Solve equations by cross multiplication method.
\[2x-y=6\]
\[x-y=2\]
Answer
676.5k+ views
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.
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.
Recently Updated Pages
Write structures of the following compounds i 2 Chloro3methylpentane class 11 chemistry CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE

A Paragraph on Pollution in about 100-150 Words

XIX+XXX A 49 B 51 C 55 D 44 class 5 maths CBSE

If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Difference between physical and chemical change class 11 chemistry CBSE

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

