Think of the second part as multiplying by $-2$, not just 2. How does that affect both $3a$ and $-4$ inside the parentheses?

Once you have expanded both products, you should have several terms. Which terms have $a$ in them, and which are just numbers? Combine each group separately.

In Desmos, type y1 = 5(2x - 3) - 2(3x - 4). This graphs the original expression as a line in terms of $x$ (you can treat $x$ as the same kind of variable as $a$).

On new lines, enter:

• y2 = 4x - 7
• y3 = 16x - 23
• y4 = 4x + 7
• y5 = 22x - 7

Look at the graphs: the correct choice will have a line that lies exactly on top of the graph of y1 for all $x$. Alternatively, use a table for each function (click the gear icon and select 'Table') and compare the $y$-values for several $x$-values (like 0, 1, and 2); the equivalent expression will always have the same $y$-values as the original.

Use the distributive property: multiply the number outside the parentheses by each term inside.

• For $5(2a - 3)$: multiply 5 by $2a$ and 5 by $-3$ to get $10a - 15$.
• For $-2(3a - 4)$: treat the factor as $-2$ and multiply $-2$ by $3a$ and $-2$ by $-4$ to get $-6a + 8$.

So the original expression becomes

Now that you have distributed both factors, rewrite the expression as a sum of terms without parentheses:

You will next combine like terms (the terms with $a$ and the constant terms).

Group the terms with $a$ together and the constant terms together:

• $10a - 6a = 4a$
• $-15 + 8 = -7$

So the simplified expression is

which is equivalent to the original expression.