Answer :
Step-by-step explanation:
To solve the system of equations using the substitution method, we start with the given system:
1. \( ax - by = 0 \)
2. \( ab^2x + a^2by - (a^2 + b^2) = 0 \)
First, solve the first equation for \( x \):
\[ ax - by = 0 \]
\[ ax = by \]
\[ x = \frac{by}{a} \]
Now substitute \( x = \frac{by}{a} \) into the second equation:
\[ ab^2\left(\frac{by}{a}\right) + a^2by - (a^2 + b^2) = 0 \]
Simplify the expression:
\[ b^3y + a^2by - (a^2 + b^2) = 0 \]
Factor out \( y \) from the terms involving \( y \):
\[ y(b^3 + a^2b) - (a^2 + b^2) = 0 \]
Rearrange to isolate \( y \):
\[ y(b^3 + a^2b) = a^2 + b^2 \]
\[ y = \frac{a^2 + b^2}{b^3 + a^2b} \]
Having found \( y \), substitute back to find \( x \):
\[ x = \frac{by}{a} \]
\[ x = \frac{b}{a} \left(\frac{a^2 + b^2}{b^3 + a^2b}\right) \]
\[ x = \frac{b(a^2 + b^2)}{a(b^3 + a^2b)} \]
Simplify the denominator:
\[ x = \frac{b(a^2 + b^2)}{ab(b^2 + a^2)} \]
Since \( a(b^2 + a^2) \) is common in the numerator and the denominator, simplify the fraction:
\[ x = \frac{b}{a} \]
Thus, the solutions to the system of equations are:
\[ x = \frac{b}{a} \]
\[ y = \frac{a^2 + b^2}{b^3 + a^2b} \]