System of Linear Equations entered :
[1] y - 2x/3 = -1
[2] y + x = 4
// To remove fractions, multiply equations by their respective LCD
Multiply equation [1] by 3
// Equations now take the shape:
[1] 3y - 2x = -3
[2] y + x = 4
Graphic Representation of the Equations :
-2x + 3y = -3 x + y = 4
Solve by Substitution :
// Solve equation [2] for the variable x
[2] x = -y + 4
// Plug this in for variable x in equation [1]
[1] 3y - 2•(-y +4) = -3
[1] 5y = 5
// Solve equation [1] for the variable y
[1] 5y = 5
[1] y = 1
// By now we know this much :
y = 1
x = -y+4
// Use the y value to solve for x
x = -(1)+4 = 3
I hope this help you
If
is an even function, then
. In this case,
.
This is because an even function is symmetric about the
. Hence the outputs of a negative x-value and a positive x-value are the same.
The correct option is the first option.
Since the graph is even, the outputs of a negative x-value and a positive x-value are the same.
Answer:
The question is incomplete. The complete question and its solution is attached below:

Micah did not explain the last step correctly. You cannot cross out a term from the numerator and denominator unless it is a factor. In other words, x² needed to be multiplied and not added in order to cross it out.