# Find the values of a and b for which the following system of equations has infinitely many solutions:

Question:

Find the values of a and b for which the following system of equations has infinitely many solutions:

(i) $(2 a-1) x-3 y=5$

$3 x+(b-2) y=3$

(ii) $2 x-(2 a+5) y=5$

$(2 b+1) x-9 y=15$

(iii) $(a-1) x+3 y=2$

$6 x+(1-2 b) y=6$

$(i v) 3 x+4 y=12$

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

$(v) 2 x+3 y=7$

$(a-b) x+(a+b) y=3 a+b-2$

$(v i) 2 x+3 y-7=0$

$\quad(a-1) x+(a+1) y=(3 a-1)$

(vii) $2 x+3 y=7$

$\quad(a-1) x+(a+2) y=3 a$

$($ viii $) x+2 y=1$

$(a-b) x+(a+b) y=a+b-2$

$(i x) 2 x+3 y=7$

$2 a x+a y=28-b y$

Solution:

(i) GIVEN:

$(2 a-1) x-3 y=5$

$3 x+(b-2) y=3$

To find: To determine for what value of k the system of equation has infinitely many solutions

We know that the system of equations

$a_{1} x+b_{1} y=c_{1}$

$a_{2} x+b_{2} y=c_{2}$

For infinitely many solution

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

Here

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

Consider

$\frac{3}{-(b-2)}=\frac{5}{3}$

$-5 b+10=9$

$-5 b=-1$

$b=\frac{1}{5}$

Again consider

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

$2 a-1=5$

$2 a=6$

$a=3$

Hence for $a=3$ and $b=\frac{1}{5}$ the system of equation has infinitely many solution.

(11) GIVEN:

$2 x-(2 a+5) y=5$

$(2 b+1) x-9 y=15$

To find: To determine for what value of k the system of equation has infinitely many solutions

We know that the system of equations

$a_{1} x+b_{1} y=c_{1}$

$a_{2} x+b_{2} y=c_{2}$

For infinitely many solution

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

Here

$\frac{2}{2 b+1}=\frac{2 a+5}{9}=\frac{5}{15}$

Consider the following

$\frac{2 a+5}{9}=\frac{5}{15}$

$30 a+75=45$

$30 a=-30$

$a=-1$

Again consider

$\frac{2}{2 b+1}=\frac{5}{15}$

$10 b+5=30$

$10 b=25$

$b=\frac{5}{2}$

Hence for $a=-1$ and $b=\frac{5}{2}$ the system of equation has infinitely many solution.

(iii) GIVEN:

$(a-1) x+3 y=2$

$6 x+(1-2 b) y=6$

To find: To determine for what value of k the system of equation has infinitely many solutions

We know that the system of equations

$a_{1} x+b_{1} y=c_{1}$

$a_{2} x+b_{2} y=c_{2}$

For infinitely many solution

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

Here

$\frac{(a-1)}{6}=\frac{3}{(1-2 b)}=\frac{2}{6}$

Consider the following

$\frac{3}{(1-2 b)}=\frac{2}{6}$

$2-4 b=18$

$-4 b=16$

$b=-4$

Again consider

$\frac{(a-1)}{6}=\frac{2}{6}$

$6 a-6=12$

$6 a=18$

$a=3$

Hence for $a=3$ and $b=-4$ the system of equation has infinitely many solution.

(iv) GIVEN:

$3 x+4 y=12$

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

To find: To determine for what value of k the system of equation has infinitely many solutions

We know that the system of equations

$a_{1} x+b_{1} y=c_{1}$

$a_{2} x+b_{2} y=c_{2}$

For infinitely many solution

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

Here

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

Consider the following

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

$24 a-24 b=20 a-4$

$4 a-24 b=-4$....$\ldots(1)$

Again consider

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

$12 a+12 b=15 a-3$

$3 a-12 b=3 \ldots \ldots(2)$

Multiplying eq. (2) by 2 and subtracting eq. (1) from eq. 2

$2 a=10$

$a=5$

Substituting the value of ‘a’ in eq. (2) we get

$15-12 b=3$

$-12 b=-12$

$b=1$

Hence for $a=5$ and $b=1$ the system of equation has infinitely many solution.

(v) GIVEN:

$2 x+3 y=7$

$(a-b) x+(a+b) y=3 a+b-2$

To find: To determine for what value of k the system of equation has infinitely many solutions

We know that the system of equations

$a_{1} x+b_{1} y=c_{1}$

$a_{2} x+b_{2} y=c_{2}$

For infinitely many solution

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

Here

$\frac{2}{(a-b)}=\frac{3}{(a+b)}=\frac{7}{3 a+b-2}$

Consider the following

$\frac{3}{(a+b)}=\frac{7}{3 a+b-2}$

$6 a+2 b-4=7 a-7 b$

$a-9 b=-4 \ldots \ldots$ (2)

Multiplying eq. (2) by 2 and subtracting eq. (1) from eq. (2)

$-14 b=-14$

$b=1$

Substituting the value of b in eq. (2) we get

$a-9=-4$

$a=5$

Hence for $a=5$ and $b=1$ the system of equation has infinitely many solution.

(vi) GIVEN:

$2 x+3 y-7=0$

$(a-1) x+(a+1) y=3 a-1$

To find: To determine for what value of k the system of equation has infinitely many solutions

Rewrite the given equations

$2 x+3 y-7=0$

$(a-1) x+(a+1) y=3 a-1$

We know that the system of equations

$a_{1} x+b_{1} y=c_{1}$

$a_{2} x+b_{2} y=c_{2}$

For infinitely many solution

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

Here

$\frac{2}{(a-1)}=\frac{3}{(a+1)}=\frac{7}{3 a-1}$

Consider the following

$\frac{3}{(a+1)}=\frac{7}{3 a-1}$

$9 a-3=7 a+7$

$2 a=10$

$a=5$

Hence for $a=5$ the system of equation have infinitely many solutions.

(vii) GIVEN :

$2 x+3 y=7$

$(a-1) x+(a+2) y=3 a$

To find: To determine for what value of k the system of equation has infinitely many solutions

We know that the system of equations

$a_{1} x+b_{1} y=c_{1}$

$a_{2} x+b_{2} y=c_{2}$

For infinitely many solution

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

Here

$\frac{2}{(a-1)}=\frac{3}{(a+2)}=\frac{7}{3 a}$

Consider the following

$\frac{3}{(a+2)}=\frac{7}{3 a}$

$9 a=7 a+14$

$2 a=14$

$a=7$

Hence for $a=7$ the system of equation have infinitely many solutions.

(viii) Given:

$x+2 y=1$

$(a-b) x+(a+b) y=a+b-2$

We know that the system of equations

$a_{1} x+b_{1} y=c_{1}$

$a_{2} x+b_{2} y=c_{2}$

has infinitely many solutions if

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

So,

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

$\Rightarrow \frac{1}{a-b}=\frac{2}{a+b}$ and $\frac{2}{a+b}=\frac{1}{a+b-2}$

$\Rightarrow a+b=2 a-2 b$ and $2 a+2 b-4=a+b$

$\Rightarrow a=3 b$ and $a+b=4$

$\Rightarrow a-3 b=0$ and $a+b=4$

Solving these two equations, we get

$-4 b=-4$

$\Rightarrow b=1$

Putting $b=1$ in $a+b=4$, we get

a = 3

(ix) Given:

$2 x+3 y=7$

$2 a x+a y=28-b y$

$\Rightarrow 2 a x+(a+b) y=28$

We know that the system of equations

$a_{1} x+b_{1} y=c_{1}$

$a_{2} x+b_{2} y=c_{2}$

has infinitely many solutions if

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

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

$\Rightarrow \frac{1}{a}=\frac{3}{a+b}=\frac{1}{4}$

$\Rightarrow \frac{1}{a}=\frac{3}{a+b}$ and $\frac{3}{a+b}=\frac{1}{4}$

Now,

$\frac{1}{a}=\frac{3}{a+b}$

$\Rightarrow a+b=3 a$

$\Rightarrow b=2 a$...$(1)$

Also, $\frac{3}{a+b}=\frac{1}{4}$

$\Rightarrow a+b=12 \quad \ldots \ldots(2)$

Solving (1) and (2), we get

$a=4$ and $b=8$