## Intermediate Algebra (6th Edition)

The solutions are $6$ and $12$.
First, note that the value of $x$ cannot be $7$ as it will make some denominators equal to zero. Multiply $(x-7)^2$ to both sides of the equation to obtain: \begin{array}{ccc} \require{cancel} &(x-7)^2(1)&=&(x-7)^2\left[\dfrac{4}{x-7}+\dfrac{5}{(x-7)^2}\right] \\&(x-7)^2&=&(x-7)^2\left(\dfrac{4}{x-7}\right)+(x-7)^2\left[\dfrac{5}{(x-7)^2}\right] \\&(x-7)^2&=&\cancel{(x-7)^2}^{x-7}\left(\dfrac{4}{\cancel{x-7}}\right)+\cancel{(x-7)^2}\left[\dfrac{5}{\cancel{(x-7)^2}}\right] \\&(x-7)^2&=&(x-7)(4)+5 \end{array} Square the binomial using the formula $(a-b)^2=a^2-2ab+b^2$, distribute $4$ to the binomial, and combine like terms to obtain: \begin{array}{ccc} \\&x^2-2(7)(x)+7^2 &= &4(x)-7(4)+5 \\&x^2-14x+49 &= &4x-28+5 \\&x^2-14x+49 &= &4x-23 \end{array} Move all terms to the left side of the equation. Note that moving a term from one side to the other changes its sign. $$x^2-14x+49-4x+23=0 \\x^2-18x+72=0$$ Factor the trinomial to obtain: $$(x-12)(x-6)=0$$ Use the Zero Factor Property by equating each factor to zero, then solve each equation to obtain: \begin{array}{ccc} \\&x-12=0 &\text{or} &x-6=0 \\&x=12 &\text{or} &x=6 \end{array} Therefore, the solutions are $6$ and $12$.