Question

How do you solve $2x-y=4$ and $4x-2y=8$? \[\]

Answer 1

Hint: We write both the equations general form$ax+by+c=0$. We recall the conditions on coefficients of two linear equations ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0,{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0$ for unique solutions$\left( \dfrac{{{a}_{1}}}{{{a}_{2}}}\ne \dfrac{{{b}_{1}}}{{{b}_{2}}} \right)$, infinite solutions $\left( \dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}} \right)$ and no solution$\left( \dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}} \right)$. We check which conditions are satisfied by the given two equations. \[\]

Complete step by step answer:
We know that the general linear equation is given by $ax+by+c=0$ where $a,b,c$are real numbers and $a\ne 0,b\ne 0$We need at least two equations to find a unique solution. If ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0,{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0$ be two linear equations then we can oblation unique solution only when ratio between coefficients of variables is not equal that is
\[\dfrac{{{a}_{1}}}{{{a}_{2}}}\ne \dfrac{{{b}_{1}}}{{{b}_{2}}}\]
 We get no solution if the ratio between coefficients of variables is equal but not equal to ratio between constants that is
\[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}\]
We get infinite solutions if the ratio between coefficients of variables is equal to ratio between constants that is
\[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}\]

We are given the following linear equations in standard form in the question
\[\begin{align}
  & 2x-y=4.....\left( 1 \right) \\
 & 4x-2y=8.....\left( 2 \right) \\
\end{align}\]
We write the above two equations in general form to have;
\[\begin{align}
  & 2x-y-4=0 \\
 & 4x-2y-8=0 \\
\end{align}\]
We find the ratio of coefficients of $x$ as $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{2}{4}=\dfrac{1}{2}$ . We find the ratio of coefficients of $y$ as $\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{-1}{-2}=\dfrac{1}{2}$. We find the ratio of constants$\dfrac{{{c}_{1}}}{{{c}_{2}}}=\dfrac{-4}{-8}=\dfrac{1}{2}$. So we have
 \[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}=\dfrac{1}{2}\]
So there will be infinite solutions for the given pair of equations.

Note: We can take $x=t$ and put $x=t$ in the equation $2x-y=4$ to have $y=4+2t$ so for every real value of $t$ we get a solution. Similarly we put $x=t$ in $4x-2y=8$ to have $y=4+2t$ . We know that every linear equation is a line in the plane and we get infinite solutions when the lines corresponding to the two equations coincide. So we can represent every point on the line as a solution in the form$\left( t,4+2t \right)$.