Let x be the volume of 34% sand soil and y the volume of 28% sand soil, both in cubic meters.
The total volume of the mix is x+y. Then:
![x+y=64](https://img.qammunity.org/2023/formulas/mathematics/college/y4fvxh58ot2url285yyyub7ztqqisoxo39.png)
There is (34/100)x sand on the first mix and (28/100)y on the second mix. Together, they must account for a total of (31/100)*64 of sand. Then:
![(34)/(100)x+(28)/(100)y=(31)/(100)\cdot64](https://img.qammunity.org/2023/formulas/mathematics/college/pivnf4re3z7a7umqwhvzw17zt91r8gaaab.png)
These two equations form a 2x2 system of equations. Solve it using the substitution method to find the volume of each mix that will be needed. To do so, isolate x from the second equation and substitute the resulting expression into the first one:
![\begin{gathered} (34)/(100)x+(28)/(100)y=(31)/(100)\cdot64 \\ \Rightarrow34x+28y=31\cdot64 \\ \Rightarrow34x=1984-28y \\ \Rightarrow x=(1984-28y)/(34) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/qnfxr8g3w6lt0enuvwnk5q1frj4iz0iwxa.png)
![\begin{gathered} x+y=64 \\ \Rightarrow(1984-28y)/(34)+y=64 \\ \Rightarrow1984-28y+34y=64\cdot34 \\ \Rightarrow1984+6y=2176 \\ \Rightarrow6y=2176-1984 \\ \Rightarrow6y=192 \\ \Rightarrow y=(192)/(6) \\ \Rightarrow y=32 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/sf4870rxcmn3ugymddt1w2pxkahtqcruqx.png)
Substitute back y=32 into the first equation and solve for x:
![\begin{gathered} x+y=64 \\ \Rightarrow x+32=64 \\ \Rightarrow x=64-32 \\ \Rightarrow x=32 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/2e595ub3t77gu41paeildm3caonetki2xl.png)
Therefore, 32 cubic meters of each kind of mix should be used.