Example Question 1

Example corresponding to question 1 in college algebra 1.

Example 1


If $\dfrac{( k + 2 ) ! \cdot ( k + 3 ) !}{ (k+1) ! \cdot ( k + 4 ) !} = \dfrac{7}{9} $, then $k =$ ?

  1. $1$
  2. $2$
  3. $3$
  4. $4$
  5. $5$

Solution


  1. One way to solve the above equation for $k$ is to first simplify the left side of the equation. We first rewrite $(k+2)!$ and $(k+4)!$ as follows

    $(k+2)!=(k+2) \cdot (k+1)!$
    $(k+4)!=(k+4) \cdot (k+3)!$


  2. Substitute in the left side of the given equation

    $\dfrac{(k+2)! \cdot ( k + 3 ) !}{ (k+1) ! \cdot ( k + 4 ) !} = \dfrac{(k+2) \cdot (k+1)! \cdot ( k + 3 ) !}{ (k+1) ! \cdot (k+4) \cdot (k+3)!} $

  3. Simplify

    $=\dfrac{k+2}{k+4}$

  4. Rewrite equation with left side simplified

    $\dfrac{k+2}{k+4}=\dfrac{7}{9}$

  5. Use cross product and solve for $k$

    $9(k+2)=7(k+4)$

    $2k=10$

    $k=5$

    Answer E