Help on these difficult math problems?

<p>well, difficult in my opinion. I had trouble with these and don't know how to solve them :(</p>

<ol>
<li><p>if x+y=7 and xy=4, then find the value of x^2+y^2.</p></li>
<li><p>if y+2q=15, q+2p=5, and p+2y=7, then find the value of p+q+y.</p></li>
<li><p>On a street with 25 houses, 10 houses have fewer than 6 rooms, 10 houses have more than 7 rooms, and 4 houses have more than 8 rooms. what is the total number of houses on the street that are either 6-,7-, or 8-room houses?</p></li>
</ol>

<p>answers:
1. 41
2. 9
3. 11</p>

<p>an explanation of how to solve one, two or all of these in the quickest way possible would be awesome!!</p>

<p>1) I just used a TI Nspire CAS to solve this…to be on the safe side.</p>

<p>2) Combine all the equations together, get 3y + 3q + 3p = 27… divide both sides by 3, get y+ q+p =9.</p>

<ol>
<li><p>(x+y)^2=x^2+2xy+y^2
plug-in: 7^2=x^2+2<em>4+y^2
x^2+y^2=7^2-2</em>4=49-8=41</p></li>
<li><p>(y+2q)+(q+2p)+(p+2y)=3(y+p+q)=15+5+7=27
y+p+q=27/3=9</p></li>
<li><p>all-(<6)=(>=6)=25-10=15 there are 15 houses with 6 or more than 6 rooms
(>=6)-(>7)=(6,7)=15-10=5 there are 5 houses with 6 or 7 rooms
(>7)-(>8)=(8)=10-4=6 there are 6 houses with 8 rooms
(6,7)+(8)=5+6=11 there are 11 houses with 6, 7 or 8 rooms</p></li>
</ol>

<p>The last Q may seem to be a little complicated. Try think of it as like, more or equal to 6 but not more than 7, thats 6 or 7. Tell me if you still don’t get it ;)</p>

<p>Hope these help!</p>

<p>The third one would seem less confusing without the “more than 7 rooms” information. It’s superfluous, and the unneeded information distracts you from the information you do need.</p>

<p>Scratch that sentence out and try the question again. From the 25, subtract the houses that have fewer than six rooms and the ones that have more than eight. The remaining houses will have six or seven or eight rooms. Easier?</p>

<p>Thanks guys! all of it made sense and definitely feeling stupid after looking at the explanations…</p>