# Problem of the Week

## Updated at Aug 31, 2020 5:40 PM

This week's problem comes from the equation category.

How would you solve the equation $$\frac{5}{4t}\times \frac{5}{t-3}=\frac{25}{16}$$?

Let's begin!

$\frac{5}{4t}\times \frac{5}{t-3}=\frac{25}{16}$

 1 Use this rule: $$\frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}$$.$\frac{5\times 5}{4t(t-3)}=\frac{25}{16}$2 Simplify  $$5\times 5$$  to  $$25$$.$\frac{25}{4t(t-3)}=\frac{25}{16}$3 Multiply both sides by $$4t(t-3)$$.$25=\frac{25}{16}\times 4t(t-3)$4 Use this rule: $$\frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}$$.$25=\frac{25\times 4t(t-3)}{16}$5 Simplify  $$25\times 4t(t-3)$$  to  $$100t(t-3)$$.$25=\frac{100t(t-3)}{16}$6 Simplify  $$\frac{100t(t-3)}{16}$$  to  $$\frac{25t(t-3)}{4}$$.$25=\frac{25t(t-3)}{4}$7 Multiply both sides by $$4$$.$100=25t(t-3)$8 Expand.$100=25{t}^{2}-75t$9 Move all terms to one side.$100-25{t}^{2}+75t=0$10 Factor out the common term $$25$$.$25(4-{t}^{2}+3t)=0$11 Factor out the negative sign.$25\times -({t}^{2}-3t-4)=0$12 Divide both sides by $$25$$.$-{t}^{2}+3t+4=0$13 Multiply both sides by $$-1$$.${t}^{2}-3t-4=0$14 Factor $${t}^{2}-3t-4$$.1 Ask: Which two numbers add up to $$-3$$ and multiply to $$-4$$?$$-4$$ and $$1$$2 Rewrite the expression using the above.$(t-4)(t+1)$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$(t-4)(t+1)=0$15 Solve for $$t$$.1 Ask: When will $$(t-4)(t+1)$$ equal zero?When $$t-4=0$$ or $$t+1=0$$2 Solve each of the 2 equations above.$t=4,-1$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$t=4,-1$Done t=4,-1