For what value of pand q respectively, do the given system of equations have infinitely many solutions?
3x+2y=7
(p+q)x+2py=28
Answers
Answer:
p = 4
q = 8
Step-by-step explanation:
Solution:-
• 3x + 2y = 7
a= 3 ,b = 2 ,c = 7
• (p+q)x + 2py = 28
a' = (p+q) , b' = 2py , c' = 28
Condition for Infinity Many Solution
a/a' = b/b' = c/c'
Now , Comparing we get
→ 3/(p+q) = 2/2p & 2/2p = 7/28
→ 6p = 2(p+q) & 14p = 56
→ 6p = 2(p+q) & p = 56/14
→ 6p = 2(p+q) & p = 4
Hence, p = 4
Substitute the value of p = 4 we get
→ 6×4 = 2(4+q)
→ 24 = 8 + 2q
→ 24-8 = 2q
→ 16 = 2q
→ q = 16/2
→ q = 8
Hence, the value of p = 4 & value of q = 8 .
p = 4
q = 8
Step-by-step explanation:
Solution:-
• 3x + 2y = 7
a= 3 ,b = 2 ,c = 7
• (p+q)x + 2py = 28
a' = (p+q) , b' = 2py , c' = 28
Condition for Infinity Many Solution
a/a' = b/b' = c/c'
Now , Comparing we get
→ 3/(p+q) = 2/2p & 2/2p = 7/28
→ 6p = 2(p+q) & 14p = 56
→ 6p = 2(p+q) & p = 56/14
→ 6p = 2(p+q) & p = 4
Hence, p = 4
Substitute the value of p = 4 we get
→ 6×4 = 2(4+q)
→ 24 = 8 + 2q
→ 24-8 = 2q
→ 16 = 2q
→ q = 16/2
→ q = 8
Hence, the value of p = 4 & value of q = 8 .
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