linear-equations se related questions ko ek jagah revise karein. Har question me bilingual content, answer feedback aur explanation available hai.
1257 questions tagged with linear-equations.
What is the solution of (2(x+y)=34) and (3(x-y)=15)?
C. (x=11,\ y=6)
दिए समीकरण (x+y=17) और (x-y=5) बनते हैं। जोड़ने पर (x=11) और फिर (y=6)। / The equations become (x+y=17) and (x-y=5). Adding gives (x=11), then (y=6).
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If (3x-4y=-2) and (5x+4y=34), what is the value of (2x+y)?
A. \(\frac{23}{2}\)
दोनों समीकरण जोड़ने पर (8x=32), इसलिए (x=4)। फिर \(y=\frac{7}{2}\), अतः \(2x+y=\frac{23}{2}\)। / Adding both equations gives (8x=32), so (x=4). Then \(y=\frac{7}{2}\), hence \(2x+y=\frac{23}{2}\).
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D. (24)
लड़के (x) और लड़कियां (y) मानें। (x+y=68) और (2x+y=112) से (x=44), इसलिए (y=24)। / Let boys be (x) and girls be (y). From (x+y=68) and (2x+y=112), (x=44), so (y=24).
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For (5x+10y=20) and (kx+2y=7) to have no solution, what is the value of (k)?
B. (k=1)
पहला समीकरण (x+2y=4) बनता है। (k=1) पर दूसरा (x+2y=7) होगा, इसलिए कोई हल नहीं। / The first equation becomes (x+2y=4). At (k=1), the second becomes (x+2y=7), so there is no solution.
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If (x=3y-4) and (2x+5y=37), what is the value of (y)?
C. \(y=\frac{45}{11}\)
(x=3y-4) को दूसरे समीकरण में रखें। (6y-8+5y=37), इसलिए \(y=\frac{45}{11}\)। / Substitute (x=3y-4) in the second equation. (6y-8+5y=37), so \(y=\frac{45}{11}\).
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What is the value of (x+2y) from (7x-2y=39) and (3x+2y=21)?
A. (9)
दोनों समीकरण जोड़ने पर (10x=60), इसलिए (x=6)। फिर \(y=\frac{3}{2}\), अतः (x+2y=9)। / Adding both equations gives (10x=60), so (x=6). Then \(y=\frac{3}{2}\), hence (x+2y=9).
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If (4a+3b=276) and (2a+5b=250), what is the value of (a)?
D. (a=45)
दूसरे समीकरण को (2) से गुणा कर पहले से घटाएं। (b=32) और फिर (a=45) मिलता है। / Multiply the second equation by (2) and subtract the first. This gives (b=32), then (a=45).
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What is the value of (y) from (4x+9y=71) and (5x-3y=8)?
B. \(y=\frac{17}{3}\)
दूसरे समीकरण को (3) से गुणा कर पहले से जोड़ें। (x=5), फिर (4(5)+9y=71) से \(y=\frac{17}{3}\)। / Multiply the second equation by (3) and add the first. (x=5), then (4(5)+9y=71) gives \(y=\frac{17}{3}\).
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Which statement is correct about (2x+3y=11) and (4x+6y=21)?
C. कोई हल नहीं हैThere is no solution
पहले समीकरण का (2) गुना (4x+6y=22) है, लेकिन दूसरा (4x+6y=21) है। इसलिए कोई हल नहीं है। / Twice the first equation is (4x+6y=22), but the second is (4x+6y=21). Therefore there is no solution.
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What is the number of solutions of (2x+3y=11) and (4x+6y=22)?
A. अनंत हलInfinitely many solutions
दूसरा समीकरण पहले का (2) गुना है। इसलिए दोनों एक ही रेखा दर्शाते हैं और अनंत हल हैं। / The second equation is (2) times the first. Therefore both represent the same line and have infinitely many solutions.
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What is the value of (x-y) from \(\frac{x+2y}{3}=8\) and \(\frac{2x-y}{5}=3\)?
D. \(\frac{21}{5}\)
दिए समीकरण (x+2y=24) और (2x-y=15) बनते हैं। हल \(x=\frac{54}{5},\ y=\frac{33}{5}\), इसलिए \(x-y=\frac{21}{5}\)। / The equations become (x+2y=24) and (2x-y=15). The solution is \(x=\frac{54}{5},\ y=\frac{33}{5}\), so \(x-y=\frac{21}{5}\).
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If (x+y=15) and (2x-3y=10), what is the value of (3x+y)?
B. (37)
(x=15-y) रखने पर (30-5y=10), इसलिए (y=4) और (x=11)। अतः (3x+y=37)। / Using (x=15-y) gives (30-5y=10), so (y=4) and (x=11). Hence (3x+y=37).
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What will be the value of (y) from (5x+3y=14) and (2x-3y=7)?
C. \(y=-\frac{1}{3}\)
दोनों समीकरण जोड़ने पर (7x=21), इसलिए (x=3)। पहले समीकरण से \(y=-\frac{1}{3}\)। / Adding both equations gives (7x=21), so (x=3). From the first equation, \(y=-\frac{1}{3}\).
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If (px+2y=18) and (3x-y=5) have solution (x=4,\ y=7), what is the value of (p)?
A. (p=1)
(x=4,\ y=7) को (px+2y=18) में रखें। (4p+14=18), इसलिए (p=1)। / Put (x=4,\ y=7) in (px+2y=18). (4p+14=18), so (p=1).
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D. (375) वर्ग सेमी(375) square cm
(l+b=40) और (l-b=10) से (l=25,\ b=15)। इसलिए क्षेत्रफल \(25\times15=375\) वर्ग सेमी है। / From (l+b=40) and (l-b=10), (l=25,\ b=15). Therefore the area is \(25\times15=375\) square cm.
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What is the value of (x) from \(\frac{x}{5}+\frac{y}{2}=7\) and \(\frac{x}{3}-\frac{y}{4}=1\)?
B. \(x=\frac{135}{13}\)
हर हटाने पर (2x+5y=70) और (4x-3y=12) मिलते हैं। विलोपन से \(x=\frac{135}{13}\) मिलता है। / After clearing denominators, (2x+5y=70) and (4x-3y=12) are obtained. Elimination gives \(x=\frac{135}{13}\).
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On solving (8x-3y=31) and (2x+3y=29), what is (y)?
C. \(y=\frac{17}{3}\)
दोनों समीकरण जोड़ने पर (10x=60), इसलिए (x=6)। दूसरे समीकरण से \(y=\frac{17}{3}\)। / Adding both equations gives (10x=60), so (x=6). From the second equation, \(y=\frac{17}{3}\).
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What is the value of (x+y) from (0.2x+0.5y=3.1) and (0.4x-0.1y=1.3)?
A. \(x+y=\frac{97}{11}\)
दशमलव हटाने पर (2x+5y=31) और (4x-y=13) मिलते हैं। हल \(x=\frac{48}{11},\ y=\frac{49}{11}\), इसलिए \(x+y=\frac{97}{11}\)। / Removing decimals gives (2x+5y=31) and (4x-y=13). The solution is \(x=\frac{48}{11},\ y=\frac{49}{11}\), so \(x+y=\frac{97}{11}\).
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For (3x+ay=7) and (6x+8y=20) to have no solution, what should be the value of (a)?
D. (a=4)
गुणांक समानुपाती करने के लिए (3:6=a:8) होना चाहिए। इससे (a=4), और स्थिरांक (7:20) समान अनुपात में नहीं हैं। / To make coefficients proportional, (3:6=a:8) must hold. This gives (a=4), and constants (7:20) are not in the same ratio.
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What is the solution of (6x+5y=2) and (4x-5y=18)?
B. (x=2,\ y=-2)
दोनों समीकरण जोड़ने पर (10x=20), इसलिए (x=2)। फिर (6(2)+5y=2) से (y=-2)। / Adding both equations gives (10x=20), so (x=2). Then (6(2)+5y=2) gives (y=-2).
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C. (12) वर्ष(12) years
मान लें माता (x) और बेटी (y) है। (x+y=56) और (x+4=3(y+4)) से (y=12)। / Let the mother be (x) and the daughter be (y). From (x+y=56) and (x+4=3(y+4)), (y=12).
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If (4x+7y=53) and (4x-3y=13), what is the value of (x-y)?
A. \(x-y=\frac{9}{4}\)
दोनों समीकरण घटाने पर (10y=40), इसलिए (y=4)। फिर \(x=\frac{25}{4}\), अतः \(x-y=\frac{9}{4}\)। / Subtracting the equations gives (10y=40), so (y=4). Then \(x=\frac{25}{4}\), hence \(x-y=\frac{9}{4}\).
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In (5x-6y=11) and (3x+2y=9), by what should the second equation be multiplied to eliminate (y)?
D. (3)
दूसरे समीकरण को (3) से गुणा करने पर (6y) मिलेगा। यह पहले समीकरण के (-6y) के साथ हट जाएगा। / Multiplying the second equation by (3) gives (6y). It cancels with (-6y) in the first equation.
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What is the value of (y) from \(\frac{2x-3y}{5}=1\) and \(\frac{x+y}{2}=6\)?
B. \(y=\frac{19}{5}\)
दिए समीकरण (2x-3y=5) और (x+y=12) बनते हैं। प्रतिस्थापन से \(y=\frac{19}{5}\) मिलता है। / The equations become (2x-3y=5) and (x+y=12). Substitution gives \(y=\frac{19}{5}\).
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If (2x+ky=15) and (x-2y=1) have solution (x=5,\ y=2), what is the value of (k)?
C. \(k=\frac{5}{2}\)
(x=5,\ y=2) को (2x+ky=15) में रखें। (10+2k=15), इसलिए \(k=\frac{5}{2}\)। / Put (x=5,\ y=2) in (2x+ky=15). (10+2k=15), so \(k=\frac{5}{2}\).
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What is the value of (y) from (7x+2y=32) and (3x-4y=-6)?
A. \(y=\frac{69}{17}\)
पहले समीकरण को (2) से गुणा कर दूसरे से जोड़ें। \(x=\frac{58}{17}\) और \(y=\frac{69}{17}\)। / Multiply the first equation by (2) and add the second. \(x=\frac{58}{17}\) and \(y=\frac{69}{17}\).
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D. (63)
दहाई अंक (x) और इकाई अंक (y) मानें। (x+y=9) और (9(x-y)=27) से (x=6,\ y=3)। / Let the tens digit be (x) and units digit be (y). From (x+y=9) and (9(x-y)=27), (x=6,\ y=3).
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What is the value of (a) for (ax+4y=10) and (3x+6y=15) to have infinitely many solutions?
B. (a=2)
अनंत हल के लिए गुणांक और स्थिरांक समान अनुपात में होने चाहिए। (4:6=10:15=2:3), इसलिए (a=2)। / For infinitely many solutions, coefficients and constants must be in the same ratio. Since (4:6=10:15=2:3), (a=2).
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What is the solution of \(\frac{x}{4}+\frac{y}{6}=5\) and \(\frac{x}{2}-\frac{y}{3}=2\)?
C. (x=12,\ y=12)
हर हटाने पर (3x+2y=60) और (3x-2y=12) बनते हैं। जोड़ने पर (x=12) और (y=12)। / After clearing denominators, the equations become (3x+2y=60) and (3x-2y=12). Adding gives (x=12) and (y=12).
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What is the value of (x+y) from (4x-3y=7) and (5x+6y=44)?
A. \(x+y=\frac{105}{13}\)
पहले समीकरण को (2) से गुणा कर दूसरे से जोड़ें। \(x=\frac{58}{13}\) और \(y=\frac{47}{13}\), अतः \(x+y=\frac{105}{13}\)। / Multiply the first equation by (2) and add the second. \(x=\frac{58}{13}\) and \(y=\frac{47}{13}\), so \(x+y=\frac{105}{13}\).
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