Question

which expression is equivalent to $9f + 6f$?
$12f + 2f$
$f + 15$
$f + 15f$
$f \cdot 15$

Explanation:

Step1: Simplify the original expression

We have the expression \(9f + 6f\). Since \(9 + 6 = 15\), by the distributive property \(af+bf=(a + b)f\), we get \(9f+6f=(9 + 6)f=15f\).

Step2: Simplify each option

• Option 1: \(12f+2f=(12 + 2)f = 14f

eq15f\)

• Option 2: \(f + 15\) is not a like - term combination with \(15f\), and \(f+15

eq15f\)

• Option 3: \(f+15f=(1 + 15)f=16f

eq15f\)? Wait, no, wait. Wait, original expression is \(9f + 6f=15f\). Wait, maybe I made a mistake. Wait, no, let's re - check. Wait, the options: Wait, the first option is \(12f + 2f\), \(12f+2f = 14f\). The second option is \(f + 15\), which is not a term with \(f\) multiplied by a constant in the same way. The third option is \(f+15f=(1 + 15)f = 16f\)? Wait, no, wait the original problem's options: Wait, maybe I misread. Wait, the original expression is \(9f+6f = 15f\). Let's check each option again:

Wait, maybe there is a mistake in my calculation. Wait, \(9f+6f=(9 + 6)f = 15f\). Now check each option:

• Option 1: \(12f+2f=(12 + 2)f=14f

eq15f\)

• Option 2: \(f + 15\) is not equal to \(15f\) (since \(f+15\) has a constant term and a linear term, while \(15f\) is just a linear term)
• Option 3: \(f + 15f=(1+15)f = 16f

eq15f\)

• Option 4: \(f\cdot15 = 15f\) (because multiplication is commutative, \(f\times15=15\times f = 15f\))

Ah! I missed the fourth option. So the fourth option \(f\cdot15\) is equal to \(15f\), which is the same as \(9f + 6f\).

Answer:

The expression equivalent to \(9f + 6f\) is \(f\cdot15\) (or \(15f\)). Among the given options, the correct one is \(f\cdot15\) (the fourth option, \(f\cdot15\)).