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Algebra / Linear equations in two variables Difficulty: Medium

If the graph of $27 x + 33 y = 297$ is shifted down $5$ units in the xy-plane, what is the y-intercept of the resulting graph?

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Explanation

Choice A is correct. When the graph of an equation in the form $A x + B y = C$ , where $A$ , $B$ , and $C$ are constants, is shifted down $k$ units in the xy-plane, the resulting graph can be represented by the equation $A x + B (y + k) = C$ . It's given that the graph of $27 x + 33 y = 297$ is shifted down $5$ units in the xy-plane. Therefore, the resulting graph can be represented by the equation $27 x + 33 (y + 5) = 297$ , or $27 x + 33 y + 165 = 297$ . Subtracting $165$ from both sides of this equation yields $27 x + 33 y = 132$ . The y-intercept of the graph of an equation in the xy-plane is the point where the line intersects the y-axis, represented by the point $left parenthesis 0 comma y right parenthesis$ . Substituting $0$ for $x$ in the equation $27 x + 33 y = 132$ yields $27 (0) + 33 y = 132$ , or $33 y equals 132$ . Dividing both sides of this equation by $33$ yields $y equals 4$ . Therefore, if the graph of $27 x + 33 y = 297$ is shifted down $5$ units, the y-intercept of the resulting graph is $left parenthesis 0 comma 4 right parenthesis$ .

Choice B is incorrect and may result from conceptual or calculation errors.

Choice C is incorrect. This is the y-intercept of the graph of $27 x + 33 y = 297$ shifted up, not down, $5$ units.

Choice D is incorrect and may result from conceptual or calculation errors.