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

you can use substitution to get

This identifies the 3rd selection as the appropriate one.
<span>1/6 of a change to roll a 7
Hope this helps!
</span>
Is the question 76,569,878 + 790 divided by 10 + 7/8?
Answer:
Step-by-step explanation:
1 bowl holds 5 fish
x bowls hold 263 fish
1/x = 5/263 Cross multiply
5x = 263 Divide by 5
5x/5 = 263/5
x = 52.6
So you actually need 53 fishbowls. One of them holds only 3 fish
Answer:
y=-3x-1
Step-by-step explanation:
The mistake that was made was in the denominator of the work. The second 2 should be the -1. This is what you should get:

Hope that clear things up.