Question

Solve the following system of equations using the substitution method:
ax – by = 0
\[a{{b}^{2}}x+{{a}^{2}}by-\left( {{a}^{2}}+{{b}^{2}} \right)=0\]

Answer 1

Hint: First of all, find the value of x from the first equation and substitute it in the second equation. Now, find the value of y from this equation and substitute this value of y in the equation of x that we got from the first equation, to find the actual value of x.

Complete step-by-step answer:
In this question, we have to solve the system of equations using the substitution method. Let us consider the equation given in the question.
\[ax-by=0.....\left( i \right)\]
\[a{{b}^{2}}x+{{a}^{2}}by-\left( {{a}^{2}}+{{b}^{2}} \right)=0.....\left( ii \right)\]
Let us consider the equation (i)
\[ax-by=0\]
By transposing ‘by’ to the other side of the above equation, we get,
\[ax=by\]
By dividing ‘a’ on both the sides of the above equation, we get,
\[\dfrac{ax}{a}=\dfrac{by}{a}\]
\[x=\dfrac{by}{a}....\left( iii \right)\]
Now, by substituting \[x=\dfrac{by}{a}\] in equation (ii), we get,
\[a{{b}^{2}}x+{{a}^{2}}by-\left( {{a}^{2}}+{{b}^{2}} \right)=0\]
\[a{{b}^{2}}\left( \dfrac{by}{a} \right)+{{a}^{2}}by-\left( {{a}^{2}}+{{b}^{2}} \right)=0\]
\[{{b}^{3}}y+{{a}^{2}}by=\left( {{a}^{2}}+{{b}^{2}} \right)\]
\[by\left( {{b}^{2}}+{{a}^{2}} \right)=\left( {{a}^{2}}+{{b}^{2}} \right)\]
\[y=\dfrac{\left( {{a}^{2}}+{{b}^{2}} \right)}{b\left( {{a}^{2}}+{{b}^{2}} \right)}\]
By canceling the like terms from the above equation, we get,
\[y=\dfrac{1}{b}\]
Now, by substituting this value of \[y=\dfrac{1}{b}\] in equation (iii), we get,
\[x=\dfrac{b}{a}\left( \dfrac{1}{b} \right)=\dfrac{1}{a}\]
So, we get the value of x and y as \[\dfrac{1}{a}\text{ and }\dfrac{1}{b}\].

Note: In this question, students can verify their answer by substituting the values of x and y in each equation and verify if LHS = RHS as follows:
\[ax-by=0.....\left( i \right)\]
\[a{{b}^{2}}x+{{a}^{2}}by-\left( {{a}^{2}}+{{b}^{2}} \right)=0.....\left( ii \right)\]
By substituting \[x=\dfrac{1}{a}\text{ and }y=\dfrac{1}{b}\] in equation (i), we get,
\[a\left( \dfrac{1}{a} \right)-b\left( \dfrac{1}{b} \right)=0\]
\[1-1=0\]
\[0=0\]
Hence, LHS = RHS
Now, by substituting \[x=\dfrac{1}{a}\text{ and }y=\dfrac{1}{b}\] in equation (ii), we get,
\[a{{b}^{2}}\left( \dfrac{1}{a} \right)+{{a}^{2}}b\left( \dfrac{1}{b} \right)-\left( {{a}^{2}}+{{b}^{2}} \right)=0\]
\[\left( {{b}^{2}}+{{a}^{2}} \right)-\left( {{a}^{2}}+{{b}^{2}} \right)=0\]
\[0=0\]
Here also, LHS = RHS
So, our answer is correct.