Question

Solve equations by cross multiplication method.
\[2x-y=6\]
\[x-y=2\]

Answer 1

Hint: Use the cross multiplication method to get the solution.
Apply \[\begin{align}
  & x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \\
 & {{b}_{1}}\searrow \,\,\,\,{{c}_{1}}\,\,\searrow \,\,\,\,{{a}_{1}}\,\,\,\searrow \,\,\,\,\,{{b}_{1}} \\
 & {{b}_{2}}\nearrow \,\,\,\,{{c}_{2}}\,\,\nearrow \,\,\,\,{{a}_{2}}\,\,\,\nearrow \,\,\,\,{{b}_{2}} \\
\end{align}\] in stages and get the equations in order accordingly.

Complete step-by-step answer:
On applying we get the equations as follows :

\[\begin{align}
  & 2x-y=6\,\,\,\,\,\,\,2x-y-6=0 \\
 & x-y=2\,\,\,\,\,\,\,\,\,\,x-y-2=0 \\
\end{align}\]

The terms of the equations are :

\[\begin{align}
  & {{a}_{1}}=2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{b}_{1}}=-1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{c}_{1}}=-6 \\
 & {{a}_{2}}=1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{b}_{2}}=-1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{c}_{2}}=-2 \\
\end{align}\]

Carry out the following operations to form the relations between the terms :

\[\begin{align}
  & x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \\
 & {{b}_{1}}\searrow \,\,\,\,{{c}_{1}}\,\,\searrow \,\,\,\,{{a}_{1}}\,\,\,\searrow \,\,\,\,\,{{b}_{1}} \\
 & {{b}_{2}}\nearrow \,\,\,\,{{c}_{2}}\,\,\nearrow \,\,\,\,{{a}_{2}}\,\,\,\nearrow \,\,\,\,{{b}_{2}} \\
\end{align}\]

The relations between the terms are found out to be :
\[\dfrac{x}{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}=\dfrac{y}{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}=\dfrac{1}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}\]

Substituting the relevant values we get :

\[\dfrac{x}{-1(-2)-(-1)(-6)}=\dfrac{y}{-6(1)-(-2)(2)}=\dfrac{1}{2(-1)-1(-1)}\]

On simplifying the above we get :

\[\dfrac{x}{+2+1(-6)}=\dfrac{y}{-6+2 \times 2}=\dfrac{1}{-2-1(-1)}\]

We get as follows :

\[\dfrac{x}{+2-6}=\dfrac{y}{-6+4}=\dfrac{1}{-2+1}\]

Thus we get :

\[\dfrac{x}{-4}=\dfrac{y}{-2}=\dfrac{1}{-1}\]

Upon restructuring the signs we get :

\[\dfrac{x}{-4}=\dfrac{y}{-2}=-1\]

Relating the x and y terms we get :

\[\begin{align}
  & \dfrac{x}{-4}=\,-1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\dfrac{y}{-2}=-1 \\
 & x=(-1) \times (-4)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y=(-1) \times (-2) \\
 & x=4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y=2 \\
\end{align}\]

Note: Take care to substitute relevant values. Make sure to cross multiply the right terms and perform the operations in the right sequence by taking care to avoid making mistakes with regard to signs.