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

$48 x - 64 y = 48 y + 24$

$r y = \frac{1}{8} - 12 x$

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

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Explanation

The correct answer is $negative 28$ . 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 for $x$ and $y$ in the other equation. The first equation in the given system, $48 x - 64 y = 48 y + 24$ , can be written in the form $A x + B y = C$ by subtracting $48 y$ from both sides of the equation to yield $48 x - 112 y = 24$ . The second equation in the given system, $r y = \frac{1}{8} - 12 x$ , can be written in the form $A x + B y = C$ by adding $12 x$ to both sides of the equation to yield $12 x + r y = \frac{1}{8}$ . The coefficient of $x$ in the second equation is $\frac{1}{4}$ times the coefficient of $x$ in the first equation. That is, $48 left parenthesis one fourth right parenthesis equals 12$ . For the lines to be parallel, the coefficient of $y$ in the second equation must also be $\frac{1}{4}$ times the coefficient of $y$ in the first equation. Therefore, $- 112 (\frac{1}{4}) = r$ , or $- 28 = r$ . Thus, if the given system has no solution, the value of $r$ is $negative 28$ .