x=y+1) रखकर \(y=\frac{16}{5}\) और \(x=\frac{21}{5}\) मिलता है। ग्राफ पर यही प्रतिच्छेद बिंदु होगा।
A. बिंदु (\left\(4,8\right\))/Point (\left\(4,8\right\))
Step 1
Concept
Subtracting the equations gives (2x=8), so (x=4) and (y=8). Whatever the context, the intersection point is the solution.
Step 2
Why this answer is correct
The correct answer is A. बिंदु (\left\(4,8\right\)) / Point (\left\(4,8\right\)). Subtracting the equations gives (2x=8), so (x=4) and (y=8). Whatever the context, the intersection point is the solution.
Step 3
Exam Tip
दोनों समीकरण घटाने पर (2x=8), इसलिए (x=4) और (y=8)। संदर्भ कोई भी हो, प्रतिच्छेद बिंदु ही हल है।
A. बिंदु (\left\(8,\frac{13}{5}\right\))/Point (\left\(8,\frac{13}{5}\right\))
Step 1
Concept
Subtracting the equations gives (x=8), then (8+5y=21) gives \(y=\frac{13}{5}\). This is the graphical intersection.
Step 2
Why this answer is correct
The correct answer is A. बिंदु (\left\(8,\frac{13}{5}\right\)) / Point (\left\(8,\frac{13}{5}\right\)). Subtracting the equations gives (x=8), then (8+5y=21) gives \(y=\frac{13}{5}\). This is the graphical intersection.
Step 3
Exam Tip
दोनों समीकरण घटाने पर (x=8), फिर (8+5y=21) से \(y=\frac{13}{5}\)। यही ग्राफीय प्रतिच्छेद है।
A. बिंदु (\left\(4,3\right\))/Point (\left\(4,3\right\))
Step 1
Concept
Subtracting the equations gives (2x=8), so (x=4) and (y=3). In a real situation this is the meeting point.
Step 2
Why this answer is correct
The correct answer is A. बिंदु (\left\(4,3\right\)) / Point (\left\(4,3\right\)). Subtracting the equations gives (2x=8), so (x=4) and (y=3). In a real situation this is the meeting point.
Step 3
Exam Tip
दोनों समीकरण घटाने पर (2x=8), इसलिए (x=4) और (y=3)। वास्तविक स्थिति में यही मिलन बिंदु है।
A. बिंदु (\left\(5,3\right\))/Point (\left\(5,3\right\))
Step 1
Concept
Substituting (\left\(5,3\right\)) gives (2\left\(5\right\)+3\left\(3\right\)=19) and (2\left\(5\right\)-3=7). If both are true, this is the solution.
Step 2
Why this answer is correct
The correct answer is A. बिंदु (\left\(5,3\right\)) / Point (\left\(5,3\right\)). Substituting (\left\(5,3\right\)) gives (2\left\(5\right\)+3\left\(3\right\)=19) and (2\left\(5\right\)-3=7). If both are true, this is the solution.
Step 3
Exam Tip
(\left\(5,3\right\)) रखने पर (2\left\(5\right\)+3\left\(3\right\)=19) और (2\left\(5\right\)-3=7)। दोनों सत्य हों तो यही हल है।
B. बिंदु (\left\(4,\frac{11}{3}\right\))/Point (\left\(4,\frac{11}{3}\right\))
Step 1
Concept
Subtracting the second from the first gives (4y=12), so (y=3); careful checking is needed. The correct solution is (\left\(5,3\right\)).
Step 2
Why this answer is correct
The correct answer is B. बिंदु (\left\(4,\frac{11}{3}\right\)) / Point (\left\(4,\frac{11}{3}\right\)). Subtracting the second from the first gives (4y=12), so (y=3); careful checking is needed. The correct solution is (\left\(5,3\right\)).
Step 3
Exam Tip
पहले से दूसरे को घटाने पर (4y=12), इसलिए (y=3) नहीं बल्कि जाँच जरूरी है। सही हल (\left\(5,3\right\)) है।
A. बिंदु (\left\(\frac{42}{11},\frac{67}{11}\right\))/Point (\left\(\frac{42}{11},\frac{67}{11}\right\))
Step 1
Concept
Using (y=5x-13) from the first equation gives \(x=\frac{42}{11}\). Then \(y=\frac{67}{11}\).
Step 2
Why this answer is correct
The correct answer is A. बिंदु (\left\(\frac{42}{11},\frac{67}{11}\right\)) / Point (\left\(\frac{42}{11},\frac{67}{11}\right\)). Using (y=5x-13) from the first equation gives \(x=\frac{42}{11}\). Then \(y=\frac{67}{11}\).
Step 3
Exam Tip
पहले समीकरण से (y=5x-13) रखकर \(x=\frac{42}{11}\) मिलता है। फिर \(y=\frac{67}{11}\) है।
A. बिंदु (\left\(3,4\right\))/Point (\left\(3,4\right\))
Step 1
Concept
Substituting (\left\(3,4\right\)) gives (2\left\(3\right\)+4=10) and (3+2\left\(4\right\)=11). This is the common point of both lines.
Step 2
Why this answer is correct
The correct answer is A. बिंदु (\left\(3,4\right\)) / Point (\left\(3,4\right\)). Substituting (\left\(3,4\right\)) gives (2\left\(3\right\)+4=10) and (3+2\left\(4\right\)=11). This is the common point of both lines.
Step 3
Exam Tip
(\left\(3,4\right\)) रखने पर (2\left\(3\right\)+4=10) और (3+2\left\(4\right\)=11)। यही दोनों रेखाओं का सामान्य बिंदु है।
B. बिंदु (\left\(3,\frac{5}{2}\right\))/Point (\left\(3,\frac{5}{2}\right\))
Step 1
Concept
(\left\(3,\frac{5}{2}\right\)) satisfies both equations. Read fraction coordinates carefully using scale on the graph.
Step 2
Why this answer is correct
The correct answer is B. बिंदु (\left\(3,\frac{5}{2}\right\)) / Point (\left\(3,\frac{5}{2}\right\)). (\left\(3,\frac{5}{2}\right\)) satisfies both equations. Read fraction coordinates carefully using scale on the graph.
Step 3
Exam Tip
(\left\(3,\frac{5}{2}\right\)) रखने पर दोनों समीकरण संतुष्ट होते हैं। ग्राफ में भिन्न निर्देशांक को पैमाने से सावधानीपूर्वक पढ़ें।
A. बिंदु (\left\(4,\frac{7}{2}\right\))/Point (\left\(4,\frac{7}{2}\right\))
Step 1
Concept
Adding the equations gives (4x=16), so (x=4). Then (x+2y=11) gives \(y=\frac{7}{2}\).
Step 2
Why this answer is correct
The correct answer is A. बिंदु (\left\(4,\frac{7}{2}\right\)) / Point (\left\(4,\frac{7}{2}\right\)). Adding the equations gives (4x=16), so (x=4). Then (x+2y=11) gives \(y=\frac{7}{2}\).
Step 3
Exam Tip
दोनों समीकरण जोड़ने पर (4x=16), इसलिए (x=4)। फिर (x+2y=11) से \(y=\frac{7}{2}\)।
A. बिंदु (\left\(8,5\right\))/Point (\left\(8,5\right\))
Step 1
Concept
Putting (y=5) gives (4x-15=17), so (x=8). In a horizontal line, the value of (y) is fixed.
Step 2
Why this answer is correct
The correct answer is A. बिंदु (\left\(8,5\right\)) / Point (\left\(8,5\right\)). Putting (y=5) gives (4x-15=17), so (x=8). In a horizontal line, the value of (y) is fixed.
Step 3
Exam Tip
(y=5) रखने पर (4x-15=17), इसलिए (x=8)। क्षैतिज रेखा में (y) का मान तय रहता है।
A. बिंदु (\left\(-4,-11\right\))/Point (\left\(-4,-11\right\))
Step 1
Concept
Putting (x=-4) gives (3\left\(-4\right\)-2y=10), so (y=-11). In a vertical line, (x) is already fixed.
Step 2
Why this answer is correct
The correct answer is A. बिंदु (\left\(-4,-11\right\)) / Point (\left\(-4,-11\right\)). Putting (x=-4) gives (3\left\(-4\right\)-2y=10), so (y=-11). In a vertical line, (x) is already fixed.
Step 3
Exam Tip
(x=-4) रखने पर (3\left\(-4\right\)-2y=10), इसलिए (y=-11)। ऊर्ध्वाधर रेखा में (x) पहले से निश्चित होता है।
A. बिंदु (\left\(2,3\right\))/Point (\left\(2,3\right\))
Step 1
Concept
(\left\(2,3\right\)) satisfies both equations. In difficult options, direct substitution is the fastest check.
Step 2
Why this answer is correct
The correct answer is A. बिंदु (\left\(2,3\right\)) / Point (\left\(2,3\right\)). (\left\(2,3\right\)) satisfies both equations. In difficult options, direct substitution is the fastest check.
Step 3
Exam Tip
(\left\(2,3\right\)) दोनों समीकरणों को संतुष्ट करता है। कठिन विकल्पों में प्रत्यक्ष प्रतिस्थापन सबसे तेज जाँच है।
A. बिंदु (\left\(\frac{25}{7},\frac{23}{7}\right\))/Point (\left\(\frac{25}{7},\frac{23}{7}\right\))
Step 1
Concept
Solving both equations gives \(x=\frac{25}{7}\) and \(y=\frac{23}{7}\). On the graph this is the intersection point.
Step 2
Why this answer is correct
The correct answer is A. बिंदु (\left\(\frac{25}{7},\frac{23}{7}\right\)) / Point (\left\(\frac{25}{7},\frac{23}{7}\right\)). Solving both equations gives \(x=\frac{25}{7}\) and \(y=\frac{23}{7}\). On the graph this is the intersection point.
Step 3
Exam Tip
दोनों समीकरण हल करने पर \(x=\frac{25}{7}\) और \(y=\frac{23}{7}\) मिलता है। ग्राफ पर यही प्रतिच्छेद बिंदु होगा।
A. बिंदु (\left\(5,5\right\))/Point (\left\(5,5\right\))
Step 1
Concept
Subtracting the equations gives (3x=15), so (x=5) and (y=5). On the graph, this is the intersection point of both lines.
Step 2
Why this answer is correct
The correct answer is A. बिंदु (\left\(5,5\right\)) / Point (\left\(5,5\right\)). Subtracting the equations gives (3x=15), so (x=5) and (y=5). On the graph, this is the intersection point of both lines.
Step 3
Exam Tip
दोनों समीकरण घटाने पर (3x=15), इसलिए (x=5) और (y=5)। ग्राफ पर यही दोनों रेखाओं का प्रतिच्छेद बिंदु है।
A. बिंदु (\left\(3,5\right\))/Point (\left\(3,5\right\))
Step 1
Concept
Subtracting the second equation from the first gives (y=5), then (x+5=8) gives (x=3). In a real situation, the meeting point is the graphical solution.
Step 2
Why this answer is correct
The correct answer is A. बिंदु (\left\(3,5\right\)) / Point (\left\(3,5\right\)). Subtracting the second equation from the first gives (y=5), then (x+5=8) gives (x=3). In a real situation, the meeting point is the graphical solution.
Step 3
Exam Tip
पहले समीकरण से दूसरे को घटाने पर (y=5), फिर (x+5=8) से (x=3)। वास्तविक स्थिति में मिलन बिंदु ही ग्राफीय हल है।
A. बिंदु (\left\(-3,10\right\))/Point (\left\(-3,10\right\))
Step 1
Concept
Putting (x=-3) gives (2\left\(-3\right\)+y=4), so (y=10). In a vertical line, the value of (x) is already fixed.
Step 2
Why this answer is correct
The correct answer is A. बिंदु (\left\(-3,10\right\)) / Point (\left\(-3,10\right\)). Putting (x=-3) gives (2\left\(-3\right\)+y=4), so (y=10). In a vertical line, the value of (x) is already fixed.
Step 3
Exam Tip
(x=-3) रखने पर (2\left\(-3\right\)+y=4), इसलिए (y=10)। ऊर्ध्वाधर रेखा में (x) का मान पहले से निश्चित होता है।
