Solve the systems.
a) $$\left . \begin{array}{rcl} 2 \log x-3\log y&=&1 \\ \log x+\log y &=&3\end{array}\right \} $$
b) $$\left . \begin{array}{rcl}\log x + \log y &=&\log 100- \log 10 \\ x+ y &=&7\end{array}\right \} $$
Development:
a) The two equations of the system are logarithmic, so we can use the elimination method.
We can multiply the second equation by $$3$$ and then add the first one, with which we eliminate $$y$$: $$$3\cdot[\log x+\log y=3] \Rightarrow 3\log x+3\log y=9$$$ $$$ + \left[\begin{array}{r} 2\log x-3\log y = 1 \\ \underline{3\log x+3\log y =9} \\ 5\log x+0 =10 \end{array}\right]$$$ The resultant equation allows us to find the value of $$x$$: $$$5\log x=10 \Rightarrow \log x=\dfrac{10}{5}=2 \Rightarrow x=10^2=100$$$ Once we know the value of $$x$$, we can replace it into the first equation to find $$y$$: $$$2\log 100-3\log y=1 \Rightarrow 2\cdot2-3\log y=1 \Rightarrow 4-3\log y=1 \Rightarrow -3\log y=1-4 \Rightarrow$$$ $$$\Rightarrow -3\log y=-3 \Rightarrow \log y=\dfrac{-3}{-3}=1 \Rightarrow y=10^1=10$$$
b) In the second case there is one of the equations that is not logarithmic, so the replacement method will be applied. But, first,wewant toget rid of the logarithms of the first equation: $$$\log x + \log y =\log 100- \log 10 \Rightarrow \log xy=\log\dfrac{100}{10} \Rightarrow xy=10$$$ The obtained equation is equivalent to the logarithmic one, so we have the equivalent system:
$$\left . \begin{array}{rcl} xy &=&10 \\ x+y &=&7\end{array}\right \} $$
We can obtain $$x$$ in terms of $$y$$ from the first equation: $$$xy=10 \Rightarrow x=\dfrac{10}{y}$$$ Now this expression can be replaced in the second equation: $$$x+y=7 \Rightarrow \dfrac{10}{y}+y=7 \Rightarrow \dfrac{10+y^2}{y}=7 \Rightarrow 10+y^2=7y \Rightarrow y^2-7y+10=0$$$ And we have an equation of second degree that we know how to solve by using the formula: $$$y=\dfrac{7\pm\sqrt{7^2-4\cdot1\cdot10}}{2\cdot1}=\dfrac{7\pm\sqrt{49-40}}{2}=\dfrac{7\pm\sqrt{9}}{2}=\dfrac{7\pm3}{2}$$$ $$$y=\dfrac{7+3}{2}=\dfrac{10}{2}=5 \\ y=\dfrac{7-3}{2}=\dfrac{4}{2}=2$$$ Both solutions are valid in the logarithmic equation, because both $$\log 5$$ and $$\log 2$$ exist. So it will be necessary to look for every value of $$x$$ that corresponds to these two values of $$y$$.
We need to replace $$y$$ into the first equation to obtain the values of $$x$$:
When $$y=5$$: $$$x=7-y \Rightarrow x=7-5=2$$$
When $$y=2$$: $$$x=7-y \Rightarrow x=7-2=5$$$
Solution:
a) $$x=100; \ y=10$$
b) The second equation has two possible solutions: $$x=2; \ y=5$$ and $$x=5; \ y=2$$.