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12 questions tagged with fraction solution.
What is the solution of (2x-y=6) and (x+3y=13)?
D. \(x=\frac{31}{7},\ y=\frac{20}{7}\)
पहले समीकरण से (y=2x-6) रखें। दूसरे में रखने पर (7x=31), इसलिए \(x=\frac{31}{7},\ y=\frac{20}{7}\)। / Use (y=2x-6) from the first equation. Substitution gives (7x=31), so \(x=\frac{31}{7},\ y=\frac{20}{7}\).
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What is the solution of (2x-y=7) and (x+2y=14)?
A. \(x=\frac{28}{5},\ y=\frac{21}{5}\)
पहले समीकरण से (y=2x-7) रखें। दूसरे में रखने पर (5x=28), इसलिए \(x=\frac{28}{5}\) और \(y=\frac{21}{5}\)। / Use (y=2x-7) from the first equation. Substitution gives (5x=28), so \(x=\frac{28}{5}\) and \(y=\frac{21}{5}\).
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Which point lies on both lines (3x-2y=4) and (x+4y=22)?
A. बिंदु (\left\(\frac{15}{4},\frac{73}{16}\right\))Point (\left\(\frac{15}{4},\frac{73}{16}\right\))
(x+4y=22) से (x=22-4y), और पहले समीकरण में रखने पर \(y=\frac{73}{16}\)। फिर \(x=\frac{15}{4}\) है। / From (x+4y=22), (x=22-4y), and substituting in the first equation gives \(y=\frac{73}{16}\). Then \(x=\frac{15}{4}\).
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Which point lies on both lines (x-3y=-8) and (3x+y=20)?
A. बिंदु (\left\(\frac{52}{10},\frac{22}{5}\right\))Point (\left\(\frac{52}{10},\frac{22}{5}\right\))
दोनों समीकरण हल करने पर \(y=\frac{22}{5}\) और \(x=\frac{26}{5}\) मिलता है। \(\frac{52}{10}\) को \(\frac{26}{5}\) भी लिख सकते हैं। / Solving both equations gives \(y=\frac{22}{5}\) and \(x=\frac{26}{5}\). \(\frac{52}{10}\) can also be written as \(\frac{26}{5}\).
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Where do the lines (6x-y=19) and (x+3y=22) meet?
A. बिंदु (\left\(\frac{79}{19},\frac{113}{19}\right\))Point (\left\(\frac{79}{19},\frac{113}{19}\right\))
पहले समीकरण से (y=6x-19) रखकर \(x=\frac{79}{19}\) मिलता है। फिर \(y=\frac{113}{19}\) है। / Using (y=6x-19) from the first equation gives \(x=\frac{79}{19}\). Then \(y=\frac{113}{19}\).
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Which point lies on both lines (x-2y=-5) and (2x+y=13)?
A. बिंदु (\left\(\frac{21}{5},\frac{23}{5}\right\))Point (\left\(\frac{21}{5},\frac{23}{5}\right\))
दोनों समीकरण हल करने पर \(x=\frac{21}{5}\) और \(y=\frac{23}{5}\) मिलता है। यह बिंदु दोनों रेखाओं पर है। / Solving both equations gives \(x=\frac{21}{5}\) and \(y=\frac{23}{5}\). This point lies on both lines.
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Where do the lines (5x-y=13) and (x+2y=16) meet?
A. बिंदु (\left\(\frac{42}{11},\frac{67}{11}\right\))Point (\left\(\frac{42}{11},\frac{67}{11}\right\))
पहले समीकरण से (y=5x-13) रखकर \(x=\frac{42}{11}\) मिलता है। फिर \(y=\frac{67}{11}\) है। / Using (y=5x-13) from the first equation gives \(x=\frac{42}{11}\). Then \(y=\frac{67}{11}\).
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Which is the solution of (x-2y=-4) and (3x+y=11)?
B. बिंदु (\left\(3,2\right\))Point (\left\(3,2\right\))
(\left\(3,2\right\)) पर (3-2\left\(2\right\)=-1) है इसलिए यह नहीं है। सही हल (\left\(\frac{18}{7},\frac{23}{7}\right\)) है। / At (\left\(3,2\right\)), (3-2\left\(2\right\)=-1), so it is not correct. The correct solution is (\left\(\frac{18}{7},\frac{23}{7}\right\)).
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What is the intersection point of (2x-3y=1) and (x+y=7)?
B. बिंदु (\left\(4,3\right\))Point (\left\(4,3\right\))
(\left\(4,3\right\)) रखने पर (2\left\(4\right\)-3\left\(3\right\)=-1), इसलिए यह नहीं है। सही हल (\left\(\frac{22}{5},\frac{13}{5}\right\)) है। / Substituting (\left\(4,3\right\)) gives (2\left\(4\right\)-3\left\(3\right\)=-1), so it is not correct. The correct solution is (\left\(\frac{22}{5},\frac{13}{5}\right\)).
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Which is the graphical solution of (x+3y=15) and (2x-y=3)?
B. बिंदु (\left\(4,3\right\))Point (\left\(4,3\right\))
(\left\(4,3\right\)) पर (4+3\left\(3\right\)=13) नहीं है इसलिए विकल्प जाँचें। सही प्रतिच्छेद (\left\(\frac{24}{7},\frac{27}{7}\right\)) है। / At (\left\(4,3\right\)), (4+3\left\(3\right\)=13), so check options carefully. The correct intersection is (\left\(\frac{24}{7},\frac{27}{7}\right\)).
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What is the graphical solution of (4x-y=11) and (x+y=7)?
C. ( (4,3) )
( (4,3) ) रखने पर (4(4)-3=13) नहीं है, इसलिए जाँच जरूरी है। सही हल ( \left\(\frac{18}{5},\frac{17}{5}\right\) ) है। / Substituting ( (4,3) ) gives (4(4)-3=13), so checking is necessary. The correct solution is ( \left\(\frac{18}{5},\frac{17}{5}\right\) ).
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What is the correct intersection point of (2x+3y=18) and (x-y=1)?
A. ( (3,2) )
( (3,2) ) रखने पर (2(3)+3(2)=12) है, इसलिए यह भी सही नहीं है। सही हल ( \( \frac{21}{5},\frac{16}{5}\) ) होता है, अतः ऐसे विकल्पों में पुनः गणना जरूरी है। / Putting ( (3,2) ) gives (2(3)+3(2)=12), so it is not correct. The correct solution is ( \(\frac{21}{5},\frac{16}{5}\) ), so recalculation is needed in such options.
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