The product would expand to
![(7t-10)(5t+x)=35t^2+7tx-50t-10x=35t^2+(7x-50)t-10x](https://img.qammunity.org/2021/formulas/mathematics/middle-school/7vuenpun0tu2p898fmhnev03is71endquw.png)
This is a trinomial, and the only way to make it a binomial is to cancel out a coefficient using our variable
.
So, we can cancel either the linear term or the constant term.
In the first case, we require
![7x-50=0 \iff 7x=50 \iff x=(50)/(7)=(49)/(7)+(1)/(7)=7(1)/(7)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/nsads4f393v7ug0s5vjifkuwaixs8b1nhk.png)
In the second case, we require
![-10x=0\iff 10x=0 \iff x=0](https://img.qammunity.org/2021/formulas/mathematics/middle-school/v4eq1itbfwfhzu4td487pvjwbcwh4t8zjv.png)
But
must be a non-zero rational number, so this solution is not feasible.