Simultaneous equations to solve unknown exponents Solved the following simultaneous equations for $a$ and $b$:
(1)
$$a + 2b = 2\\3a + b = 10$$
$$3a + 6b = 6\\3a + b = 10$$
$$5b = -4$$
$$b = \frac{-4}{5}\\a = \frac{18}{5}$$
(2)
How do we use the answers/information in (1) to solve the following equations for $x$ and $y$?
$$2^{x+2y} = 2^{20}\\5^{5x+y} = 25^{x+50}$$
In general how do these types of equations relate?
 A: No worries, here is the answer for your corrected question.
$$2^{x+2y} = 2^{20} \; \; (i)$$
$$5^{5x+y} = 25^{x+50} \; \; (ii)$$
As the bases are the same, from the first equation we can derive that $x+2y=20$, so $x=20-2y$. Now we can sub this into equation $ii$
$$5^{5(20-2y)+y} = 25^{(20-2y)+50}$$
$$5^{100-9y} = 25^{70-2y}$$
We also need to recognize that $25 = 5^2$, and that $(a^m)^n=a^{mn}$, so:
$$5^{100-9y} = 5^{140-4y}$$
As both sides have the same base we only need to consider the exponents:
$$100-9y=140-4y$$
I'll let you take it from here...
A: Note that $25^{x+50}=(5^2)^{x+50}=5^{2x+100}$ so the system can be written as
$$2^{x+2y} = 2^{20}\\5^{5x+y} = 5^{2x+100}$$
or
$$2^{x+2y} = 2^{20}\\5^{3x+y} = 5^{100}$$
and comparing exponents we obtain
$$x+2y=20$$
$$3x+y=100$$
This is very similar to the system
$$a + 2b = 2\\3a + b = 10$$
but the RHS is multiplied by $10$. So multiplying by $10$ we have
$$10a + 2(10b) =x+2y= 20\\3(10a) + 10b =3x+y= 100$$
where $x=10a$ and $y=10b$ and you have already found $a$ and $b$.
