Question

Find the values of a and b for which the following system of linear equations has infinite number of solutions \[2x-3y=7\]; \[(a+b)x-(a+b-3)y=4a+b\]
A. a = -5; b=-1
B. a= 1; b= 5
C. a= -1; b= -5
D. a= 5; b= 1

Answer 1

Hint: Given system of linear equations are \[2x-3y=7\]; \[(a+b)x-(a+b-3)y=4a+b\]
The system of equations has infinitely many solutions when the lines are coincident.
Complete step-by-step answer:
To solve systems of an equation in two or three variables, first, we need to determine
whether the equation is dependent, independent, consistent, or inconsistent. If a pair of the linear equations have unique or infinite solutions, then the system of equation is said to be a
consistent pair of linear equations. Thus, suppose we have two equations in two variables as
follows:
$a_1$x + $b_1$y = $c_1$ ——- (1)
$a_2$x + $b_2$y = $c_2$ ——- (2)
The given equations are consistent and dependent and have infinitely many solutions, if and only if,
\[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}\]
We have our two equations; \[2x-3y=7\]; \[(a+b)x-(a+b-3)y=4a+b\]
Now writing in the form \[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}\]
\[\dfrac{2}{(a+b)}=\dfrac{-3}{-(a+b-3)}=\dfrac{7}{4a+b}\]
Now equating any two terms results in \[\dfrac{2}{(a+b)}=\dfrac{7}{4a+b}\]
The terms appeared are \[8a+2b=7a+7b\]
By solving we get the value of a and b as \[a=5b\]. . . . . . . . . . . . (1)
Now equating the other two terms we get \[\dfrac{-3}{-(a+b-3)}=\dfrac{2}{(a+b)}\]
\[\Rightarrow \]\[2a+2b-6=3a+3b\]
\[\Rightarrow \]\[a+b=-6\]. . . . . . . . . . …(2)
Substituting the value of (1) in equation (2) we get
\[\Rightarrow \]\[5b+b=-6\]
\[\Rightarrow \]\[b=-1\]
\[\Rightarrow \]\[a=-5\]
The answer is option A.
Note: If the equations are parallel lines, there will be an infinite number of solutions for the given system of linear equations. Thus, the given equations can also be plotted on a graph to see that they give us parallel lines.