Multiply 4 by both $2x$ and $-3$, and multiply 5 by both $3x$ and $2$. Write out each of the four products separately so you do not miss any terms.

After distributing, you should have two $x$-terms and two constant terms. Add the $x$-terms together and then add the constants, being careful with the negative sign on $-3$.

Type y1 = 4(2x - 3) + 5(3x + 2) into Desmos so it graphs or defines the original expression.

One at a time, type each option as a new function (for example, y2 = 7x + 14, then y3 = 23x - 2, etc.). The correct choice will be the one whose graph lies exactly on top of the graph of $y1$ for all $x$ (or always gives the same table values as $y1$ for several $x$-values you test).

The expression $4(2x - 3) + 5(3x + 2)$ has numbers outside parentheses multiplying binomials. To simplify it, use the distributive property: $a(b + c) = ab + ac$ and $a(b - c) = ab - ac$.

First distribute 4 over $(2x - 3)$:

• $4 \cdot 2x = 8x$
• $4 \cdot (-3) = -12$ So $4(2x - 3) = 8x - 12$.

Then distribute 5 over $(3x + 2)$:

• $5 \cdot 3x = 15x$
• $5 \cdot 2 = 10$ So $5(3x + 2) = 15x + 10$.

Now rewrite the whole expression as:

Group the $x$-terms and the constant terms:

• $x$-terms: $8x + 15x = 23x$
• Constants: $-12 + 10 = -2$

So the expression simplifies to $23x - 2$, which matches answer choice B) 23x - 2.