let x, y, z be the three test scores
![(x+y+z)/(3)=92](https://img.qammunity.org/2023/formulas/mathematics/college/4seb72pcqow50frxpoqdcw6gljxqifjsv7.png)
y is the median, y = 90
If the range is 6, then;
z - x = 6
Make z subject of formula from z - x = 6
z = 6+ x
substitute y =90 and z = 6+ x into our first equation and then solve for x
![(x+90+6+x)/(3)=\text{ 92}](https://img.qammunity.org/2023/formulas/mathematics/college/oq8an87u1kzx3m4a8iwsdhlqridzxj3n2y.png)
![(2x+96)/(3)=92](https://img.qammunity.org/2023/formulas/mathematics/college/rl3bm941f14lewn14unhze2gduuiuu939m.png)
Multiply bothside by 3
![2x\text{ + 96 =276}](https://img.qammunity.org/2023/formulas/mathematics/college/ibvjw6f1rdukmm0d7acc1t4tl91p63bxdr.png)
subtract 96 from bothside
![2x\text{ =180}](https://img.qammunity.org/2023/formulas/mathematics/college/r5rbso0vcjp83rxlc1he25rhzam664pxe4.png)
Divide bothside by 2
![x=(180)/(2)](https://img.qammunity.org/2023/formulas/mathematics/college/fb684ce9jql8lvwmpljh8fwxii0kc2x89m.png)
![x=90](https://img.qammunity.org/2023/formulas/mathematics/high-school/b3gysuczch2cy3du01rnrwhaomq73y1mdz.png)
substitute x =90 into z = 6+ x
z = 6 + 90 =96
The three test scores are;
90, 90, 96