<span>A. (-3, 5)
the equation is xy=-15
or x=-15/y
put y=5
x=-15/5
x=-3</span>
The properties that were used to derive the properties of logarithms are:
1. a^x · a^y = a^(x+y)
2. a^x / a^y = a^(x - y)
3. a^0 = 1
4. a^(-x) = 1 / x
5. (a^x)^y = a^(<span>xy)</span><span>
</span>
Answer:
In a system, the substitution method is one of the 3 main ways to solve a system and can be very efficient at times.
<u>Skills needed: Systems, Algebra</u>
Step-by-step explanation:
1) Let's say we are given two equations below:

We can use substitution here by substituting in for
in the second equation. This means we put in
for
in the 2nd equation so we only have
variables in the equation, allowing us to solve for
.
2) Solving it out:

We essentially substitute in that value as seen in step 1. Steps 2 and 3 are just simplifying the left side and allowing for us to solve. Step 4 is where we divide by -16 on both sides to solve for x. Step 5 and 6 show us solving for y using the value for x. We get 