Question

For each of the following system of equations find whether it has unique solution, infinite number of solutions or no solutions.

$\frac{5x}{6}+\frac{y}{2}=5;\frac{x}{3}+\frac{y}{5}=2$

Answer 1

Step 1: Given Equations

Understanding and Applying

The given system of equations may be written as ,

$\frac{5}{6}x+\frac{y}{2}-5=0,\frac{x}{3}+\frac{y}{5}-2=0$

The given system of equations is of the form

$$\begin{gathered} a_{1}x+b_{1}y+c_1=0 \\ {a}_{2}x+b_{2}y+c_2=0 \end{gathered}$$

Step 2: Given values

$$\begin{gathered} where{a}_1=\frac{5}{6},b_1=\frac{1}{2},c_1=-5 \\ {a}_2=\frac{1}{3},b_2=\frac{1}{5},c_2=-2 \end{gathered}$$

Step 3:Equating the coefficients of the equation

We have

$$\begin{gathered} \frac{a_1}{a_2}=\frac{{\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}},\frac{b_1}{b_2}=\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}},\frac{c_1}{c_2}=\frac{-5}{-2} \\ \frac{a_1}{a_2}=\frac{15}{6}=\frac{5}{2}\left|\frac{{\displaystyle \raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$b$}\right.}}{{\displaystyle \raisebox{1ex}{$c$}\!\left/ \!\raisebox{-1ex}{$d$}\right.}}=\frac{ad}{bc}\right| \\ \frac{b_1}{b_2}=\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}}=\frac{5}{2}, \\ \frac{c_1}{c_2}=\frac{5}{2}, \\ \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2} \end{gathered}$$

Step 4: Result

Clearly,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=\frac{5}{2}$

Therefore,The given system of equation is dependent and has infinite numbers of solutions.