Checking each equation one by one.
First, take the equation 5x+5y=10
Now, take 5 commons on both sides of the equation.
![5(x+y)=5(2)](https://img.qammunity.org/2023/formulas/mathematics/college/mu4fmv9n70gtdo55zb1yv7k25mcevscv31.png)
Now, cancel out 5 from both sides of the equation.
![\begin{gathered} x+y=2 \\ x+y-2=0 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/bnmmsny3k3qtk5jf9w09g2ygwc8opmen0p.png)
Hence, proved this equation is not equivalent to equation x+y-10=0.
Now, check the second equation.
![\begin{gathered} 2x+2y=5 \\ 2(x+y)=5 \\ x+y=(5)/(2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/9tfwijhowllbnzilv7z7we3qolzc17oovp.png)
This equation is not equivalent to the equation x+y-10=0.
Now, check the equation third.
![\begin{gathered} x+y=7 \\ x+y-7=0 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/zvyw7y7yxtmzef0n82s2dvmouznkgpxt0q.png)
It is clear that this equation is also not equivalent to the equation x+y-10=0.
Now, check the last equation.
![7x+7y=70](https://img.qammunity.org/2023/formulas/mathematics/college/chcpgssy3zsp9o4xa16k1ziscisww5xihu.png)
Take the 7 commons from both sides of the equation.
![\begin{gathered} 7(x+y)=7*(10) \\ (x+y)=10 \\ x+y-10=0 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/c98usc17lvkom2v7c6bgk0i02ef1px1lbd.png)
Hence, proved this equation is equivalent to the equation x+y-10=0.
Therefore, the given option 7x+7y=70 is correct.