Question 1070040
<pre><b> 
{{{system(-2^(10-x)+y = 0, y = 8^(x+2))}}}

The second equation is solved for y, so substitute
8<sup>x+2</sup> for y in the first equation:

{{{-2^(10-x)+8^(x+2)=0}}}

Add 2<sup>10-x</sup> to both sides:

{{{8^(x+2)=2^(10-x)}}}

Write 8 as 2<sup>3</sup>

{{{(2^3)^(x+2)=2^(10-x)}}}

Remove the parentheses on the left by multiplying the
exponents:

{{{2^(3x+6)=2^(10-x)}}}

Now we use the principle that 
if B<sup>P</sup> = B<sup>Q</sup>, then P = Q

{{{3x+6=10-x}}}
{{{4x=4}}}
{{{x=1}}}

Substitute in 

{{{y = 8^(x+2)}}}
{{{y = 8^(1+2)}}}
{{{y = 8^3}}}
{{{y=512}}}

Answer: (x,y) = (1,512)

Edwin</pre>