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College Algebra Questions

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  • Started 2 years ago by lee

  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 2 years ago #

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