# Problem of the Week

## Updated at Jan 25, 2021 2:54 PM

To get more practice in equation, we brought you this problem of the week:

How can we solve the equation $$\frac{{(3-x)}^{2}}{4(x-3)}=\frac{1}{4}$$?

Check out the solution below!

$\frac{{(3-x)}^{2}}{4(x-3)}=\frac{1}{4}$

 1 Multiply both sides by $$4(x-3)$$.${(3-x)}^{2}=\frac{1}{4}\times 4(x-3)$2 Cancel $$4$$.${(3-x)}^{2}=x-3$3 Expand.$9-6x+{x}^{2}=x-3$4 Move all terms to one side.$9-6x+{x}^{2}-x+3=0$5 Simplify  $$9-6x+{x}^{2}-x+3$$  to  $$12-7x+{x}^{2}$$.$12-7x+{x}^{2}=0$6 Factor $$12-7x+{x}^{2}$$.1 Ask: Which two numbers add up to $$-7$$ and multiply to $$12$$?$$-4$$ and $$-3$$2 Rewrite the expression using the above.$(x-4)(x-3)$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$(x-4)(x-3)=0$7 Solve for $$x$$.1 Ask: When will $$(x-4)(x-3)$$ equal zero?When $$x-4=0$$ or $$x-3=0$$2 Solve each of the 2 equations above.$x=4,3$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$x=4,3$8 Check solutionWhen $$x=3$$, the original equation $$\frac{{(3-x)}^{2}}{4(x-3)}=\frac{1}{4}$$ does not hold true.We will drop $$x=3$$ from the solution set.9 Therefore,$$x=4$$Done x=4