Question

Miss Heather writes the equation $2x+3y=2$ on the board and asks her students to find the solution of the equation. Three students come up with the following answers.

Student A $\to (1,0)$
Student B $\to (2,-\frac{2}{3})$
Student C $\to (0,\frac{2}{3})$

Choose the correct option.

Answer 1

The given equation is $2x+3y=2.$

Note that this is a linear equation in two variables, $x$ and $y.$

Let us first check the answers given by each of the three students.

$\underset{–}{\text{Student A:}}$

On substituting $x=1$ and $y=0,$ we get:

$2x+3y=2\left(1\right)+3\left(0\right)=2,$ which is true.

Thus, $(1,0)$ is a solution of the given equation.

$\underset{–}{\text{Student B:}}$

On substituting $x=2$ and $y=-\frac{2}{3},$ we get:

$2x+3y=2\left(2\right)+3(-\frac{2}{3})=4-2=2,$ which is true.

Thus, $(2,-\frac{2}{3})$ is a solution of the given equation.

$\underset{–}{\text{Student C:}}$

On substituting $x=0$ and $y=\frac{2}{3},$ we get:

$2x+3y=2\left(0\right)+3\left(\frac{2}{3}\right)=2,$ which is true.

Thus, $(0,\frac{2}{3})$ is a solution of the given equation.

A linear equation in two variables has infinitely many solutions.

Thus, all three students are correct.