# Recitation 1

The first recitation is designed to help students diagnose themselves, refresh the rudiments of elementary algebra and prepare them for the course.

Exercise: What ate the dimensions of a rectangular carpet with perimeter 48 feet and area 108 square feet?

Proof. Let the dimensions be $x$ and $y$. Thus by the conditions given in the problem, we have $2(x+y)=48$ and $xy=108$.By the first equation, we have $y=24-x$. Plug it into the second one, we have $x(24-x)=108$, which is equivalent to $x^2-24x+108=0$. Solve the quadratics equation, we got $x_1 = 6, x_2 = 18$. The correspondent $y_1 = 18, y_2 = 6$.

Exercise: Factor: $x^2-5x+6$.

Proof. This is when you find two numbers, 2 and 3, whose sum is the negation of the second coefficient and whose product is the third coefficient. Those two numbers are the roots you are looking for. So the answer is $(x-2)(x-3)$.

Exercise: Factor using the roots from the quadratic formula: $x^2+2x-7$.

Proof. Use the quadratic formula to get the roots $x=\frac{-2\pm\sqrt{32}}{2}=-1\pm 2\sqrt{2}$. Hence the answer is $(x+1+2\sqrt{2})(x+1-2\sqrt{2})$.

Exercise: Factor $p(x)=x^3+x^2-5x+3$.

Proof. First notice that $p(1)=0$. So $x-1$ is a factor of it, which leave $x^2+2x-3$ for further factoring. Again this formula is zero for $x=1$. By factoring out $x-1$ once again, we got the third factor $x+3$.

Exercise: Solve the system of equations$$\begin{cases} 3x + y = 5 \\ 2x - 4y = 8 \end{cases}$$

Proof. The first equation gives us $y=5-3x$. Plug it in the second one, we have $2x-4(5-3x)=8$. Simplify it and solve the linear equation, we have $x=2$. Thus $y=5-3\times 2=-1$.

Exercise: Simplify by putting over the common denominator:$$\frac{x}{x^2+5x+6}-\frac{x-1}{x^2+3x+2}$$

Proof. To find the common denominator, we first factor the denominators of both fractions. Note that $x^2+5x+6=(x+2)(x+3)$ and $x^2+3x+2=(x+1)(x+2)$. So the common denominator can be $(x+1)(x+2)(x+3)$. Hence\begin{aligned}\frac{x}{x^2+5x+6}-\frac{x-1}{x^2+3x+2} &= \frac{x(x+1)-(x-1)(x+3)}{(x+1)(x+2)(x+3)} \\ &= \frac{-x+3}{(x+1)(x+2)(x+3)}.\end{aligned}