The guts of the table are ...
![\left[\begin{array}{cccc}x&y&x+y=85&y=2x+4\\19&65&\text{false}&\text{false}\\25&60&\text{true}&\text{false}\\27&58&\text{true}&\text{true}\\32&53&\text{true}&\text{false}\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7Dx%26y%26x%2By%3D85%26y%3D2x%2B4%5C%5C19%2665%26%5Ctext%7Bfalse%7D%26%5Ctext%7Bfalse%7D%5C%5C25%2660%26%5Ctext%7Btrue%7D%26%5Ctext%7Bfalse%7D%5C%5C27%2658%26%5Ctext%7Btrue%7D%26%5Ctext%7Btrue%7D%5C%5C32%2653%26%5Ctext%7Btrue%7D%26%5Ctext%7Bfalse%7D%5Cend%7Barray%7D%5Cright%5D%20%20)
Of course, you know the answer after figuring the third line of the table.
The two numbers are 27 and 58.
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If the second number is odd, the result after subtracting 4 will not be divisible by 2. This lets you reject the 1st and 4th choices immediately. As for the second choice, twice 25 is 10 less than 60, not 4 less, so that choice can also be discarded.
Starting from the system of equations
![x+y=85\\y=2x+4](https://tex.z-dn.net/?f=x%2By%3D85%5C%5Cy%3D2x%2B4)
you can use substitution to get
![x+(2x+4)=85\\\\3x=81\qquad\text{subtract 4}\\\\x=27\qquad\text{divide by 3}](https://tex.z-dn.net/?f=x%2B%282x%2B4%29%3D85%5C%5C%5C%5C3x%3D81%5Cqquad%5Ctext%7Bsubtract%204%7D%5C%5C%5C%5Cx%3D27%5Cqquad%5Ctext%7Bdivide%20by%203%7D)
This identifies the 3rd selection as the appropriate one.