Check which of the following are the solutions of the equation 5x – 4y = 20.
Question:

Check which of the following are the solutions of the equation 5x – 4y = 20.

(i) $(4,0)$

(ii) $(0,5)$

(iii) $\left(-2, \frac{5}{2}\right)$

(iv) $(0,-5)$

(v) $\left(2, \frac{-5}{2}\right)$

Solution:

The equation given is 5x – 4y = 20.
(i) (4, 0)
Putting the value in the given equation we have

LHS : $5(4)-4(0)=20$

RHS : 20

LHS = RHS

Thus, (4, 0) is a solution of the given equation.
(ii) (0, 5)
Putting the value in the given equation we have

LHS : $5(0)-4(5)=0-20=-20$

RHS : 20

$\mathrm{LHS} \neq \mathrm{RHS}$

Thus, (0, 5) is not a solution of the given equation.

(iii) $\left(-2, \frac{5}{2}\right)$

Putting the value in the given equation we have

LHS : $5(-2)-4\left(\frac{5}{2}\right)=-10-10=-20$

RHS : 20

LHS $\neq$ RHS

Thus, $\left(-2, \frac{5}{2}\right)$ is not a solution of the given equation.

(iv) (0, –5)
Putting the value in the given equation we have

LHS : $5(0)-4(-5)=0+20=20$

RHS : 20

LHS = RHS

Thus, (0, –5) is a solution of the given equation.

(v) $\left(2, \frac{-5}{2}\right)$

Putting the value in the given equation we have

LHS : $5(2)-4\left(\frac{-5}{2}\right)=10+10=20$

RHS : 20

$\mathrm{LHS}=\mathrm{RHS}$

Thus, $\left(2, \frac{-5}{2}\right)$ is a solution of the given equation.