Question 115·Easy·Systems of Two Linear Equations in Two Variables

Look at how $4x - 3y$ changes to $12x - 9y$. By what number do you multiply $4$ to get $12$? Do you multiply $-3$ by the same number to get $-9$?

Once you find the multiplier that turns the first equation’s left side into the second equation’s left side, apply that same multiplier to the constant $5$ on the right side. That result is $k$.

Rewrite each equation to solve for $y$:

• First equation: $4x - 3y = 5$ becomes $y = \frac{4x - 5}{3}$.
• Second equation: $12x - 9y = k$ becomes $y = \frac{12x - k}{9}$.

In Desmos, type these as:

• y = (4x - 5)/3
• y = (12x - k)/9 (Desmos will create a slider for $k$).

Look at the two lines on the graph. Adjust the slider for $k$ until the two lines lie exactly on top of each other (they coincide everywhere, not just intersect at one point).

When the lines perfectly overlap, note the value of $k$ shown by the slider. That value of $k$ is the one that makes the system have infinitely many solutions.

If a system of two linear equations has infinitely many solutions, then both equations represent the same line.

That means one equation must be a constant multiple of the other: every term on the left and the constant on the right are all multiplied by the same number.

Compare the coefficients in

and

From $4x$ to $12x$, the coefficient is multiplied by $3$ because $12 \div 4 = 3$.

Check the $y$ term: $-3$ to $-9$ is also multiplied by $3$ because $-9 \div (-3) = 3$.

So the entire first equation must be multiplied by $3$ to match the left side of the second equation.

Since we multiply the whole first equation by $3$, we must also multiply the constant term on the right side by $3$.

So the constant on the right, originally $5$, becomes $3 \cdot 5$.

Therefore, $k$ must equal $3 \cdot 5$ for the two equations to represent the same line.

Calculate $3 \cdot 5 = 15$, so $k = 15$.

The correct answer choice is C) 15.