Problem of the Week

Updated at Apr 22, 2019 5:15 PM

This week's problem comes from the equation category.

How can we solve the equation $$\frac{4m}{5(2+{m}^{2})}=\frac{4}{15}$$?

Let's begin!

$\frac{4m}{5(2+{m}^{2})}=\frac{4}{15}$

 1 Multiply both sides by $$5(2+{m}^{2})$$.$4m=\frac{4}{15}\times 5(2+{m}^{2})$2 Use this rule: $$\frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}$$.$4m=\frac{4\times 5(2+{m}^{2})}{15}$3 Simplify  $$4\times 5(2+{m}^{2})$$  to  $$20(2+{m}^{2})$$.$4m=\frac{20(2+{m}^{2})}{15}$4 Simplify  $$\frac{20(2+{m}^{2})}{15}$$  to  $$\frac{4(2+{m}^{2})}{3}$$.$4m=\frac{4(2+{m}^{2})}{3}$5 Multiply both sides by $$3$$.$12m=4(2+{m}^{2})$6 Divide both sides by $$4$$.$3m=2+{m}^{2}$7 Move all terms to one side.$3m-2-{m}^{2}=0$8 Multiply both sides by $$-1$$.${m}^{2}-3m+2=0$9 Factor $${m}^{2}-3m+2$$.1 Ask: Which two numbers add up to $$-3$$ and multiply to $$2$$?$$-2$$ and $$-1$$2 Rewrite the expression using the above.$(m-2)(m-1)$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$(m-2)(m-1)=0$10 Solve for $$m$$.1 Ask: When will $$(m-2)(m-1)$$ equal zero?When $$m-2=0$$ or $$m-1=0$$2 Solve each of the 2 equations above.$m=2,1$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$m=2,1$Donem=2,1
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}