Let a denotes the weight of a plate
Let b denotes the weight of a blocks
We are given
[tex]\begin{gathered} a+15b=140 \\ a+5b=90 \end{gathered}[/tex]We want to find a
Solution
The above equation is a simultaneous equation
and we want to find a
[tex]\begin{gathered} a+15b=140\text{ eqn(1)} \\ a+5b=90\text{ eqn(2)} \end{gathered}[/tex]we multiply eqn(2) by 3 throughout, so that the new system of equations will be
[tex]\begin{gathered} a+15b=140\ldots\ldots\ldots(1) \\ 3a+15b=270\ldots\ldots\ldots(2) \end{gathered}[/tex]Now, eqn(2) - eqn(1)
[tex]\begin{gathered} (3a-a)+(15b-15b)=270-140 \\ 2a+0=130 \\ 2a=130 \\ a=\frac{130}{2} \\ a=65 \end{gathered}[/tex]The plate weigh 65g