# What is the solution of the following?: x + y < 3, x – y < 4

Apr 4, 2017

The solution is $\left\{x , y | x < \frac{7}{2} \cap y \in \mathbb{R}\right\}$, or $x \in \left(- \infty , \frac{7}{2}\right)$ and $y \in \left(- \infty , \infty\right)$.

#### Explanation:

Let's start with the first inequality: $x + y < 3$. We can add a value $a$ to the left-hand side of this inequality as long as we add another value b≥a to the right-hand side of this inequality. After looking at the second inequality, we notice that we can set $a = x - y$ and $b = 4$ as $x - y < 4$ from the second inequality.

This becomes $x + y + x - y < 3 + 4$. The $y$'s cancel out, and we get $2 x < 7$. We can divide both sides by $2$ to get $x < \frac{7}{2}$.

Now, $x + y < 3$ and $x - y < 4$. Since $x < \frac{7}{2}$, $y$ can have any value. To see this, solve for $y$ to get $y < 3 - x$ and $y > x - 4$. In the first inequality, as $x \to - \infty$ (allowed since $x < \frac{7}{2}$), $y < \infty$. In the second inequality, as $x \to - \infty$, $y > - \infty$.

So our solution is $\left\{x , y | x < \frac{7}{2} \cap y \in \mathbb{R}\right\}$, or $x \in \left(- \infty , \frac{7}{2}\right)$ and $y \in \left(- \infty , \infty\right)$.