Question 125·Medium·Math

After expanding, rewrite each side so it looks like $(\text{number})x + (\text{number})$. Identify the coefficient of $x$ and the constant term on each side.

If two linear expressions are equal for every real $x$, their $x$-coefficients must be equal, and their constant terms must be equal. Turn this idea into two separate equations in $p$ and $q$.

Use the simpler equation (the one with only $p$) to find $p$ first, then plug that value into the other equation to find $q$.

From the algebra, you have two equations: $4p = 6$ (from matching x-coefficients) and $p + q = 15$ (from matching constants).

In Desmos, graph y = 4x and y = 6. The x-coordinate of their intersection gives the value of $p$.

Once you have $p$, graph y = x + p_value and y = 15 (replacing p_value with your result). The x-coordinate of the intersection is the value of $q$.

Plug your $p$ and $q$ values into y1 = p*(4x+1)+q and y2 = 3*(2x+5). The two lines should overlap completely if you found the correct values.

Start by expanding each side of the equation:

Left side:

Right side:

So the equation becomes:

Think of both sides as "coefficient of $x$" plus "constant term":

• Left side: coefficient of $x$ is $4p$, constant term is $p+q$.
• Right side: coefficient of $x$ is $6$, constant term is $15$.

Because the equation is true for all real $x$, these matching parts must be equal:

• Coefficient of $x$: $4p = 6$
• Constant term: $p + q = 15$

Solve the first equation $4p = 6$:

Now you know the value of $p$, and you will use it in the constant-term equation $p + q = 15$ to find $q$.

Substitute $p = \frac{3}{2}$ into $p + q = 15$:

Subtract $\frac{3}{2}$ from both sides:

So $p = \frac{3}{2}$ and $q = \frac{27}{2}$, which corresponds to choice D.