Drag the tiles to the correct boxes to complete the pairs.Model each problem as an equation, and then match it to its solution.TilesThe sum of the squaresof two positive integers is185. If one integer is 3less than the other, findthe larger integer.The numerator of afraction is 1 more thantwice its denominator. If 4is added to both thenumerator and thedenominator, the fractionreduces to . Find thedenominator.The difference of apositive integer and itsinverse is . Find theinteger.

For the first problem, the integers are 8 and 11.
The second problem has missing information.  So does the third.  They cannot be worked without the information that's missing.

Explanation:
Let x be the larger integer and y be the smaller integer.  We know that the smaller integer, y, is 3 less than the larger integer, x, so:
y=x-3

The sum of their squares is 185.  This means:
x²+y²=185
x²+(x-3)²=185
x²+(x-3)(x-3)=185

Multiplying the binomials we get:
x²+x*x-3*x-3*x-3(-3)=185
x²+x²-3x-3x+9=185

Combining like terms we get:
2x²-6x+9=185

When solving quadratics, we want the function to equal 0; subtract 185 from both sides:
2x²-6x+9-185=185-185
2x²-6x-176=0

Use the quadratic formula:
[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ \\=\frac{--6\pm \sqrt{(-6)^2-4(2)(-176)}}{2(2)}=\frac{6\pm \sqrt{36--1408}}{4} \\ \\=\frac{6\pm \sqrt{1444}}{4}=\frac{6\pm 38}{4}=\frac{6-38}{4}\text{ or }\frac{6+38}{4}=\frac{-32}{4}\text{ or }\frac{44}{4}=-8\text{ or }11[/tex]

Since both integers must be positive, x = 11.  This means that y=11-3=8.