Problem of the Week

Updated at Jan 22, 2024 8:44 AM

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

How would you solve 45z−33=−1115\frac{\frac{4}{5}z-3}{3}=-\frac{11}{15}?

Check out the solution below!



45z−33=−1115\frac{\frac{4}{5}z-3}{3}=-\frac{11}{15}

1
Simplify  45z\frac{4}{5}z  to  4z5\frac{4z}{5}.
4z5−33=−1115\frac{\frac{4z}{5}-3}{3}=-\frac{11}{15}

2
Simplify  4z5−33\frac{\frac{4z}{5}-3}{3}  to  −1+4z53-1+\frac{\frac{4z}{5}}{3}.
−1+4z53=−1115-1+\frac{\frac{4z}{5}}{3}=-\frac{11}{15}

3
Simplify  4z53\frac{\frac{4z}{5}}{3}  to  4z5×3\frac{4z}{5\times 3}.
−1+4z5×3=−1115-1+\frac{4z}{5\times 3}=-\frac{11}{15}

4
Simplify  5×35\times 3  to  1515.
−1+4z15=−1115-1+\frac{4z}{15}=-\frac{11}{15}

5
Regroup terms.
4z15−1=−1115\frac{4z}{15}-1=-\frac{11}{15}

6
Add 11 to both sides.
4z15=−1115+1\frac{4z}{15}=-\frac{11}{15}+1

7
Simplify  −1115+1-\frac{11}{15}+1  to  415\frac{4}{15}.
4z15=415\frac{4z}{15}=\frac{4}{15}

8
Multiply both sides by 1515.
4z=415×154z=\frac{4}{15}\times 15

9
Cancel 1515.
4z=44z=4

10
Divide both sides by 44.
z=1z=1

Done
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