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(Nod)

(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

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- Jan 30th 2010, 03:00 AMdecoy808solve simultaneously
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(Nod)

(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 - Jan 30th 2010, 03:15 AMe^(i*pi)
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__: - Jan 30th 2010, 05:27 AMSoroban
Hello, decoy808!

If I read it correctly, the second one has no solution . . .

Quote:

$\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 ansystem of equations: . $\displaystyle \begin{array}{ccc}x - y &=& 4 \\ x - y &=& \text{-}4\end{array}$*inconsistent*