Write each equation in the form $Ax + By = C$. How can you use $A$ and $B$ from each equation to make the slopes equal so that the lines are parallel?

You should get one equation from the equal-slope condition and another from $a + b = 7$. How can you use these two equations together—perhaps by turning them into a quadratic—to find possible values of $a$?

Once you find the possible values of $a$, check what happens to the system for each one. For which value do the equations describe parallel but different lines, instead of the exact same line?

In Desmos, define $b = 7 - a$ (type b = 7 - a) so that the condition $a + b = 7$ is always satisfied. Then enter the two equations a x + 4 y = 12 and 3 x + b y = 9 so both lines are graphed.

Create a slider for $a$, or manually set $a$ equal to each answer choice (1, 3, 4, 6). For each value of $a$, Desmos will automatically update $b = 7 - a$ and the two lines.

For each tested value of $a$, observe the two graphs: you want the value where the lines are parallel and distinct (they never cross). That value of $a$ is the correct answer.

A system of two linear equations in $x$ and $y$ has no solution when the equations represent parallel but different lines.

For equations of the form $A_1x + B_1y = C_1$ and $A_2x + B_2y = C_2$:

• The lines are parallel if their slopes are equal, which happens when $A_1/B_1 = A_2/B_2$.
• They are not the same line if the constants are not in the same ratio: $C_1/C_2 \ne A_1/A_2$ (or $C_1/C_2 \ne B_1/B_2$).

Here the equations are:

So for no solution, we need the slopes (from $x$ and $y$ coefficients) to match, but the constants to not match that ratio.

The slope of a line written as $Ax + By = C$ is $-A/B$.

• For $ax + 4y = 12$, the slope is $-a/4$.
• For $3x + by = 9$, the slope is $-3/b$.

Setting the slopes equal (for parallel lines):

Drop the negative signs and cross-multiply:

So we now have two equations involving $a$ and $b$:

• $a + b = 7$
• $ab = 12$.

Treat $a$ and $b$ as the two roots of a quadratic. For a quadratic $t^2 - St + P = 0$:

• The sum of the roots is $S$.
• The product of the roots is $P$.

Here, $a + b = 7$ and $ab = 12$, so $a$ and $b$ are roots of

This quadratic will give the possible values of $a$.

Factor the quadratic:

So the roots are $t = 3$ and $t = 4$, meaning $(a, b)$ is either $(3, 4)$ or $(4, 3)$.

• If $a = 3$, $b = 4$, the equations are

which are parallel with the same left side but different constants, so there is no solution.

• If $a = 4$, $b = 3$, the equations are

which simplify to the same line $y = -x + 3$, giving infinitely many solutions, not no solution.

Therefore, only $a = 3$ fits the condition that the system has no solution, so the correct value of $a$ is $3$ (choice B).