# solve simultaneously

• Jan 30th 2010, 03:00 AM
decoy808
solve 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

(x=10/y)
(y = 10/ x)

(b)
e^x+y = e^2x+4
x-y = 4
• Jan 30th 2010, 03:15 AM
e^(i*pi)
Quote:

Originally Posted by decoy808
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

(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

$log(xy) = 2$

$xy = 100$

$x = \frac{100}{y}$

$y - \frac{100}{y} = 15 \: \: \longrightarrow \: \: y^2-15y-100=0$

Solve the quadratic for y and then put any valid solutions back into the first equation to check for x

Spoiler:
$(y-20)(y+5) = 0 \: \: ,\: \therefore \: \: y = 20 \text{or} y=-5$

y = -5 is not in the domain so discard it.

$
log(x) = 2-log(20)$

$
x = 10^{2-log(20)} = 5$

Therefore $x = 5, y = 20$.
• Jan 30th 2010, 05:27 AM
Soroban
Hello, decoy808!

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

Quote:

$(b)\;\;\begin{array}{cccc}e^{x+y} &=& e^{2x+4} & [1] \\
x-y &=& 4& [2] \end{array}$

Equation [1] gives us: . $x+y \:=\:2x+4 \quad\Rightarrow\quad x - y \:=\:-4$

With [2], we have an inconsistent system of equations: . $\begin{array}{ccc}x - y &=& 4 \\ x - y &=& \text{-}4\end{array}$