Pre Calculus Homework Help

  We feel how you feel when you have to deal with those hard Calculus problems. We can help with that extra edge that you need! We can provide you with step-by-step solutions. See some examples below:

Problem 1. Solve the following system of equations:

\[ 3x+2y=11 \,\,\,\,\,\,\,\,\,(1) \] \[ 5x-4y=11 \,\,\,\,\,\,\,\,\, (2) \]

Solution. We have to find $x$ and $y$ in such a way that (1) and (2) are satisfied simultaneously. The method consists on eliminating one of the variables using both equation, multiplying them by the right constants. More precisely, if we multiply the first equation by 2 we obtain

\[ 6x+4y=22 \,\,\,\,\,\,\,\,\,(3) \] \[ 5x-4y=11 \,\,\,\,\,\,\,\,\, (4) \]

and now we sum both equations (3) and (4) to get $11x=33$, which means that $x=3$. Now that we have $x$, we can find $y$ simply replacing any of the equations (1) or (2). We choose (1), so we get $9+2y=11$, which implies that $2y=2$, and then $y=1$. Summarizing

\[ x=3 \quad , \quad y=1 \]

is the solution. $\Box$

Problem 2. Rationalize the denominator

\[\displaystyle{ \frac{\sqrt{2}}{\sqrt 3-\sqrt 2}} \]

Solution. We have the factor $\sqrt 3-\sqrt 2$ in the denominator, so we multiply both the numerator and denominator by the conjugate $\sqrt 3+\sqrt 2$, to obtain:

\[\displaystyle \frac{\sqrt{2}}{\sqrt 3-\sqrt 2}= \frac{\sqrt{2} (\sqrt 3+\sqrt 2)}{(\sqrt 3-\sqrt 2)(\sqrt 3+\sqrt 2)} = \frac{\sqrt 2 \sqrt 3 + (\sqrt 2)^2}{(\sqrt 3)^2-(\sqrt 2)^2} \] \[\displaystyle = \frac{\sqrt 2 \sqrt 3 + (\sqrt 2)^2}{3-2}=\sqrt 6-2 \]

This way, we have found an equivalent expression with no radicals at the denominator, i.e.

\[ \displaystyle \frac{\sqrt{2}}{\sqrt 3-\sqrt 2}=\sqrt 6-2 \]

but this last expression is rationalized. $\Box$