Search any question & find its solution
Question:
Answered & Verified by Expert
If the system of equations $2 x+3 y=7$ and $2 a x+(a+b) y=28$ has infinitely many solutions, then which one of the following is correct?
Options:
Solution:
2258 Upvotes
Verified Answer
The correct answer is:
$b=2 a$
The given system of equations has infinitely many solution, then
$\frac{2}{2 a}=\frac{3}{a+b}=\frac{7}{28}$
$\Rightarrow \quad a=4$ and $12=a+b$
$\Rightarrow \mathrm{a}=4$ and $b=8 \Rightarrow b=2 a$
$ALTERNATE SOLUTION$ : Given equations are $2 x+3 y=7$
$2 a x+(a+b) y=28$
Matrix form by these equations is $\left[\begin{array}{cc}2 & 3 \\ 2 a & (a+b)\end{array}\right]$
As we know if value of determinant is zero then system of equations have infinitely many solutions.
So, $\left|\begin{array}{cc}2 & 3 \\ 2 a & a+b\end{array}\right|=0$
$\Rightarrow 2 a+2 b-3 \times 2 a=0$
$\Rightarrow 2 a+2 b-6 a=0$
$\Rightarrow \quad 2 b-4 a=0 \Rightarrow b=2 a$
$\frac{2}{2 a}=\frac{3}{a+b}=\frac{7}{28}$
$\Rightarrow \quad a=4$ and $12=a+b$
$\Rightarrow \mathrm{a}=4$ and $b=8 \Rightarrow b=2 a$
$ALTERNATE SOLUTION$ : Given equations are $2 x+3 y=7$
$2 a x+(a+b) y=28$
Matrix form by these equations is $\left[\begin{array}{cc}2 & 3 \\ 2 a & (a+b)\end{array}\right]$
As we know if value of determinant is zero then system of equations have infinitely many solutions.
So, $\left|\begin{array}{cc}2 & 3 \\ 2 a & a+b\end{array}\right|=0$
$\Rightarrow 2 a+2 b-3 \times 2 a=0$
$\Rightarrow 2 a+2 b-6 a=0$
$\Rightarrow \quad 2 b-4 a=0 \Rightarrow b=2 a$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.