Question

Find the value of a and b for which the given system of linear equations has an infinite number of solutions.

$(a+b)x-2by=5a+2b+1$ and $3x-y=14$

• A. $a=2,b=-9$
• B. $a=7,b=-13$
• C. $a=0,b=4$
• D. $a=5,b=1$

Answer 1

The correct option is D $a=5,b=1$

Comparing $(a+b)x-2by=5a+2b+1$ with $a_{1}x+b_{1}y+c_1=0$,

we get,

$a_1=a+b,b_1=-2b,c_1=-(5a+2b+1)$

Comparing $3x-y=14$ with $a_{2}x+b_{2}y+c_2=0$,

we get,

$a_2=3,b_2=-1,c_2=-14$

Given system of equations have infinite solution
$\Rightarrow \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

$\Rightarrow \frac{(a+b)}{3}=\frac{-2b}{-1}=\frac{-(5a+2b+1)}{-14}$

Now from $\frac{a_1}{a_2}=\frac{b_1}{b_2}$

$\Rightarrow \frac{(a+b)}{3}=\frac{-2b}{-1}$

$\Rightarrow a+b=6b$

$\Rightarrow a-5b=0.....\left(3\right)$

Now from $\frac{b_1}{b_2}=\frac{c_1}{c_2}$

$\Rightarrow \frac{-2b}{-1}=\frac{-(5a+2b+1)}{-14}$

$\Rightarrow 28b=5a+2b+1$

$\Rightarrow 5a+2b-28b=-1$

$\Rightarrow 5a-26b=-1......\left(4\right)$

Now, for subtracting equation 3 and 4

$\Rightarrow a-5b=0$

$-5a-(-26b)=-(-1)$

$-5a+26b=1$

Now will have:

$\Rightarrow a-5b=0$

$-5a+26b=1$

Now multiplying eq3 from 5

$\Rightarrow 5a-25b=0$

$-5a+26b=1$

$\Rightarrow b=1$

Now substituting $b=1$ in equation3 will get

$a-5\left(1\right)=0$

$\Rightarrow a-5=0$

$\Rightarrow a=5$

$\Rightarrow a=5,b=1$