What is the value of k,for Which the simultaneous equation 3x-(k+1)y=20 and (k+2)x-10y=40 have infinitely many solutions?
A)7
B)-7
C)4
D)-4
Answers
Answer:
Step-by-step explanation:
=>Condition for infinitely many Solutions-
=>\frac{a_1}{a_2}
a
2
a
1
=\frac{b_1}{b_2}
b
2
b
1
=\frac{c_1}{c_2}
c
2
c
1
=>\frac{3}{k+2}
k+2
3
=\frac{k+1}{10}
10
k+1
=\frac{20}{40}
40
20
1.On taking first and second -
=>\frac{3}{k+2}
k+2
3
=\frac{k+1}{10}
10
k+1
=>30=(k+1)(k+2)30=(k+1)(k+2)
=>30=k^2+3k+230=k
2
+3k+2
=>k^2+3k-28=0k
2
+3k−28=0
=>k^2+7k-4k-28=0k
2
+7k−4k−28=0
=>k(k+7)-4(k+7)=0k(k+7)−4(k+7)=0
=>(k-4)(k+7)=0(k−4)(k+7)=0
K=4 and K=-7
2.Now taking second and third-
=>\frac{k+1}{10}
10
k+1
=\frac{20}{40}
40
20
=>\frac{k+1}{10}
10
k+1
=\frac{1}{2}
2
1
=>2k+2=102k+2=10
=>K=\frac{8}{2}
2
8
=>K=4K=4
Since K=4 is common in both solutions-
\huge\orange{\fbox{\pink{\text{K=4}}}}
K=4
COPIED