Question

for each expression, select all equivalent expressions from the list.
(a) $9x - 2x + 6x$
$\square 13x$
$\square 13 + x$
$\square 7x - 6x$
$\square 7x + 6x$
(b) $6(1 + 6y)$
$\square 6 + 6y$
$\square 6 + 36y$
$\square 6 \cdot 1 + 6 \cdot 6y$
$\square 6 + 1 \cdot 6 + 6y$

Explanation:

Step1: Simplify \(9x - 2x + 6x\)

Combine like terms: \(9x - 2x + 6x=(9 - 2 + 6)x = 13x\)? Wait, no, \(9-2 = 7\), \(7 + 6=13\)? Wait, no, \(9x-2x = 7x\), then \(7x+6x = 13x\)? Wait, no, the original is \(9x - 2x+6x\). Let's recalculate: \(9x-2x = 7x\), then \(7x + 6x=13x\)? Wait, no, wait the options: the first column for (a) is \(13x\), \(13 + x\), \(7x - 6x\), \(7x + 6x\). Wait, let's do \(9x-2x+6x\): combine \(x\) terms. \(9x-2x = 7x\), then \(7x+6x = 13x\)? Wait, no, \(9 - 2+6=13\), so \(13x\)? Wait, but also, \(7x + 6x=13x\) (since \(7x+6x=(7 + 6)x = 13x\)). Wait, let's check each option:

• \(13x\): \(9x-2x+6x=(9 - 2 + 6)x = 13x\), so this is equivalent.
• \(13 + x\): This is a constant term plus \(x\), not like terms with \(13x\), so not equivalent.
• \(7x - 6x\): \(7x-6x=x\), not \(13x\), so no.
• \(7x + 6x\): \(7x+6x = 13x\), same as simplified \(9x-2x+6x\), so this is equivalent.

Wait, maybe I made a mistake. Let's re-express \(9x - 2x+6x\):

\(9x-2x = 7x\), then \(7x+6x = 13x\). So \(7x + 6x = 13x\), and \(13x\) is also \(13x\). So the equivalent expressions for (a) are \(13x\) and \(7x + 6x\).

Step2: Simplify \(6(1 + 6y)\)

Use distributive property: \(6\times1+6\times6y = 6 + 36y\)? Wait, no, \(6(1 + 6y)=6\times1+6\times6y = 6 + 36y\)? Wait, the options for (b) are \(6 + 6y\), \(6 + 36y\), \(6\cdot1+6\cdot6y\), \(6 + 1\cdot6+6y\). Wait, distributive property: \(a(b + c)=ab + ac\), so \(6(1 + 6y)=6\times1+6\times6y=6 + 36y\), and also \(6\cdot1+6\cdot6y\) is the expanded form, which is equal to \(6 + 36y\). Let's check each option:

• \(6 + 6y\): This is \(6+6y\), but \(6(1 + 6y)=6 + 36y\), so no.
• \(6 + 36y\): \(6(1 + 6y)=6 + 36y\), so this is equivalent.
• \(6\cdot1+6\cdot6y\): This is the expanded form of \(6(1 + 6y)\) (distributive property), so it's equal to \(6 + 36y\), hence equivalent.
• \(6 + 1\cdot6+6y\): \(6 + 6+6y=12 + 6y\), which is not equal to \(6 + 36y\), so no.

Wait, maybe I messed up the (b) part. Let's redo (b):

\(6(1 + 6y)\) expanded is \(6\times1 + 6\times6y=6 + 36y\). So:

• \(6 + 6y\): No, because \(6\times6y = 36y\), not \(6y\).
• \(6 + 36y\): Yes, that's the expansion.
• \(6\cdot1+6\cdot6y\): Yes, that's the distributive form, equal to \(6 + 36y\).
• \(6 + 1\cdot6+6y\): \(6 + 6+6y = 12 + 6y\), no.

So for (a): \(9x - 2x + 6x\) simplifies to \(13x\), and \(7x + 6x = 13x\), so check \(13x\) and \(7x + 6x\).

For (b): \(6(1 + 6y)\) expands to \(6 + 36y\) and \(6\cdot1+6\cdot6y\) (which is \(6 + 36y\)).

Wait, but the original (a) expression is \(9x - 2x + 6x\), let's compute again: \(9x-2x=7x\), \(7x+6x=13x\). So:

• \(13x\): Yes, because \(9x-2x+6x=13x\).
• \(7x + 6x\): \(7x+6x=13x\), so yes.

For (b):

• \(6 + 36y\): \(6(1 + 6y)=6 + 36y\), yes.
• \(6\cdot1+6\cdot6y\): \(6\times1+6\times6y=6 + 36y\), yes.

Answer:

(a) Equivalent expressions: \(13x\), \(7x + 6x\)
(b) Equivalent expressions: \(6 + 36y\), \(6\cdot1+6\cdot6y\)

(Note: The checkboxes for (a) should be on \(13x\) and \(7x + 6x\); for (b) on \(6 + 36y\) and \(6\cdot1+6\cdot6y\))