Question

solve by comparison method and please shoe the steps​

Answer 1

Answer:

$\tt \red{\left\{\begin{matrix}x=-\frac{3\left(3-q\right)}{p\left(q-p\right)}\text{, }y=-\frac{3\left(p-3\right)}{q\left(q-p\right)}\text{, }&q\neq 0\text{ and }p\neq q\text{ and }p\neq 0\\x=-\frac{qy-3}{p}\text{, }y\in \mathrm{R}\text{, }&\left(q=3\text{ or }q=0\right)\text{ and }p=3\\x\in \mathrm{R}\text{, }y=1\text{, }&p=0\text{ and }q=3\end{matrix}\right.}$

$\tt \pink{\left. \begin{cases} { px+qy=3 } \\ { p ^ { 2 } x+q ^ { 2 } y=9 } \end{cases} \right.}$

$\tt \blue{px+qy=3,p^{2}x+q^{2}y=9 }$

$\tt \red{\left(\begin{matrix}p&q\\p^{2}&q^{2}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}3\\9\end{matrix}\right) }$

$\tt \orange{inverse(\left(\begin{matrix}p&q\\p^{2}&q^{2}\end{matrix}\right))\left(\begin{matrix}p&q\\p^{2}&q^{2}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}p&q\\p^{2}&q^{2}\end{matrix}\right))\left(\begin{matrix}3\\9\end{matrix}\right) }$

$\tt \green{\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}p&q\\p^{2}&q^{2}\end{matrix}\right))\left(\begin{matrix}3\\9\end{matrix}\right) }$

$\tt \purple{\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}p&q\\p^{2}&q^{2}\end{matrix}\right))\left(\begin{matrix}3\\9\end{matrix}\right) }$

$\tt \red{\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{q^{2}}{pq^{2}-qp^{2}}&-\frac{q}{pq^{2}-qp^{2}}\\-\frac{p^{2}}{pq^{2}-qp^{2}}&\frac{p}{pq^{2}-qp^{2}}\end{matrix}\right)\left(\begin{matrix}3\\9\end{matrix}\right) }$

$\blue{\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{q}{p\left(q-p\right)}&-\frac{1}{p\left(q-p\right)}\\-\frac{p}{q\left(q-p\right)}&\frac{1}{q\left(q-p\right)}\end{matrix}\right)\left(\begin{matrix}3\\9\end{matrix}\right) }$

$\orange{\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{q}{p\left(q-p\right)}\times 3+\left(-\frac{1}{p\left(q-p\right)}\right)\times 9\\\left(-\frac{p}{q\left(q-p\right)}\right)\times 3+\frac{1}{q\left(q-p\right)}\times 9\end{matrix}\right) }$

$\blue{\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3\left(q-3\right)}{p\left(q-p\right)}\\\frac{3\left(3-p\right)}{q\left(q-p\right)}\end{matrix}\right) }$

$\red{\boxed{ \boxed{\boxed{ \boxed{\boxed{ \boxed{\boxed{ \boxed{ \tt \pink{ x=\frac{3\left(q-3\right)}{p\left(q-p\right)},y=\frac{3\left(3-p\right)}{q\left(q-p\right)} }}}}}}}}}}$