We are given two lengths of the rectangle:
RU=3x-6
UT=x+9
These two lengths are shown in the following diagram:
Since this is a rectangle, the lengths of RU and UT must be equal:
![RU=UT](https://img.qammunity.org/2023/formulas/mathematics/college/m5sy2c6jwckv1csvk3fo8ptoa9milqqyib.png)
Thus
![3x-6=x+9](https://img.qammunity.org/2023/formulas/mathematics/college/41qnb30k1gf4shy095fq1tce3avler5s5t.png)
We need to solve this equation for x.
We start by subtracting x to both sides of the equation:
![\begin{gathered} 3x-x-6=9 \\ 2x-6=9 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/e8b58187yhzd6inudkrlct1eqnh52nvbm3.png)
Now, add 6 to both sides:
![\begin{gathered} 2x=9+6 \\ 2x=15 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/gs879yz6mh0awf7kvoe5hxb8w430inzkc7.png)
Finally, divide both sides by 2:
![\begin{gathered} (2x)/(2)=(15)/(2) \\ x=7.5 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/vgw5a125n1tc33xe4krapvuucn14sl2b0i.png)
We have the value of x: x=7.5
Now we have to find the length of QS. Since QS and RT are diagonals of the same rectangle, they have to be equal:
![RT=QS](https://img.qammunity.org/2023/formulas/mathematics/college/bqqyqzpt18yy51gqdk5dk4xm842ha8v9yd.png)
This means that we can find RT by adding RU and UT, and the result will be equal to QS:
![QS=RU+TU](https://img.qammunity.org/2023/formulas/mathematics/college/fy96rmzrfl7myd6h3ceuvpp8g5elotlepw.png)
substituting the given expressions for RU and TU:
![QS=3x-6+x+9](https://img.qammunity.org/2023/formulas/mathematics/college/54152vswf0kkbc8jqh70z5fw883qco3t7p.png)
And now, substitute x=7.5 and solve for QS:
![QS=3(7.5)-6+7.5+9](https://img.qammunity.org/2023/formulas/mathematics/college/pqal28tsmne8eo3vqbcp2s7nxomt9m583p.png)
![\begin{gathered} QS=22.5-6+7.5+9 \\ QS=33 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/o3wir2189pphz8iqwx66bekm916dta523m.png)
Answer:
x=7.5 and QS=33