Question

The equation $3y^2+y+4=2(6x^2+y+2)$ where $y=2x$ is satisfied by:

no value of $x$

all values of $x$

$x=0$ only

all integral values of $x$ only

all rational values of $x$ only

Answer: C

Concepts:substitution quadratic cancellation linear equation

Difficulty rating: 1280

Small Hint:

Substitute $y=2x$ on both sides before expanding

Big Hint:

The $12x^2$ and constant terms cancel

Solution:

Substituting $y=2x$ gives $12x^2+2x+4=12x^2+4x+4.$ Cancelling the common quadratic and constant terms leaves $2x=4x,$ so $x=0.$ This value satisfies both equations.

Thus, the correct answer is C.

1956 AMC 12 Problem 41

Answer: C

Difficulty rating: 1280

Small Hint:

Big Hint:

Solution:

Problem 41 in Other Years