Question

Find the condition that the pair of linear equations 2x + 3y = 15 and 2ax + (a + b)y = 7b + a has
infinitely many solutions.
24. Solve : ax +by = −a, bx + ay + b = 0
25. Solve : a(x + y) + b(x − y) = a
2 + b
2 − ab and a(x + y) − b(x − y) = a
2 + b
2 + ab

Answer 1

1.

The given linear equations are

$\quad\quad 2x+3y=15$

$\quad\quad 2ax+(a+b)y=7b+a$

For infinitely many solutions, we must have the coefficient determinant being zero.

$\quad D=0$

$\Rightarrow \left|\begin{array}{cc}2&3\\2a&a+b\end{array}\right|=0$

$\Rightarrow 2(a+b)-3(2a)=0$

$\Rightarrow 2a+2b-6a=0$

$\Rightarrow 4a=2b$

$\Rightarrow \boxed{\color{blue}{2a=b}}$

This is the required condition.

2.

The given linear equations are

$\quad\quad ax+by=-a\quad...(1)$

$\quad\quad bx+ay=-b\quad...(2)$

Multiplying $(1)$ by $b$ and $(2)$ by $a$, we get

$\quad\quad abx+b^{2}y=-ab$

$\quad\quad abx+a^{2}y=-ab$

On subtraction, we have

$\quad (b^{2}-a^{2})y = 0$

$\Rightarrow y=0$

Putting $y=0$ in $(1)$, we get

$\quad ax+0=-a$

$\Rightarrow x=-1$

$\therefore$ the required solution is

$\quad\quad \boxed{\color{blue}{x=-1,\:y=0}}$.

3.

The given linear equations are

$\quad\quad a(x+y)+b(x-y)=a^{2}+b^{2}-ab$

$\quad\quad a(x+y)-b(x-y)=a^{2}+b^{2}+ab$

On addition, we get

$\quad 2a(x+y)=2(a^{2}+b^{2})$

$\Rightarrow ax+ay=a^{2}+b^{2}\quad...(1)$

On subtraction, we get

$\quad 2b(x-y)=-2ab$

$\Rightarrow x-y=-a\quad...(2)$

Multiplying $(1)$ by $1$ and $(2)$ by $2$, we have

$\quad\quad ax+ay=a^{2}+b^{2}$

$\quad\quad ax-ay=-a^{2}$

On addition, we get

$\quad 2ax=b^{2}$

$\Rightarrow x=\frac{b^{2}}{2a}$

Putting $x=\frac{b^{2}}{2a}$ in $(2)$, we get

$\quad \frac{b^{2}}{2a}-y=-a$

$\Rightarrow y=\frac{b^{2}}{2a}+a$

$\Rightarrow y=\frac{b^{2}+2a^{2}}{2a}$

$\therefore$ the required solution is

$\quad \boxed{\color{blue}{x=\frac{b^{2}}{2a},\:y=\frac{b^{2}+2a^{2}}{2a}}}$.