Question

If the system of equations $3x+4y=2$ and $(a+b)x+2(a-b)y=5a-1$ has infinitely many solutions then $a$ & $b$ satisfy the equation

• A. $a-5b=0$
• B. $5a-b=0$
• C. $a+5b=0$
• D. $5a+b=0$

Answer 1

The correct option is A $a-5b=0$

$3x+4y=2$ and $(a+b)x+2(a-b)y=5a-1$ have infinite solutions,
then,
$a+b=3k$ (1)
$2(a-b)=4k$ (2)
Multiply (1) by 4 and (2) by 3 and subtract, we get,
$4a+4b-6(a-b)=0$
$4a+4b-6a+6b=0$
$2a=10b$
$a=5b$
or $a-5b=0$