यदि (x+y=15) और (2x-3y=10), तो (3x+y) का मान क्या है?
If (x+y=15) and (2x-3y=10), what is the value of (3x+y)?
#linear equations
#substitution
#expression value
#hard
#class 10
A (35)
B (37)
C (39)
D (41)
Explanation opens after your attempt
Step 1
Concept
Using (x=15-y) gives (30-5y=10), so (y=4) and (x=11). Hence (3x+y=37).
Step 2
Why this answer is correct
The correct answer is B. (37). Using (x=15-y) gives (30-5y=10), so (y=4) and (x=11). Hence (3x+y=37).
Step 3
Exam Tip
(x=15-y) रखने पर (30-5y=10), इसलिए (y=4) और (x=11)। अतः (3x+y=37)।
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यदि (px+2y=18) और (3x-y=5) का हल (x=4,\ y=7) है, तो (p) का मान क्या है?
If (px+2y=18) and (3x-y=5) have solution (x=4,\ y=7), what is the value of (p)?
#linear equations
#parameter
#substitution
#hard
#class 10
A (p=1)
B (p=2)
C (p=3)
D (p=4)
Explanation opens after your attempt
Step 1
Concept
Put (x=4,\ y=7) in (px+2y=18). (4p+14=18), so (p=1).
Step 2
Why this answer is correct
The correct answer is A. (p=1). Put (x=4,\ y=7) in (px+2y=18). (4p+14=18), so (p=1).
Step 3
Exam Tip
(x=4,\ y=7) को (px+2y=18) में रखें। (4p+14=18), इसलिए (p=1)।
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समीकरणों \(\frac{2x-3y}{5}=1\) और \(\frac{x+y}{2}=6\) से (y) का मान क्या है?
What is the value of (y) from \(\frac{2x-3y}{5}=1\) and \(\frac{x+y}{2}=6\)?
#linear equations
#transformed equations
#substitution
#hard
#class 10
A \(y=\frac{16}{5}\)
B \(y=\frac{19}{5}\)
C \(y=\frac{21}{5}\)
D \(y=\frac{24}{5}\)
Explanation opens after your attempt
Correct Answer
B. \(y=\frac{19}{5}\)
Step 1
Concept
The equations become (2x-3y=5) and (x+y=12). Substitution gives \(y=\frac{19}{5}\).
Step 2
Why this answer is correct
The correct answer is B. \(y=\frac{19}{5}\). The equations become (2x-3y=5) and (x+y=12). Substitution gives \(y=\frac{19}{5}\).
Step 3
Exam Tip
दिए समीकरण (2x-3y=5) और (x+y=12) बनते हैं। प्रतिस्थापन से \(y=\frac{19}{5}\) मिलता है।
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यदि (2x+ky=15) और (x-2y=1) का हल (x=5,\ y=2) है, तो (k) का मान क्या है?
If (2x+ky=15) and (x-2y=1) have solution (x=5,\ y=2), what is the value of (k)?
#linear equations
#parameter
#substitution
#hard
#class 10
A (k=2)
B (k=3)
C \(k=\frac{5}{2}\)
D \(k=\frac{7}{2}\)
Explanation opens after your attempt
Correct Answer
C. \(k=\frac{5}{2}\)
Step 1
Concept
Put (x=5,\ y=2) in (2x+ky=15). (10+2k=15), so \(k=\frac{5}{2}\).
Step 2
Why this answer is correct
The correct answer is C. \(k=\frac{5}{2}\). Put (x=5,\ y=2) in (2x+ky=15). (10+2k=15), so \(k=\frac{5}{2}\).
Step 3
Exam Tip
(x=5,\ y=2) को (2x+ky=15) में रखें। (10+2k=15), इसलिए \(k=\frac{5}{2}\)।
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यदि (3(x+y)+2(x-y)=31) और (2(x+y)-(x-y)=13), तो (x) का मान क्या है?
If (3(x+y)+2(x-y)=31) and (2(x+y)-(x-y)=13), what is the value of (x)?
#linear equations
#transformation
#substitution
#hard
#class 10
A \(x=\frac{34}{7}\)
B \(x=\frac{40}{7}\)
C \(x=\frac{44}{7}\)
D \(x=\frac{47}{7}\)
Explanation opens after your attempt
Correct Answer
B. \(x=\frac{40}{7}\)
Step 1
Concept
Let (x+y=s) and (x-y=d). Solving gives \(s=\frac{57}{7}\) and \(d=\frac{23}{7}\), so \(x=\frac{40}{7}\).
Step 2
Why this answer is correct
The correct answer is B. \(x=\frac{40}{7}\). Let (x+y=s) and (x-y=d). Solving gives \(s=\frac{57}{7}\) and \(d=\frac{23}{7}\), so \(x=\frac{40}{7}\).
Step 3
Exam Tip
(x+y=s) और (x-y=d) मानें। हल करने पर \(s=\frac{57}{7}\) और \(d=\frac{23}{7}\), इसलिए \(x=\frac{40}{7}\)।
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समीकरणों \(\frac{x}{3}+\frac{y}{5}=4\) और (x-y=6) से (x) का मान क्या है?
What is the value of (x) from \(\frac{x}{3}+\frac{y}{5}=4\) and (x-y=6)?
#linear equations
#fraction equations
#substitution
#hard
#class 10
A \(x=\frac{39}{4}\)
B \(x=\frac{37}{4}\)
C \(x=\frac{41}{4}\)
D \(x=\frac{35}{4}\)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{39}{4}\)
Step 1
Concept
Multiply the first equation by (15) to get (5x+3y=60). Using (x=y+6) gives \(x=\frac{39}{4}\).
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{39}{4}\). Multiply the first equation by (15) to get (5x+3y=60). Using (x=y+6) gives \(x=\frac{39}{4}\).
Step 3
Exam Tip
पहले समीकरण को (15) से गुणा कर (5x+3y=60) बनाएं। (x=y+6) रखने पर \(x=\frac{39}{4}\)।
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यदि (2x+my=34) का हल (x=7,\ y=4) है, तो (m) का मान क्या होगा?
If (x=7,\ y=4) is a solution of (2x+my=34), what will be the value of (m)?
#linear equations
#parameter
#substitution
#hard
#class 10
A (m=4)
B (m=5)
C (m=6)
D (m=7)
Explanation opens after your attempt
Step 1
Concept
Substitute (x=7,\ y=4) in the equation. (14+4m=34), so (m=5).
Step 2
Why this answer is correct
The correct answer is B. (m=5). Substitute (x=7,\ y=4) in the equation. (14+4m=34), so (m=5).
Step 3
Exam Tip
(x=7,\ y=4) को समीकरण में रखें। (14+4m=34), इसलिए (m=5)।
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समीकरणों (4x+3y=50) और (2x-5y=-6) को हल करने पर (y) का मान क्या है?
On solving (4x+3y=50) and (2x-5y=-6), what is the value of (y)?
#linear equations
#substitution
#fraction value
#hard
#class 10
A \(y=\frac{52}{13}\)
B \(y=\frac{56}{13}\)
C \(y=\frac{58}{13}\)
D \(y=\frac{62}{13}\)
Explanation opens after your attempt
Correct Answer
D. \(y=\frac{62}{13}\)
Step 1
Concept
Use \(x=\frac{5y-6}{2}\) from the second equation. Substitution gives (13y=62), so \(y=\frac{62}{13}\).
Step 2
Why this answer is correct
The correct answer is D. \(y=\frac{62}{13}\). Use \(x=\frac{5y-6}{2}\) from the second equation. Substitution gives (13y=62), so \(y=\frac{62}{13}\).
Step 3
Exam Tip
दूसरे समीकरण से \(x=\frac{5y-6}{2}\) रखें। पहले में रखने पर (13y=62), इसलिए \(y=\frac{62}{13}\)।
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कौन-सा क्रमित युग्म (5x+2y=41) और (3x-y=10) को संतुष्ट करता है?
Which ordered pair satisfies (5x+2y=41) and (3x-y=10)?
#linear equations
#substitution
#ordered pair
#hard
#class 10
A \(x=\frac{55}{11},\ y=\frac{80}{11}\)
B \(x=\frac{60}{11},\ y=\frac{70}{11}\)
C \(x=\frac{65}{11},\ y=\frac{68}{11}\)
D \(x=\frac{61}{11},\ y=\frac{73}{11}\)
Explanation opens after your attempt
Correct Answer
D. \(x=\frac{61}{11},\ y=\frac{73}{11}\)
Step 1
Concept
Use (y=3x-10) from the second equation. Substitution gives (11x=61), so \(x=\frac{61}{11},\ y=\frac{73}{11}\).
Step 2
Why this answer is correct
The correct answer is D. \(x=\frac{61}{11},\ y=\frac{73}{11}\). Use (y=3x-10) from the second equation. Substitution gives (11x=61), so \(x=\frac{61}{11},\ y=\frac{73}{11}\).
Step 3
Exam Tip
दूसरे समीकरण से (y=3x-10) रखें। पहले में रखने पर (11x=61), इसलिए \(x=\frac{61}{11},\ y=\frac{73}{11}\)।
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यदि (kx+3y=25) और (x-y=2) का हल (x=5,\ y=3) है, तो (k) का मान क्या है?
If (kx+3y=25) and (x-y=2) have solution (x=5,\ y=3), what is the value of (k)?
#linear equations
#parameter
#substitution
#hard
#class 10
A \(k=\frac{16}{5}\)
B \(k=\frac{14}{5}\)
C \(k=\frac{18}{5}\)
D (k=4)
Explanation opens after your attempt
Correct Answer
A. \(k=\frac{16}{5}\)
Step 1
Concept
Substitute the given solution in (kx+3y=25). (5k+9=25), so \(k=\frac{16}{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(k=\frac{16}{5}\). Substitute the given solution in (kx+3y=25). (5k+9=25), so \(k=\frac{16}{5}\).
Step 3
Exam Tip
दिए हल को (kx+3y=25) में रखें। (5k+9=25), इसलिए \(k=\frac{16}{5}\)।
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समीकरणों \(\frac{x}{2}+\frac{y}{3}=7\) और \(\frac{x}{3}-\frac{y}{2}=1\) का हल क्या है?
What is the solution of \(\frac{x}{2}+\frac{y}{3}=7\) and \(\frac{x}{3}-\frac{y}{2}=1\)?
#linear equations
#fraction equations
#substitution
#elimination
#hard
#class 10
A \(x=\frac{66}{13},\ y=\frac{138}{13}\)
B \(x=\frac{138}{13},\ y=\frac{66}{13}\)
C \(x=\frac{132}{13},\ y=\frac{72}{13}\)
D \(x=\frac{144}{13},\ y=\frac{60}{13}\)
Explanation opens after your attempt
Correct Answer
B. \(x=\frac{138}{13},\ y=\frac{66}{13}\)
Step 1
Concept
Clear the denominators using the LCM first. The solution is \(x=\frac{138}{13},\ y=\frac{66}{13}\).
Step 2
Why this answer is correct
The correct answer is B. \(x=\frac{138}{13},\ y=\frac{66}{13}\). Clear the denominators using the LCM first. The solution is \(x=\frac{138}{13},\ y=\frac{66}{13}\).
Step 3
Exam Tip
हरों का लघुत्तम समापवर्त्य लेकर समीकरणों को सरल करें। हल \(x=\frac{138}{13},\ y=\frac{66}{13}\) मिलता है।
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समीकरणों (x+3y=21) और (3x-y=11) को हल करने पर (2x+y) का मान क्या है?
On solving (x+3y=21) and (3x-y=11), what is the value of (2x+y)?
#linear-equations
#substitution
#expression-value
#medium
#class-10
A (14)
B (13)
C (12)
D (11)
Explanation opens after your attempt
Step 1
Concept
Use (y=3x-11) from the second equation. Substitution gives \(x=\frac{27}{5},\ y=\frac{16}{5}\), so (2x+y=14).
Step 2
Why this answer is correct
The correct answer is A. (14). Use (y=3x-11) from the second equation. Substitution gives \(x=\frac{27}{5},\ y=\frac{16}{5}\), so (2x+y=14).
Step 3
Exam Tip
दूसरे समीकरण से (y=3x-11) रखें। पहले में रखने पर \(x=\frac{27}{5},\ y=\frac{16}{5}\), इसलिए (2x+y=14)।
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समीकरणों (6x+5y=47) और (2x-y=5) से (y) का मान क्या है?
What is the value of (y) from (6x+5y=47) and (2x-y=5)?
#linear-equations
#substitution
#value-of-y
#medium
#class-10
A (y=2)
B (y=3)
C (y=5)
D (y=4)
Explanation opens after your attempt
Step 1
Concept
Use (y=2x-5) from the second equation. Substitution gives (16x=72), so \(x=\frac{9}{2}\) and (y=4).
Step 2
Why this answer is correct
The correct answer is D. (y=4). Use (y=2x-5) from the second equation. Substitution gives (16x=72), so \(x=\frac{9}{2}\) and (y=4).
Step 3
Exam Tip
दूसरे समीकरण से (y=2x-5) रखें। पहले में रखने पर (16x=72), इसलिए \(x=\frac{9}{2}\) और (y=4)।
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यदि (x+4y=26) और (3x-2y=8), तो सही हल कौन-सा है?
If (x+4y=26) and (3x-2y=8), which is the correct solution?
#linear-equations
#substitution
#solution-pair
#medium
#class-10
A (x=6,\ y=5)
B (x=5,\ y=6)
C (x=8,\ y=4)
D (x=4,\ y=7)
Explanation opens after your attempt
Correct Answer
A. (x=6,\ y=5)
Step 1
Concept
Use (x=26-4y) from the first equation. Substitution gives (14y=70), so (y=5,\ x=6).
Step 2
Why this answer is correct
The correct answer is A. (x=6,\ y=5). Use (x=26-4y) from the first equation. Substitution gives (14y=70), so (y=5,\ x=6).
Step 3
Exam Tip
पहले समीकरण से (x=26-4y) रखें। दूसरे में रखने पर (14y=70), इसलिए (y=5,\ x=6)।
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यदि (x=3,\ y=2) समीकरण (2x+ky=16) को संतुष्ट करता है, तो (k) का मान क्या है?
If (x=3,\ y=2) satisfies (2x+ky=16), what is the value of (k)?
#linear-equations
#parameter
#substitution
#medium
#class-10
A (k=3)
B (k=4)
C (k=6)
D (k=5)
Explanation opens after your attempt
Step 1
Concept
Substituting (x=3,\ y=2) gives (6+2k=16). Therefore (k=5).
Step 2
Why this answer is correct
The correct answer is D. (k=5). Substituting (x=3,\ y=2) gives (6+2k=16). Therefore (k=5).
Step 3
Exam Tip
(x=3,\ y=2) रखने पर (6+2k=16) मिलता है। इसलिए (k=5)।
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यदि (3x+y=22) और (x+2y=19), तो (x-y) का मान क्या है?
If (3x+y=22) and (x+2y=19), what is the value of (x-y)?
#linear-equations
#substitution
#expression-value
#medium
#class-10
A (2)
B (0)
C (-1)
D (-2)
Explanation opens after your attempt
Step 1
Concept
Use (y=22-3x) from the first equation. Substitution gives (x=5,\ y=7), so (x-y=-2).
Step 2
Why this answer is correct
The correct answer is D. (-2). Use (y=22-3x) from the first equation. Substitution gives (x=5,\ y=7), so (x-y=-2).
Step 3
Exam Tip
पहले समीकरण से (y=22-3x) रखें। दूसरे में रखने पर (x=5,\ y=7), इसलिए (x-y=-2)।
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समीकरणों (4x-7y=-19) और (2x+y=13) से (x) का मान ज्ञात कीजिए।
Find the value of (x) from (4x-7y=-19) and (2x+y=13).
#linear-equations
#substitution
#value-of-x
#medium
#class-10
A (x=3)
B (x=4)
C (x=5)
D (x=6)
Explanation opens after your attempt
Step 1
Concept
Use (y=13-2x) from the second equation. Substitution gives (18x=72), so (x=4).
Step 2
Why this answer is correct
The correct answer is B. (x=4). Use (y=13-2x) from the second equation. Substitution gives (18x=72), so (x=4).
Step 3
Exam Tip
दूसरे समीकरण से (y=13-2x) रखें। पहले में रखने पर (18x=72), इसलिए (x=4)।
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समीकरणों \(\frac{x}{2}+\frac{y}{3}=5\) और (x-y=3) का हल क्या है?
What is the solution of \(\frac{x}{2}+\frac{y}{3}=5\) and (x-y=3)?
#linear-equations
#fraction-equation
#substitution
#medium
#class-10
A \(x=\frac{36}{5},\ y=\frac{21}{5}\)
B \(x=\frac{21}{5},\ y=\frac{36}{5}\)
C (x=6,\ y=3)
D (x=7,\ y=4)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{36}{5},\ y=\frac{21}{5}\)
Step 1
Concept
Multiplying the first equation by (6) gives (3x+2y=30). Using (x=y+3) gives \(y=\frac{21}{5}\) and \(x=\frac{36}{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{36}{5},\ y=\frac{21}{5}\). Multiplying the first equation by (6) gives (3x+2y=30). Using (x=y+3) gives \(y=\frac{21}{5}\) and \(x=\frac{36}{5}\).
Step 3
Exam Tip
पहले समीकरण को (6) से गुणा करने पर (3x+2y=30) मिलता है। (x=y+3) रखने पर \(y=\frac{21}{5}\) और \(x=\frac{36}{5}\)।
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यदि (3x+my=23) और (x-y=1) का हल (x=5,\ y=4) है, तो (m) का मान क्या है?
If (3x+my=23) and (x-y=1) have solution (x=5,\ y=4), what is the value of (m)?
#linear-equations
#parameter
#substitution
#medium
#class-10
A (m=2)
B (m=3)
C (m=4)
D (m=5)
Explanation opens after your attempt
Step 1
Concept
Put (x=5,\ y=4) in (3x+my=23). Then (15+4m=23), so (m=2).
Step 2
Why this answer is correct
The correct answer is A. (m=2). Put (x=5,\ y=4) in (3x+my=23). Then (15+4m=23), so (m=2).
Step 3
Exam Tip
(x=5,\ y=4) को (3x+my=23) में रखें। (15+4m=23), इसलिए (m=2)।
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यदि (x=2y+1) और (5x-3y=33), तो सही हल कौन-सा है?
If (x=2y+1) and (5x-3y=33), which is the correct solution?
#linear-equations
#substitution
#solution-pair
#medium
#class-10
A (x=7,\ y=3)
B (x=9,\ y=4)
C (x=11,\ y=5)
D (x=13,\ y=6)
Explanation opens after your attempt
Correct Answer
B. (x=9,\ y=4)
Step 1
Concept
Substituting (x=2y+1) gives (10y+5-3y=33). This gives (y=4) and (x=9).
Step 2
Why this answer is correct
The correct answer is B. (x=9,\ y=4). Substituting (x=2y+1) gives (10y+5-3y=33). This gives (y=4) and (x=9).
Step 3
Exam Tip
(x=2y+1) रखने पर (10y+5-3y=33) मिलता है। इससे (y=4) और (x=9)।
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यदि (2x+y=23) और (x+3y=19), तो (x-2y) का मान क्या है?
If (2x+y=23) and (x+3y=19), what is the value of (x-2y)?
#linear-equations
#substitution
#expression-value
#medium
#class-10
A (2)
B (3)
C (5)
D (4)
Explanation opens after your attempt
Step 1
Concept
Use (y=23-2x) from the first equation. Substitution gives (x=10,\ y=3), so (x-2y=4).
Step 2
Why this answer is correct
The correct answer is D. (4). Use (y=23-2x) from the first equation. Substitution gives (x=10,\ y=3), so (x-2y=4).
Step 3
Exam Tip
पहले समीकरण से (y=23-2x) रखें। दूसरे में रखने पर (x=10,\ y=3), इसलिए (x-2y=4)।
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समीकरणों (4x+5y=39) और (2x-y=9) का हल कौन-सा है?
Which is the solution of (4x+5y=39) and (2x-y=9)?
#linear-equations
#substitution
#solution-pair
#medium
#class-10
A (x=6,\ y=3)
B (x=5,\ y=4)
C (x=7,\ y=2)
D (x=4,\ y=5)
Explanation opens after your attempt
Correct Answer
A. (x=6,\ y=3)
Step 1
Concept
Use (y=2x-9) from the second equation. Substitution gives (14x=84), so (x=6,\ y=3).
Step 2
Why this answer is correct
The correct answer is A. (x=6,\ y=3). Use (y=2x-9) from the second equation. Substitution gives (14x=84), so (x=6,\ y=3).
Step 3
Exam Tip
दूसरे समीकरण से (y=2x-9) रखें। पहले में रखने पर (14x=84), इसलिए (x=6,\ y=3)।
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समीकरणों (2x+5y=0) और (3x-y=17) को हल करने पर (y) कितना है?
On solving (2x+5y=0) and (3x-y=17), what is (y)?
#linear-equations
#substitution
#negative-value
#medium
#class-10
A (y=2)
B (y=-1)
C (y=1)
D (y=-2)
Explanation opens after your attempt
Step 1
Concept
Use (y=3x-17) from the second equation. Substitution gives (17x=85), so (x=5,\ y=-2).
Step 2
Why this answer is correct
The correct answer is D. (y=-2). Use (y=3x-17) from the second equation. Substitution gives (17x=85), so (x=5,\ y=-2).
Step 3
Exam Tip
दूसरे समीकरण से (y=3x-17) रखें। पहले में रखने पर (17x=85), इसलिए (x=5,\ y=-2)।
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यदि (3x+2y=25) और (x-y=1), तो (x+y) का मान क्या है?
If (3x+2y=25) and (x-y=1), what is the value of (x+y)?
#linear-equations
#substitution
#expression-value
#medium
#class-10
A (8)
B (9)
C \(\frac{49}{5}\)
D \(\frac{51}{5}\)
Explanation opens after your attempt
Correct Answer
C. \(\frac{49}{5}\)
Step 1
Concept
Using (x=y+1) gives (5y+3=25), so \(y=\frac{22}{5}\) and \(x=\frac{27}{5}\). Hence \(x+y=\frac{49}{5}\).
Step 2
Why this answer is correct
The correct answer is C. \(\frac{49}{5}\). Using (x=y+1) gives (5y+3=25), so \(y=\frac{22}{5}\) and \(x=\frac{27}{5}\). Hence \(x+y=\frac{49}{5}\).
Step 3
Exam Tip
(x=y+1) रखने पर (5y+3=25), इसलिए \(y=\frac{22}{5}\) और \(x=\frac{27}{5}\)। अतः \(x+y=\frac{49}{5}\)।
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समीकरणों (2x-y=6) और (x+3y=13) का हल क्या है?
What is the solution of (2x-y=6) and (x+3y=13)?
#linear-equations
#substitution
#fraction-solution
#medium
#class-10
A \(x=\frac{20}{7},\ y=\frac{31}{7}\)
B \(x=\frac{30}{7},\ y=\frac{19}{7}\)
C \(x=\frac{32}{7},\ y=\frac{18}{7}\)
D \(x=\frac{31}{7},\ y=\frac{20}{7}\)
Explanation opens after your attempt
Correct Answer
D. \(x=\frac{31}{7},\ y=\frac{20}{7}\)
Step 1
Concept
Use (y=2x-6) from the first equation. Substitution gives (7x=31), so \(x=\frac{31}{7},\ y=\frac{20}{7}\).
Step 2
Why this answer is correct
The correct answer is D. \(x=\frac{31}{7},\ y=\frac{20}{7}\). Use (y=2x-6) from the first equation. Substitution gives (7x=31), so \(x=\frac{31}{7},\ y=\frac{20}{7}\).
Step 3
Exam Tip
पहले समीकरण से (y=2x-6) रखें। दूसरे में रखने पर (7x=31), इसलिए \(x=\frac{31}{7},\ y=\frac{20}{7}\)।
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यदि (kx+2y=20) और (x+y=8) का हल (x=4,\ y=4) है, तो (k) का मान क्या है?
If (kx+2y=20) and (x+y=8) have solution (x=4,\ y=4), what is the value of (k)?
#linear-equations
#parameter
#substitution
#medium
#class-10
A (k=3)
B (k=4)
C (k=5)
D (k=6)
Explanation opens after your attempt
Step 1
Concept
Substituting (x=4,\ y=4) gives (4k+8=20). Therefore (k=3).
Step 2
Why this answer is correct
The correct answer is A. (k=3). Substituting (x=4,\ y=4) gives (4k+8=20). Therefore (k=3).
Step 3
Exam Tip
(x=4,\ y=4) रखने पर (4k+8=20) मिलता है। इसलिए (k=3)।
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समीकरणों (4x+3y=26) और (2x-y=8) का हल क्या है?
What is the solution of (4x+3y=26) and (2x-y=8)?
#linear-equations
#substitution
#solution-pair
#medium
#class-10
A (x=5,\ y=2)
B (x=4,\ y=3)
C (x=6,\ y=1)
D (x=3,\ y=4)
Explanation opens after your attempt
Correct Answer
A. (x=5,\ y=2)
Step 1
Concept
Use (y=2x-8) from the second equation. Substitution gives (10x=50), so (x=5,\ y=2).
Step 2
Why this answer is correct
The correct answer is A. (x=5,\ y=2). Use (y=2x-8) from the second equation. Substitution gives (10x=50), so (x=5,\ y=2).
Step 3
Exam Tip
दूसरे समीकरण से (y=2x-8) रखें। पहले में रखने पर (10x=50), इसलिए (x=5,\ y=2)।
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यदि (6x-y=21) और (2x+3y=17), तो सही हल कौन-सा है?
If (6x-y=21) and (2x+3y=17), which is the correct solution?
#linear-equations
#substitution
#solution-pair
#medium
#class-10
A (x=3,\ y=4)
B (x=5,\ y=1)
C (x=4,\ y=3)
D (x=2,\ y=5)
Explanation opens after your attempt
Correct Answer
C. (x=4,\ y=3)
Step 1
Concept
From the first equation use (y=6x-21). Substituting in the second gives (20x=80), so (x=4,\ y=3).
Step 2
Why this answer is correct
The correct answer is C. (x=4,\ y=3). From the first equation use (y=6x-21). Substituting in the second gives (20x=80), so (x=4,\ y=3).
Step 3
Exam Tip
पहले समीकरण से (y=6x-21) रखें। दूसरे में रखने पर (20x=80), इसलिए (x=4,\ y=3)।
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समीकरणों (3x+5y=31) और (x+y=9) को हल करने पर (x) का मान क्या है?
On solving (3x+5y=31) and (x+y=9), what is the value of (x)?
#linear-equations
#substitution
#value-of-x
#medium
#class-10
A (x=7)
B (x=6)
C (x=5)
D (x=4)
Explanation opens after your attempt
Step 1
Concept
Using (x=9-y) gives (27-3y+5y=31). Thus (y=2) and (x=7).
Step 2
Why this answer is correct
The correct answer is A. (x=7). Using (x=9-y) gives (27-3y+5y=31). Thus (y=2) and (x=7).
Step 3
Exam Tip
(x=9-y) रखने पर (27-3y+5y=31) मिलता है। इसलिए (y=2) और (x=7)।
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समीकरणों (2x+3y=19) और (4x-y=3) का हल क्या है?
What is the solution of (2x+3y=19) and (4x-y=3)?
#linear-equations
#substitution
#solution-pair
#medium
#class-10
A (x=3,\ y=4)
B (x=2,\ y=5)
C (x=5,\ y=2)
D (x=1,\ y=6)
Explanation opens after your attempt
Correct Answer
B. (x=2,\ y=5)
Step 1
Concept
Use (y=4x-3) from the second equation. Substitution gives (14x=28), so (x=2,\ y=5).
Step 2
Why this answer is correct
The correct answer is B. (x=2,\ y=5). Use (y=4x-3) from the second equation. Substitution gives (14x=28), so (x=2,\ y=5).
Step 3
Exam Tip
दूसरे समीकरण से (y=4x-3) रखें। पहले में रखने पर (14x=28), इसलिए (x=2,\ y=5)।
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