Find the values of $ a $ and $ b $ for which the following system of equations has infinitely many solutions:
$3x+4y=12$
$(a+b)x+2(a-b)y=5a-1$

Given: 

The given system of equations is:

$3x+4y=12$
$(a+b)x+2(a-b)y=5a-1$

To do: 

We have to determine the values of $a$ and $b$ so that the given system of equations has infinitely many solutions.

Solution:

The given system of equations can be written as:

$3x+4y-12=0$
$(a+b)x+2(a-b)y-(5a-1)=0$

The standard form of system of equations of two variables is $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y-c_{2}=0$.

Comparing the given system of equations with the standard form of equations, we have,

$a_1=3, b_1=4, c_1=-12$ and $a_2=(a+b), b_2=2(a-b), c_2=-(5a-1)$

The condition for which the given system of equations has infinitely many solutions is

$\frac{a_{1}}{a_{2}} \ =\frac{b_{1}}{b_{2}} =\frac{c_{1}}{c_{2}} \ $

$\frac{3}{a+b}=\frac{4}{2(a-b)}=\frac{-12}{-(5a-1)}$

$\frac{3}{a+b}=\frac{2}{a-b}=\frac{12}{5a-1}$

$\frac{3}{a+b}=\frac{12}{5a-1}$ and $\frac{2}{a-b}=\frac{12}{5a-1}$

$3\times(5a-1)=12\times(a+b)$ and $2\times(5a-1)=12\times(a-b)$

$15a-3=12a+12b$ and $10a-2=12a-12b$

$15a-12a=12b+3$ and $12a-10a=12b-2$

$3a=3(4b+1)$ and $2a=2(6b-1)$

$a=4b+1$ and $a=6b-1$

This implies,

$4b+1=6b-1$

$6b-4b=1+1$

$2b=2$

$b=1$

$\Rightarrow a=4(1)+1=4+1=5$

The values of $a$ and $b$ for which the given system of equations has infinitely many solutions is $5$ and $1$ respectively.    

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Updated on: 10-Oct-2022

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