Question

If the system of equations $2x+3y=7$and $2ax+(a+b)y=28$ has infinitely many solutions, then ______.

• A. $a=2b$
• B. $b=2a$
• C. $a+2b=0$
• D. $2a+b=0$

Answer 1

The correct option is B

$b=2a$

Explanation for the correct option:

Step1: Write the given equations in standard form of linear equations in two variables

As we know that, if $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$ , where $a_1,b_1,c_1,a_2,b_2,c_2$ are all real numbers and $a_1^2+b_1^2\ne 0$ , $a_2^2+b_2^2\ne 0$, is called a pair of linear equations in two variables.

It is given that,

$2x+3y=7$

$2ax+\left(a+b\right)y=28$

Step 2: Compare the standard and given equations to find the value of coefficients

Comparing the equation $2x+3y=7$ with equation,

$a_{1}x+b_{1}y+c_2=0$.

Where,

$a_1=2$,

$b_1=3$

$c_1=-7$

Comparing $2ax+\left(a+b\right)y=28$ with the equation,

$a_{2}x+b_{2}y+c_2=0$.

Where,

$a_2=2a$,

$b_2=\left(a+b\right)$

$c_2=-28$

Step 3: Apply the condition for equations to have infinitely many solutions

For infinitely many solutions,

$\Rightarrow$ $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

$\Rightarrow$ $\frac{2}{2a}=\frac{3}{a+b}=\frac{7}{28}$

$\Rightarrow$ $\frac{2}{2a}=\frac{3}{a+b}$

$\Rightarrow$ $6a=2a+2b$

$\Rightarrow$ $4a=2b$

$\Rightarrow$ $2a=b$

So, If the system of equations $2x+3y=7$and $2ax+(a+b)y=28$ has infinitely many solutions, then $b$ is equal to $2a$.

Hence, the Option $B$ is correct answer.