Question

Find three solutions of the equation: $2x+y=7$

• A. $(0,7)$, $(\frac{7}{2},0)$ and $(3,1)$
• B. $(0,6)$, $(\frac{7}{2},0)$ and $(3,1)$
• C. $(0,7)$, $(\frac{5}{2},0)$ and$(3,1)$
• D. $(0,7)$, $(\frac{7}{2},0)$ and $(2,1)$

Answer 1

The correct option is A $(0,7)$, $(\frac{7}{2},0)$ and $(3,1)$

The given equation is $2x+y=7$ ... (1)

Putting $x=0$ in (1), we get,

$2\times 0+y=7$

$\Rightarrow x=\frac{7}{2}$

$\therefore (\frac{7}{2},0)$ is a solution of (1).

Further, putting $y=1$ in (1), we get

$2x+1=7$

$\Rightarrow 2x=7-1$

$\Rightarrow 2x=6$

$\Rightarrow x=3$

$\therefore (3,1)$ is a solution of (1).

Hence, $(0,7);(\frac{7}{2},0)$ and $(3,1)$ are three solutions of the given equation.