First of all, you need to know what a 'solution' is, so you'll know it
when you see it.
A 'solution' to an equation is a number, or a set of numbers, that
you can write in place of the variables (the letters), and when you
do that, the equation will have only numbers in it, and it'll be a true
statement.
Your equation has two variables in it ... 'x' and 'y' . In order to find
just a single set of numbers for what both of them must be, you
would need two equations.
The way it stands now, with only one equation, there are actually
an infinite number of solutions. Each solution is a pair of numbers ...
one for 'x' and one for 'y' ... and if you write them into the equation
in place of 'x' and 'y', then the equation is a true statement.
I'll show you how to tell if a pair of numbers is a solution or not.
Here's what that looks like:
Say I give you two pairs of numbers, and I tell you that both of them
are solutions to your equation. The 'solutions' I give you are
x=0
y=1
and
x=2
y=3 .
You don't trust me, and you say to me "Wait just a minute there, dude !
Not so fast. I'll need to check them out and see if those are really solutions
to my equation."
You take the first pair and write it into your equation:
x=1, y=0
9x - 7y = -7
9(0) - 7(1) = -7
0 - 7 = -7
-7 = -7
OK. That's a true statement.
So x=0, y=1 is a solution.
Now check the other one I gave you:
x=2, y=3
9x - 7y = -7
9(2) - 7(3) = -7
18 - 21 = -7
-3 = -7
This is NOT a true statement.
So x=2, y=3 is NOT a solution to your equation.
I pulled a fast one on you. If I was charging you money for solutions,
then you would not pay me for this one, because it's not a solution.
Answer:
18
Step-by-step explanation:
Answer:
Step-by-step explanation:
A=1 and 2
B=2 and 4
C=1,2,7 and 14
D=1,2,4,and 8
Randomly assigned group. the art class isn't random and only consists of a select type of people--it is not a representative sample; maybe more left handed people enjoy art than do right handed people! therefore, even though the randomly assigned group is smaller, it is the better estimate.
