## College Algebra Questions

(2 posts)

Tags:

1. lee
Key Master

For those in my online college algebra class who have trouble with the following problems, here are detailed solutions.

Question 1.

Solution. The LHS is
\begin{align*}
\frac{6}{7}y+\frac{1}{6}(y-7)&=\frac{6}{7}y+\frac{1}{6}y-\frac{7}{6}\\
&=\frac{36}{42}y+\frac{7}{42}y-\frac{7}{6}\\
&=\frac{43}{42}y-\frac{7}{6}
\end{align*}
The RHS is
$$\frac{y+1}{4}=\frac{y}{4}+\frac{1}{4}$$
Since LHS=RHS,
$$\frac{43}{42}y-\frac{7}{6}=\frac{y}{4}+\frac{1}{4}$$
Adding $\frac{7}{6}$ to and subtracting $\frac{y}{4}$ from each side result
$$\frac{43}{42}y-\frac{y}{4}=\frac{7}{6}+\frac{1}{4}$$
which simplifies to
$$\frac{65}{84}y=\frac{17}{12}$$
Solving it for $y$ we obtain
$$y=\frac{17}{12}\cdot\frac{84}{65}=\frac{119}{65}$$

Question 2.

Solution. The LHS is
\begin{align*}
4x-\frac{x}{5}+\frac{x+1}{5}&=4x-\frac{x}{4}+\frac{x}{5}+\frac{1}{5}\\
&=4x+\frac{-5x+4x}{20}+\frac{1}{5}\\
&=4x-\frac{x}{20}+\frac{1}{5}\\
&=\frac{79x}{20}+\frac{1}{5}
\end{align*}
Since LHS=RHS,
$$\frac{79x}{20}+\frac{1}{5}=3x$$
Subtracting $\frac{1}{5}$ and $3x$ from each side results
$$\frac{79x}{20}-3x=-\frac{1}{5}$$
which simplifies to
$$\frac{19x}{20}=-\frac{1}{5}$$
Dividing each side by $\frac{19}{20}$ (or equivalently multiplying each side by $\frac{20}{19}$) we obtain
$$x=-\frac{1}{5}\cdot\frac{20}{19}=-\frac{4}{19}$$

Question 3.

Solution. Multiplying each equation by $4x$ we obtain
$$32=5+4x$$
Subtracting $4x$ from each side results
$$27=4x$$
Hence $x=\frac{27}{4}$.

Question 4.

Solution. The LHS is
$$(t-6)^2=t^2-12t+36$$
Here the formula $(a-b)^2=a^2-2ab+b^2$ is used. The RHS is
\begin{align*}
(t+6)^2+48&=(t^2+12t+36)+48\\
&=t^2+12t+84
\end{align*}
Since LHS=RHS,
$$t^2-12t+36=t^2+12t+84$$
Subtracting $t^2$, $12t$ and $36$ from each side results
$$-24t=48$$
Hence $t=-2$.

Posted 1 year ago #
2. octavinsu
Member

Posted 18 hours ago #