Question

$3(2x+y)=7xy;3(x+3y)=11xy$ using elimination method.

Answer 1

The given system of equations is
$3(2x+y)=7xy$ (1)
$3(x+3y)=11xy$ (2)
Observe that the given system is not linear because of the occurrence of xy term. Also, note that if x = 0, then y = 0 and vice versa. So, (0, 0) is a solution for the system and any other solution would have both $x\ne 0$ and $y\ne 0$.
Thus, we consider the case where $x\ne 0,y\ne 0$.
Dividing both sides of each equation by xy, we get
$\frac{6}{y}+\frac{3}{x}=7,i.e.,\frac{3}{x}+\frac{6}{y}=7$ (3)
and $\frac{9}{x}+\frac{3}{y}=11$ (4)
Let $a=\frac{1}{x}$ and $b=\frac{1}{y}$
Equations (3) and (4) become
$3a+6b=7$ (5)
$9a+3b=11$ (6)
which is a linear system in a and b.
To eliminate b, we have $\left(6\right)\times 2\to 18a+6b=22$ (7)
Subtracting (7) from (5) we get, -15a=15. That is, a=1.
Substituting a = 1 in (5) we get, $b=\frac{2}{3}$. Thus, a=1 and $b=\frac{2}{3}$.
When a=1, we have $\frac{1}{x}=1$, Thus x=1.
When $a=\frac{2}{3}$, we have $\frac{1}{y}=\frac{2}{3}$, Thus $y=\frac{3}{2}$.
Thus, the system has two solutions $(1,\frac{3}{2})$ and $(0,0)$.