Question

In the following systems of equations determine whether the system has a unique solution, no solution or infinitely many solutions. in case there is a unique solution, find it:
\[\begin{align}
  & 3x-5y=20 \\
 & 6x-10y=40 \\
\end{align}\]

Answer 1

Hint: We recall that in system of linear equations with two variables that the system has unique solution when the ratios of coefficients of variables is equal, infinite solution when the ratios of coefficients of variables is equal to the coefficient of constant terms and no solution when the ratios of coefficients of variables is equal to the coefficient of constant terms.

Complete step by step answer:
We know that a linear equation with two variables $x,y$ with coefficient of $x$ as $a$ , coefficient of $y$ as $b$ and constant term is given by;
\[ax+by+c=0\]
We know that we need 2 equations to find the solution. The two equations simultaneously are called a system. We write the system equations in general as
\[\begin{align}
  & {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0 \\
 & {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0 \\
\end{align}\]
We have the ratio of coefficient of $x$ as $\dfrac{{{a}_{1}}}{{{a}_{2}}}$, ratio of coefficient of $y$ as $\dfrac{{{b}_{1}}}{{{b}_{2}}}$ and ratio of constant term as $\dfrac{{{c}_{1}}}{{{c}_{2}}}$. We know we can get a unique solution when
\[\dfrac{{{a}_{1}}}{{{a}_{1}}}\ne \dfrac{{{b}_{1}}}{{{b}_{2}}}\]
We get infinite solutions when
\[\dfrac{{{a}_{1}}}{{{a}_{1}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}\]
We get no solutions when
\[\dfrac{{{a}_{1}}}{{{a}_{1}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}\]
We are give in the following equations in the question
\[\begin{align}
  & 3x-5y=20 \\
 & \Rightarrow 2x-5y-20=0....\left( 1 \right) \\
 & 6x-10y=40 \\
 & \Rightarrow 6x-10y-40=0......\left( 2 \right) \\
\end{align}\]
So we have ${{a}_{1}}=2,{{a}_{2}}=4,{{b}_{1}}=1,{{b}_{2}}=2,{{c}_{1}}=-5,{{c}_{2}}=-10$. We find the ratios of coefficients and constant term as
\[\begin{align}
  & \Rightarrow \dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{3}{6}=\dfrac{1}{2} \\
 & \Rightarrow \dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{-5}{-10}=\dfrac{1}{2} \\
 & \Rightarrow \dfrac{{{c}_{1}}}{{{c}_{2}}}=\dfrac{20}{40}=\dfrac{1}{2} \\
\end{align}\]
We see that
\[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}\]

So there are infinitely many solutions to the given system of equations.

Note: We note coordinate geometry each linear equation in two variables represent a line in the plane. If two equations have infinitely many solutions they represent coinciding lines. We can solve linear equations in two variables either by elimination or substitution. If we take $x=t$ in either of the equations and then we have $y=\dfrac{20-3t}{5}$. We find one solution for every real value of $t$ and the solutions $\left( \dfrac{20-3t}{5},t \right)$ will represent the points on the coinciding line.