$$ 2 x+3 y=7 $$ $$ 10 x+15 y=35 $$ 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) $$ \left(r,-\frac{3 r}{2}+\frac{7}{2}\right) $$

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UpStudy AI · Accepted Answer

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:

$$
\textbf{B)}
$$
Option B is the correct choice.

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