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Algebra / Linear equations in one variable Difficulty: Hard

$\frac{12 x + 28}{4} - \frac{s}{13} = r (x - 8)$ $\frac{12 x + 28}{4} - \frac{s}{13} = r (x - 8)$

In the given equation, $s$ $s$ and $r$ $r$ are constants, and $s greater than 0$ $s greater than 0$ . If the equation has infinitely many solutions, what is the value of $s$ $s$ ?

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Explanation

The correct answer is $403$ $403$ . For a linear equation in one variable to have infinitely many solutions, the coefficients of the variable must be equal on both sides of the equation and the constant terms must also be equal on both sides of the equation. The given equation can be rewritten as $\frac{4 (3 x + 7)}{4} - \frac{s}{13} = r (x - 8)$ $\frac{4 (3 x + 7)}{4} - \frac{s}{13} = r (x - 8)$ , or $3 x + 7 - \frac{s}{13} = r (x - 8)$ $3 x + 7 - \frac{s}{13} = r (x - 8)$ . Applying the distributive property to the right-hand side of this equation yields $3 x + 7 - \frac{s}{13} = r x - 8 r$ $3 x + 7 - \frac{s}{13} = r x - 8 r$ . For this equation to have infinitely many solutions, the coefficients of $x$ $x$ must be equal, so it follows that $3 equals r$ $3 equals r$ . Additionally, the constant terms must be equal, which means $7 - \frac{s}{13} = - 8 r$ $7 - \frac{s}{13} = - 8 r$ . Substituting $3$ $3$ for $r$ $r$ in this equation yields $7 - \frac{s}{13} = - 8 (3)$ $7 - \frac{s}{13} = - 8 (3)$ , or $7 - \frac{s}{13} = - 24$ $7 - \frac{s}{13} = - 24$ . Adding $StartFraction s Over 13 EndFraction$ $StartFraction s Over 13 EndFraction$ to both sides of this equation yields $7 = - 24 + \frac{s}{13}$ $7 = - 24 + \frac{s}{13}$ . Adding $24$ $24$ to both sides of this equation yields $31 = \frac{s}{13}$ $31 = \frac{s}{13}$ . Multiplying both sides of this equation by $13$ $13$ yields $403 equals s$ $403 equals s$ . Therefore, if the equation has infinitely many solutions, the value of $s$ $s$ is $403$ $403$ .