1

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?

Alessio K
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Harry B
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    Hint: $$ a^x = a^y \quad \text{if and only if} \quad x = y $$ Also, you have to make a sanity check. For example for expressions like $$ 2^x = 3^y $$ there are no integer solutions because the left-hand-side is always even and the right-hand-side is always odd. – Matti P. Sep 21 '21 at 09:19
  • A thousand apologies i mistyped the second equation in question (2) your Hint focused my mind. – Harry B Sep 21 '21 at 10:00
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    The second equation can be turned to $5^{3x+y}=5^{100}$. –  Sep 21 '21 at 10:24

2 Answers2

1

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...

LogLogLogLogx
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    You are not exploiting the resolution of the first system. Better rewrite as $$\begin{cases}x+2y=20,\3x+y=100.\end{cases}$$ –  Sep 21 '21 at 10:27
  • @IAmAGuest they both spit out the same answer champion... – LogLogLogLogx Sep 21 '21 at 10:30
  • Of course. But the fact is that you don't answer the question as asked. (Which the OP also missed.) Äres' answer is better. –  Sep 21 '21 at 12:48
1

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 $b$.

Alessio K
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