The guts of the table are ...
[tex] \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] [/tex]
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
[tex]x+y=85\\y=2x+4[/tex]
you can use substitution to get
[tex]x+(2x+4)=85\\\\3x=81\qquad\text{subtract 4}\\\\x=27\qquad\text{divide by 3}[/tex]
This identifies the 3rd selection as the appropriate one.
