Today I did a practice SAT problem.

If the function ƒ is defined by ƒ(x) = 2x + 3, and if ƒ(a) = 11, what is the value of a?

A. 4
B. 7
C. 11
D. 17
E. 25

This is the correct answer and the explanation

Here's Why:

If the function ƒ is defined by ƒ(x) = 2x + 3, and ƒ(a) = 11, then ƒ(a) = 2a + 3 = 11. Therefore, a = 11-3/2=4.

I don't understand the rules for functions. Why do you divide 11-3 by 2? Is there a specific formula you follow for them? Thanks for your help!

What they were doing is following the rules for 'solving equations'.

the notation
f(x) = 2x + 3 can be written also as
y = 2x + 3, it says the same thing
so when they say f(a) it means that whenever you saw an x has now been replaced by a,
so the 2x + 3 becomes 2a + 3
but you were told that this equals 11
so all you are doing is solving
2a + 3 = 11

Now recall your elementary rules of equations, "whatever you do to one side you must do to the other side"
so lets add -3 to both sides

2a + 3 - 3 = 11 - 3 (I wanted to get rid of +3)
2a = 8 ( I now want to get rid of the 2 in front of a)
divide both sides by 2
2a/2 = 8/2
a = 4

To find the value of a in the given problem, we need to solve the equation 2a + 3 = 11.

The reason we divide 11 - 3 by 2 is that we are isolating the variable a. When solving equations, we want to get the variable by itself on one side of the equation, and in this case, we want to isolate a.

Here are the steps to solve the equation:

1. Start with 2a + 3 = 11.

2. Subtract 3 from both sides of the equation to isolate the term with a:
2a + 3 - 3 = 11 - 3,
which simplifies to 2a = 8.

3. Now, we want to get rid of the coefficient of 2 in front of a. To do this, we divide both sides of the equation by 2:
(2a)/2 = 8/2,
which simplifies to a = 4.

So, the value of a is 4.

In general, there is no specific formula for solving equations involving functions. The approach is to apply algebraic operations to manipulate the equation and isolate the variable you are solving for. Dividing both sides by the same number is a common operation used to remove coefficients or simplify equations.