if the following system equations have infinitely many solutions value off k will be2x+3y=5 and 4x+ky=10
Answers
Step-by-step explanation:
SOLUTION:-
Given Equation:
2x +3y= 5.........(1)
4x+ ky=10........(2)
⚫Both the equation are in the form of:
a1x + b1y= c1 & a2x + b2y= c2
Where,
a1 & a2 are the coefficient of x
b1 & b2 are the coefficient of y
c1 & c2 are the constants.
⚫For the system of linear equations to have infinitely many solutions we must have:
\frac{a1}{a2} = \frac{b1}{b2} = \frac{c1}{c2} ..............(3)
a2
a1
=
b2
b1
=
c2
c1
..............(3)
According to the question:
a1=2
a2=4
b1=3
b2=k
c1=5
c2=10
Putting the above values in equation (3) & solving the extreme left and extreme right portion of the equality, we get:
\begin{gathered}\frac{2}{4} = \frac{3}{k} \\ \\ = > 2k = 12 \\ \\ = > k = \frac{12}{2} \\ \\ = > k = 6\end{gathered}
4
2
=
k
3
=>2k=12
=>k=
2
12
=>k=6
The value of k for which the system of equations has infinitely many solution is k= 6.
Answer:
Step-by-step explanation:
Given equations are 2x+3y−5=0 and 4x+ky−10=0
Then 2x+3y−5=0⇒2x+3y=5
4x+ky−10⇒4x+ky=10
Therefore ,
2/4=3/k=5/10 [Since, they have an infinite number of solutions]
1/2=3/k=1/2
1/2=3/k
⇒k=6