\[ 2 x+3 y=7 \] \[ 10 x+15 y=35 \] For each real number \( r \), which of the following points lies on the graph of each equation in the \( x y \)-plane for the given system? A) \( \left(\frac{r}{5}+7,-\frac{r}{5}+35\right) \) B) \( \left(-\frac{3 r}{2}+\frac{7}{2}, r\right) \) C) \( \left(r, \frac{2 r}{3}+\frac{7}{3}\right) \) D) \( \left(r,-\frac{3 r}{2}+\frac{7}{2}\right) \)

We start with the given system of equations: \[ 2x + 3y = 7 \] \[ 10x + 15y = 35 \] Notice that the second equation is exactly \(5\) times the first: \[ 5\,(2x+3y)=10x+15y \quad \text{and} \quad 5\,(7)=35. \] Thus, the system represents a single line: \[ 2x + 3y = 7. \] We now check each option to see which point satisfies the line for every real number \( r \). **Option A:** \[ \left(\frac{r}{5} + 7,\; -\frac{r}{5} + 35\right) \] Substitute \(x = \frac{r}{5}+7\) and \(y = -\frac{r}{5}+35\) into the equation: \[ 2\left(\frac{r}{5}+7\right) + 3\left(-\frac{r}{5}+35\right) = \frac{2r}{5} + 14 - \frac{3r}{5} + 105 = -\frac{r}{5} + 119. \] For the point to lie on the line, this expression must equal \(7\) for all \(r\): \[ -\frac{r}{5} + 119 = 7 \quad \Longrightarrow \quad -\frac{r}{5} = -112 \quad \Longrightarrow \quad r = 560. \] Since the equation holds only when \( r = 560 \) and not for every \( r \), Option A is not valid. **Option B:** \[ \left(-\frac{3r}{2}+\frac{7}{2},\; r\right) \] Let \( x = -\frac{3r}{2} + \frac{7}{2} \) and \( y = r \). Substitute into the line: \[ 2\left(-\frac{3r}{2}+\frac{7}{2}\right) + 3r = -3r + 7 + 3r = 7. \] The equation is satisfied for all \( r \). Thus, Option B is correct. **Option C:** \[ \left(r,\; \frac{2r}{3}+\frac{7}{3}\right) \] Let \( x = r \) and \( y = \frac{2r}{3} + \frac{7}{3} \). Substitute into the line: \[ 2r + 3\left(\frac{2r}{3}+\frac{7}{3}\right) = 2r + 2r + 7 = 4r + 7. \] For the point to lie on the line, this must equal \(7\): \[ 4r + 7 = 7 \quad \Longrightarrow \quad 4r = 0 \quad \Longrightarrow \quad r = 0. \] It only works when \( r = 0 \), so Option C is not valid. **Option D:** \[ \left(r,\; -\frac{3r}{2}+\frac{7}{2}\right) \] Let \( x = r \) and \( y = -\frac{3r}{2}+\frac{7}{2} \). Substitute into the line: \[ 2r + 3\left(-\frac{3r}{2}+\frac{7}{2}\right) = 2r - \frac{9r}{2} + \frac{21}{2} = \frac{4r-9r+21}{2} = \frac{-5r+21}{2}. \] For the point to lie on the line, we require: \[ \frac{-5r+21}{2} = 7 \quad \Longrightarrow \quad -5r+21 = 14 \quad \Longrightarrow \quad -5r = -7 \quad \Longrightarrow \quad r = \frac{7}{5}. \] The equality holds only for \( r = \frac{7}{5} \), so Option D is not valid. Thus, the only option that works for all real numbers \( r \) is: \[ \textbf{B)} \]