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Algebra / Systems of two linear equations in two variables Difficulty: Hard

$4 x - 9 y = 9 y + 5$

$h y = 2 + 4 x$

In the given system of equations, $h$ is a constant. If the system has no solution, what is the value of $h$ ?

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Explanation

Choice D is correct. A system of two linear equations in two variables, $x$ and $y$ , has no solution if the lines represented by the equations in the xy-plane are distinct and parallel. The graphs of two lines in the xy-plane represented by equations in the form $A x + B y = C$ , where $A$ , $B$ , and $C$ are constants, are parallel if the coefficients for $x$ and $y$ in one equation are proportional to the corresponding coefficients in the other equation. The first equation in the given system can be written in the form $A x + B y = C$ by subtracting $9 y$ from both sides of the equation to yield $4 x - 18 y = 5$ . The second equation in the given system can be written in the form $A x + B y = C$ by subtracting $4 x$ from both sides of the equation to yield $- 4 x + h y = 2$ . The coefficient of $x$ in this second equation, $negative 4$ , is $negative 1$ times the coefficient of $x$ in the first equation, $4$ . For the lines to be parallel, the coefficient of $y$ in the second equation, $h$ , must also be $negative 1$ times the coefficient of $y$ in the first equation, $negative 18$ . Thus, $h = - 1 (- 18)$ , or $h equals 18$ . Therefore, if the given system has no solution, the value of $h$ is $18$ .

Choice A is incorrect. If the value of $h$ is $negative 9$ , then the given system would have one solution, rather than no solution.

Choice B is incorrect. If the value of $h$ is $0$ , then the given system would have one solution, rather than no solution.

Choice C is incorrect. If the value of $h$ is $9$ , then the given system would have one solution, rather than no solution.