Solve: $4 \log_5 x- \log_5y =1 \quad \&\quad 5\log_5 x+ 3\log_5 y =14$ Find $x,y$ given:
$$4 \log_5 x- \log_5y =1 \quad \&\quad 
5\log_5 x+ 3\log_5 y =14$$
How do I help my son this assignment? Not so good in maths myself 
 A: hint: Put $a = \log_5x, b = \log_5y\implies 4a-b=1 = 5a+3b=14\implies 5a+3(4a-1)=14...$ . Can you finish it ? point is solve for $a$, then for $x$ since $x = 5^a$...
A: multiplying the first equation by $3$ and adding to the second we get
$$17\log_5 x=17$$ thus we get
$$\log_5 x=1$$ can you go on?
A: $$4 \log_5 x- \log_5y =1 \quad \&\quad 
5\log_5 x+ 3\log_5 y =14$$
$$\log_5x^4y^{-1}=1 \quad \&\quad \log_5x^5y^3=14$$
$$\implies x^4y^{-1}=5 \quad \&\quad x^5y^3=5^{14} $$
$$\implies  x^{12}y^{-3}=5 \quad \&\quad x^5y^3=5^{14}$$
$$\implies x^{17}=5^{17}\implies x=5 \quad \&\quad 5^5y^3=5^{14}\implies y=5^3$$
A: $$
\begin{align}
4 \log_5 x- \log_5y &=1\tag 1 \\ 
5\log_5 x+ 3\log_5 y &=14\tag 2
\end{align}
$$
Multiplying $(1)$ by $3$ and adding $(1)+(2)$ we have
$$17\log_5 x=17\quad\Longrightarrow\quad \log_5 x=1\quad\Longrightarrow\quad x=5$$
Multiplying $(1)$ by $-5$  and $(2)$ by $4$ and adding $(1)+(2)$ we have
$$17\log_5 y=51\quad\Longrightarrow\quad \log_5 y=51/17=3\quad\Longrightarrow\quad y=5^3=125$$
