Answer:
It is not a solution.
Explanation:
First, as we know what x is equal to, substitute the second equation into the first:
3x + 2y = 27
3(y + 1) + 2y = 27
3y + 3 + 2y = 27
5y + 3 = 27
Subtract 3 from both sides:
5y + 3 - 3 = 27 - 3
5y = 24
Divide both sides by 5:
![(5y)/(5) = (24)/(5)](https://img.qammunity.org/2022/formulas/mathematics/high-school/ci9iojdk8fezwys7u2d783xztd5mzwcbwm.png)
So
![y= (24)/(5)](https://img.qammunity.org/2022/formulas/mathematics/high-school/4q615mxveaj1sw2xz3v9nebeieuhd0ekax.png)
At this point we know that (5, 6) is not a solution to the system because this should be the y value (6). To confirm this we will calculate x by substituting in the known y in the second equation:
x = y + 1
x =
![(24)/(5) + 1](https://img.qammunity.org/2022/formulas/mathematics/high-school/uus5yghdkfa9u1tc9b4u8ai7wjbn0ng2k7.png)
![x = (29)/(5)](https://img.qammunity.org/2022/formulas/mathematics/high-school/yaz3vrrajtmk11e8kexlhd1ozvajpq76an.png)
Further proving that (5,6) is not a solution.
Hope this helps!