Answer:
x = 3 and y = 1
Step-by-step explanation:
Given equations
y=3x-8 and y=4-x
We will substitute y from second equation in the first and get:
4-x = 3x-8 first we will add (+x) to both sides and get
4-x+x = 3x+x-8 => 4 = 4x-8 now we will add (+8) to the both sides and get
4+8 = 4x-8+8 => 12 = 4x now we will change the sides of equation and get
4x = 12 => x = 12/4 = 3 => x=3 now we will replace x=3 in the second equation
y= 4-x = 4-3=1 => y=1
Finally we have (x , y) = ( 3 , 1)
God with you!!!
Relations are subsets of products <span><span>A×B</span><span>A×B</span></span> where <span>AA</span> is the domain and <span>BB</span> the codomain of the relation.
A function <span>ff</span> is a relation with a special property: for each <span><span>a∈A</span><span>a∈A</span></span> there is a unique <span><span>b∈B</span><span>b∈B</span></span> s.t. <span><span>⟨a,b⟩∈f</span><span>⟨a,b⟩∈f</span></span>.
This unique <span>bb</span> is denoted as <span><span>f(a)</span><span>f(a)</span></span> and the 'range' of function <span>ff</span> is the set <span><span>{f(a)∣a∈A}⊆B</span><span>{f(a)∣a∈A}⊆B</span></span>.
You could also use the notation <span><span>{b∈B∣∃a∈A<span>[<span>⟨a,b⟩∈f</span>]</span>}</span><span>{b∈B∣∃a∈A<span>[<span>⟨a,b⟩∈f</span>]</span>}</span></span>
Applying that on a relation <span>RR</span> it becomes <span><span>{b∈B∣∃a∈A<span>[<span>⟨a,b⟩∈R</span>]</span>}</span><span>{b∈B∣∃a∈A<span>[<span>⟨a,b⟩∈R</span>]</span>}</span></span>
That set can be labeled as the range of relation <span>RR</span>.
C never, there are always different methods to solve a problem but don't forget that the equation always stays the same!
The answer is 4x+28 it was pretty simple you just had to add up all the sides
-(z + (-17)) = 12 + 17
-(z - 17) = 12 + 17
Distribute -1:
-z + 17 = 12 + 17
Subtract 17 to both sides:
-z = 12
Multiply -1 to both sides:
z = -12
