if someone could help i have an answer for the first one does it look okey. but am unsure how to tackle the 2nd one. many thanks
(a)
Log (x) + Log (y) = 2
y - x = 15
answer?
(x=10/y)
(y = 10/ x)
(b)
e^x+y = e^2x+4
x-y = 4
a) Simultaneous equations should not be solved in terms of each other
$\displaystyle log(xy) = 2$
$\displaystyle xy = 100$
$\displaystyle x = \frac{100}{y}$
$\displaystyle y - \frac{100}{y} = 15 \: \: \longrightarrow \: \: y^2-15y-100=0$
Solve the quadratic for y and then put any valid solutions back into the first equation to check for x
Spoiler:
Hello, decoy808!
If I read it correctly, the second one has no solution . . .
$\displaystyle (b)\;\;\begin{array}{cccc}e^{x+y} &=& e^{2x+4} & [1] \\
x-y &=& 4& [2] \end{array}$
Equation [1] gives us: .$\displaystyle x+y \:=\:2x+4 \quad\Rightarrow\quad x - y \:=\:-4$
With [2], we have an inconsistent system of equations: . $\displaystyle \begin{array}{ccc}x - y &=& 4 \\ x - y &=& \text{-}4\end{array}$