Question

Please help, I'm late on this and I can't put words with numbers well enough to make anything readable!!!

Answer 1

so, what we do is, we multiply the value by a power of 10, that moves the recurring numbers to the left of the dot, in this case we want moved the "15", so we'll need two zeros, so we'll be using 100.

$\bf 0.151515\overline{15}~\hspace{7em}x=0.151515\overline{15} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{array}{|lll|ll} \cline{1-3} &&\\ 100\cdot x&=&15.1515\overline{15}\\ &&15+0.1515\overline{15}\\ &&15+x \\&&\\ \cline{1-3} \end{array}\implies 100x=15+x \\\\\\ 99x=15\implies x=\cfrac{15}{99}\implies \stackrel{simplified}{x=\cfrac{5}{33}}$

Answer 2

We are presented with the repeating decimal 0.15151515...

Let $x=0.151515...$

$100x-x=15.151515... - 0.151515... = 15$

Why did I use 100? Since there are 2 numbers that are repeated (1 and 5), we use 100, which is 10 to the power of 2. If there was 3 repeating numbers, we would be using 1000, 10 to the power of 3.

$100x-x$ is the same as $99x$

Thus, $99x=15$

Use equation operations to solve for x, which is the fraction we desire.

Divide both sides of the equation by 99.

$x = \dfrac{15}{99}$

Simplify (divide both numerator and denominator by 3)

$x=\dfrac{5}{33}$