if the pair of linear equations 2x + 3y = 7
and 2px + py = 28-qy has infinitely many
solutions, find the values of p and q.
Share
if the pair of linear equations 2x + 3y = 7
and 2px + py = 28-qy has infinitely many
solutions, find the values of p and q.
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Verified answer
Answer:p=4 and q=8
Step-by-step explanation:
2x+3y = 7
2x+3y-7=0
2px+py=28-qy
2px+py+qy=28
2px+(p+q)y-28=0
HERE:
a1=2 , b1=3 , c1=(-7)
a2=2p , b2=(p+q) , c2= (-28)
for infinite many solutions,
a1/a2 = b1/b2 = c1/c2
therefore,
2/2p = 3/(p+q) = (-7)/(-28)
2/2p = (-7)/(-28)
2×(-28) =(-7)(2p)
(-56) =(-14p)
56=14p
56/14 =p
4=p
p=4. (1)
3/(p+q)=(-7)/(-28)
3/(4+q)=7/28. [from (1)]
3/(4+q)=1/4
3×4=4+q
12=4+q
12-4=q
8=q
q=8