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

Answer 1

The equation given is 5x – 4y = 20.
(i) (4, 0)
Putting the value in the given equation we have
$$\begin{gathered} \mathrm{LHS}:5\left(4\right)-4\left(0\right)=20 \\ \mathrm{RHS}:20 \\ \mathrm{LHS}=\mathrm{RHS} \end{gathered}$$
Thus, (4, 0) is a solution of the given equation.
(ii) (0, 5)
Putting the value in the given equation we have
$$\begin{gathered} \mathrm{LHS}:5\left(0\right)-4\left(5\right)=0-20=-20 \\ \mathrm{RHS}:20 \\ \mathrm{LHS}\ne \mathrm{RHS} \end{gathered}$$
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
$$\begin{gathered} \mathrm{LHS}:5\left(-2\right)-4\left(\frac{5}{2}\right)=-10-10=-20 \\ \mathrm{RHS}:20 \\ \mathrm{LHS}\ne \mathrm{RHS} \end{gathered}$$
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
$$\begin{gathered} \mathrm{LHS}:5\left(0\right)-4\left(-5\right)=0+20=20 \\ \mathrm{RHS}:20 \\ \mathrm{LHS}=\mathrm{RHS} \end{gathered}$$
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
$$\begin{gathered} \mathrm{LHS}:5\left(2\right)-4\left(\frac{-5}{2}\right)=10+10=20 \\ \mathrm{RHS}:20 \\ \mathrm{LHS}=\mathrm{RHS} \end{gathered}$$
Thus, $\left(2,\frac{-5}{2}\right)$ is a solution of the given equation.