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2. Write four solutions for each of the following equations:

(i) $2x + y = 7$ (ii) $\pi x +y = 9$ (iii) $x = 4y$

Answers (1)

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(i) Given : $2x + y = 7$

Putting x=0, we have , $y=7-2\times 0=7$ means $(0,7)$ is a solution.

Putting x=1, we have , $y=7-2\times 1=5$ means $(1,5)$ is a solution.

Putting x=2, we have , $y=7-2\times 2=3$ means $(2,3)$ is a solution.

Putting x=3, we have , $y=7-2\times 3=1$ means $(3,1)$ is a solution.

The four solutions are : $(0,7),(1,5),(2,3),(3,1)$ .

(ii) Given : $\pi x +y = 9$

Putting x=0, we have , $y=9-\pi \times 0=9$ means $(0,9)$ is a solution.

Putting x=1, we have , $y=9-\pi \times 1=9-\pi$ means $(1,9-\pi )$ is a solution.

Putting x=2, we have , $y=9-\pi \times 2=9-2\pi$ means $(2,9-2\pi )$ is a solution.

Putting x=3, we have , $y=9-\pi \times 3=9-3\pi$ means $(3,9-3\pi )$ is a solution.

The four solutions are : $(0,9),(1,9-\pi ),(2,9-2\pi ),(3,9-3\pi )$ .

(iii) Given : $x = 4y$

Putting x=0, we have , $y=\frac{0}{4}=0$ means $(0,0)$ is a solution.

Putting x=1, we have , $y=\frac{1}{4}$ means $(1,\frac{1}{4})$ is a solution.

Putting x=2, we have , $y=\frac{2}{4}=\frac{1}{2}$ means $(2,\frac{1}{2})$ is a solution.

Putting x=3, we have , $y=\frac{3}{4}$ means $(3,\frac{3}{4})$ is a solution.

The four solutions are : $(0,0)$ , $(1,\frac{1}{4})$ , $(2,\frac{1}{2})$ and $(3,\frac{3}{4})$ .

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