It was given that the length of stretch is directly proportional to the weight. Assuming the length is represented by l and the weight by w, we have the following relationship:
![\begin{gathered} l\propto w \\ \therefore \\ l=kw \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/jjgsfb62oips4w2kjvcz6kh1f4lxodaq6j.png)
where k is a constant.
The question gives the following parameters:
![\begin{gathered} when\text{ } \\ w=24 \\ l=12 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/326to5g7ydpj9uguvxbkb24ne23j33qx5b.png)
Therefore, we can calculate the constant to be:
![\begin{gathered} 12=k\cdot24 \\ k=(12)/(24) \\ k=(1)/(2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/jjufs54k92i6eh3luax86l7h2x0nvh57xw.png)
Therefore, the relationship will be:
![l=(1)/(2)w](https://img.qammunity.org/2023/formulas/mathematics/college/uno46drvf7mtbnjn6pxjh4y9yhyxry4trz.png)
If the weight is increased to 43 pounds, the length will be:
![\begin{gathered} l=(1)/(2)\cdot43 \\ l=21.5\approx22\text{ inches} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/b6sxgizp4n4is6q73gju90ij3qwldsi29o.png)
The spring will stretch 22 inches.