Question

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

$$\begin{gathered} \left(i\right)\left(2a-1\right)x-3y=5 \\ 3x+\left(b-2\right)y=3 \\ \left(ii\right)2x-\left(2a+5\right)y=5 \\ \left(2b+1\right)x-9y=15 \\ \left(iii\right)\left(a-1\right)x+3y=2 \\ 6x+\left(1-2b\right)y=6 \\ \left(iv\right)3x+4y=12 \\ \left(a+b\right)x+2\left(a-b\right)y=5a-1 \\ \left(v\right)2x+3y=7 \\ \left(a-b\right)x+\left(a+b\right)y=3a+b-2 \\ \left(vi\right)2x+3y-7=0 \\ \left(a-1\right)x+\left(a+1\right)y=\left(3a-1\right) \\ \left(vii\right)2x+3y=7 \\ \left(a-1\right)x+\left(a+2\right)y=3a \end{gathered}$$
$$\begin{gathered} \left(viii\right)x+2y=1 \\ \left(a-b\right)x+\left(a+b\right)y=a+b-2 \end{gathered}$$
$$\begin{gathered} \left(ix\right)2x+3y=7 \\ 2ax+ay=28-by \end{gathered}$$

Answer 1

(i) GIVEN:

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

We know that the system of equations

For infinitely many solution

Here

Consider

Again consider

Hence forand the system of equation has infinitely many solution.

(ii) GIVEN:

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

We know that the system of equations

For infinitely many solution

Here

Consider the following

Again consider

Hence for and the system of equation has infinitely many solution.

(iii) GIVEN:

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

We know that the system of equations

For infinitely many solution

Here

Consider the following

Again consider

Hence forand the system of equation has infinitely many solution.

(iv) GIVEN:

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

We know that the system of equations

For infinitely many solution

Here

Consider the following

…… (1)

Again consider

…… (2)

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

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

Hence forand the system of equation has infinitely many solution.

(v) GIVEN:

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

We know that the system of equations

For infinitely many solution

Here

Consider the following

…… (1)

Again consider the following

…… (2)

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

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

Hence forand the system of equation has infinitely many solution.

(vi) GIVEN:

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

Rewrite the given equations

We know that the system of equations

For infinitely many solution

Here

Consider the following

Hence for the system of equation have infinitely many solutions.

(vii) GIVEN :

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

We know that the system of equations

For infinitely many solution

Here

Consider the following

Hence for the system of equation have infinitely many solutions.
(viii) Given:
$$\begin{gathered} x+2y=1 \\ \left(a-b\right)x+\left(a+b\right)y=a+b-2 \end{gathered}$$
We know that the system of equations

has infinitely many solutions if

So,
$$\begin{gathered} \frac{1}{a-b}=\frac{2}{a+b}=\frac{1}{a+b-2} \\ \Rightarrow \frac{{\displaystyle 1}}{{\displaystyle a-b}}=\frac{{\displaystyle 2}}{{\displaystyle a+b}}\mathrm{and}\frac{{\displaystyle 2}}{{\displaystyle a+b}}=\frac{{\displaystyle 1}}{{\displaystyle a+b-2}} \\ \Rightarrow a+b=2a-2b\mathrm{and}2a+2b-4=a+b \\ \Rightarrow a=3b\mathrm{and}a+b=4 \\ \Rightarrow a-3b=0\mathrm{and}a+b=4 \end{gathered}$$
Solving these two equations, we get
$$\begin{gathered} -4b=-4 \\ \Rightarrow b=1 \end{gathered}$$
Putting b = 1 in a + b = 4, we get
a = 3

(ix) Given:
$$\begin{gathered} 2x+3y=7 \\ 2ax+ay=28-by \end{gathered}$$
$\Rightarrow 2ax+\left(a+b\right)y=28$
We know that the system of equations

has infinitely many solutions if

$$\begin{gathered} \therefore \frac{2}{2a}=\frac{3}{a+b}=\frac{7}{28} \\ \Rightarrow \frac{1}{a}=\frac{3}{a+b}=\frac{1}{4} \\ \Rightarrow \frac{1}{a}=\frac{3}{a+b}\mathrm{and}\frac{3}{a+b}=\frac{1}{4} \end{gathered}$$
Now,
$\frac{1}{a}=\frac{3}{a+b}$
⇒ a + b = 3a
⇒ b = 2a .....(1)
Also, $\frac{3}{a+b}=\frac{1}{4}$
⇒ a + b = 12 .....(2)
Solving (1) and (2), we get
a = 4 and b = 8