Linear Equations B762A6

1. **State the problem:** We have two linear equations: $$5x + b = 3x - 9$$ $$5y + c = 3y - 9$$ where $b$ and $c$ are constants, and $b$ is 2 less than $c$, i.e., $$b = c - 2$$ We want to find the relationship between $x$ and $y$. 2. **Solve each equation for $x$ and $y$ respectively:** For the first equation: $$5x + b = 3x - 9$$ Subtract $3x$ from both sides: $$5x - 3x + b = -9$$ $$2x + b = -9$$ Subtract $b$ from both sides: $$2x = -9 - b$$ Divide both sides by 2: $$x = \frac{-9 - b}{2}$$ For the second equation: $$5y + c = 3y - 9$$ Subtract $3y$ from both sides: $$5y - 3y + c = -9$$ $$2y + c = -9$$ Subtract $c$ from both sides: $$2y = -9 - c$$ Divide both sides by 2: $$y = \frac{-9 - c}{2}$$ 3. **Express $x$ in terms of $y$ and $c$ using $b = c - 2$:** Substitute $b = c - 2$ into $x$: $$x = \frac{-9 - (c - 2)}{2} = \frac{-9 - c + 2}{2} = \frac{-7 - c}{2}$$ 4. **Find $x - y$ to determine their difference:** $$x - y = \frac{-7 - c}{2} - \frac{-9 - c}{2} = \frac{-7 - c + 9 + c}{2} = \frac{2}{2} = 1$$ 5. **Interpretation:** Since $x - y = 1$, it means $$x = y + 1$$ So, $x$ is 1 more than $y$. **Answer:** (D) $x$ is 1 more than $y$.