Question

If (x^2 - 6x) is vertical to (1/2x + 42) what is x?​

Answer 1

Check the picture below.

`x^2-6x=\cfrac{1}{2}x+42\implies x^2-6x=\cfrac{x}{2}+42\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{2}}{2(x^2-6x)=2\left( \cfrac{x}{2}+42 \right)} \\\\\\ 2x^2-12x=x+84\implies 2x^2-13x=84 \\\\\\ 2x^2-13x-84=0 \implies (x+4)(2x-21)=0 \\\\[-0.35em] ~\dotfill\\\\ x+4=0\implies x=-4 ~~ \bigotimes \\\\[-0.35em] ~\dotfill\\\\ 2x-21=0\implies 2x=21\implies x=\cfrac{21}{2} ~~ \textit{\LARGE \checkmark} \\\\[-0.35em] \rule{34em}{0.25pt}`

`\measuredangle 1+\cfrac{1}{2}x+42~~ = ~~180\implies \measuredangle 1+\cfrac{1}{2}\cdot \cfrac{21}{2}+42=180 \\\\\\ \measuredangle 1+\cfrac{21}{4}+42=180\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{4}}{4\left( \measuredangle 1+\cfrac{21}{4}+42 \right)=4(180)} \\\\\\ 4\measuredangle 1+21+168=720\implies 4\measuredangle 1+189=720\implies 4\measuredangle 1=531 \\\\\\ \measuredangle 1=\cfrac{531}{4}\implies {\Large \begin{array}{llll} \measuredangle 1=132(3)/(4) \end{array}}`