How many solutions will a pair of linear equations have, if the equations are $7x+4y-16=0$ and $14x+6y-32=0$?

The given equations are:$7x+4y-16=0$$14x+6y-32=0$ Comparing with the standard forms $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$:
$$a_1=7,b_1=4,c_1=-16$$
$$a_2=14,b_2=6,c_2=-32$$
Let us check the ratios:
$$\frac{a_1}{a_2}=\frac{7}{14}=\frac{1}{2}$$
$$\frac{b_1}{b_2}=\frac{4}{6}=\frac{2}{3}$$
Since $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$, the pair of linear equations represents intersecting lines and has exactly one unique solution.