Gabriella Sellers

2022-06-20

How to cancel out a negative in a denominator?
The question was to make $y$ the subject in $x=5-3y$ (i.e. solve for $y$). My working was this:
$\begin{array}{rl}x& =5-3y\\ -3y& =x-5\\ y& =\left(x-5\right)/\left(-3\right)\end{array}$
But this was apparently wrong and the correct answer was $\left(5-x\right)/3$. Does anyone know how to get that and how to cancel out the $-3$ I had in the denominator of the equation?

nuvolor8

$\begin{array}{rl}y\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}& =\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\frac{\left(x-5\right)}{\left(-3\right)}\\ \\ & =\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\frac{-1\left(-x+5\right)}{-1\left(3\right)}\\ \\ & =\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\frac{-1}{-1}×\frac{\left(-x+5\right)}{3}\\ \\ & =\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\overline{)\frac{-1}{-1}}×\frac{\left(-x+5\right)}{3}\\ \\ & =\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\frac{\left(-x+5\right)}{3}\\ \\ & =\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\frac{\left(5-x\right)}{3}\end{array}$

deceptie3j