Midterm 2: Version A Answer Key
[latexpage]
-
\(x+2y=-5\) \(x\) \(y\) 0 \(-\dfrac{5}{2}\) −5 0 \(x-y=-2\) \(x\) \(y\) 0 2 −2 0 
- \(\begin{array}{ll}
\\ \\ \\ \\ \\ \\ \\ \\
\begin{array}{rrrrrl}
&(4x&-&3y&=&13)(2) \\
&(5x&-&2y&=&\phantom{1}4)(-3) \\ \\
&8x&-&6y&=&\phantom{-}26 \\
+&-15x&+&6y&=&-12 \\
\midrule
&&&-7x&=&\phantom{-}14 \\ \\
&&&x&=&\dfrac{14}{-7}\text{ or }-2
\end{array}
& \hspace{0.25in}
\begin{array}{rrcrl}
\\
5x&-&2y&=&\phantom{+1}4 \\
5(-2)&-&2y&=&\phantom{+1}4 \\
-10&-&2y&=&\phantom{+1}4 \\
+10&&&&+10 \\
\midrule
&&-2y&=&\phantom{+}14 \\ \\
&&y&=&\dfrac{14}{-2}\text{ or }-7
\end{array}
\end{array}\)
\((-2,-7)\) - \(\begin{array}{rrrrrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\
&x&-&2y&=&-5 \\
&(2x&+&y&=&\phantom{-}5)(2) \\ \\
&x&-&2y&=&-5 \\
+&4x&+&2y&=&10 \\
\midrule
&&&5x&=&5 \\
&&&x&=&1 \\ \\
\therefore &2(1)&+&y&=&\phantom{-}5 \\
&-2&&&&-2 \\
\midrule
&&&y&=&\phantom{-}3
\end{array}\) - \(\begin{array}{ll}
\\ \\ \\ \\ \\ \\ \\
\begin{array}{rrrrrrrl}
\\ \\ \\ \\ \\ \\
&(x&+&y&+&2z&=&0)(-2) \\ \\
&-2x&-&2y&-&4z&=&0 \\
+&2x&&&+&z&=&1 \\
\midrule
&&&-2y&-&3z&=&1 \\ \\
&&&(-2y&-&3z&=&1)(3) \\
&&&(3y&+&4z&=&0)(2) \\ \\
&&&-6y&-&9z&=&3 \\
&&+&6y&+&8z&=&0 \\
\midrule
&&&&&-z&=&3 \\
&&&&&z&=&-3 \\
\end{array}
& \hspace{0.25in}
\begin{array}{rrcrl}
\\ \\
3y&+&4z&=&0 \\
3y&+&4(-3)&=&0 \\
3y&-&12&=&0 \\
&&3y&=&12 \\
&&y&=&4 \\ \\
2x&+&z&=&1 \\
2x&+&(-3)&=&1 \\
&&2x&=&4 \\
&&x&=&2 \\
\end{array}
\end{array}\)
\((2,4,-3)\) - \(28-\{5x-\cancel{\left[6x-3(5-2x)\right]^0}1\}+5x^2\)
\(28-\{5x-1\}+5x^2\)
\(28-5x+1+5x^2\)
\(5x^2-5x+29\) - \(\begin{array}{rrrcrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\
&a&-&3&& \\
\times &a&-&3&& \\
\midrule
&a^2&-&3a&& \\
+&&-&3a&+&9 \\
\midrule
&a^2&-&6a&+&9 \\ \\
&a^2&-&6a&+&9 \\
\times&&&&&4a^2 \\
\midrule
&4a^4&-&24a^3&+&36a^2 \\
\end{array}\) - \(\begin{array}{rrrrrlrrrr}
\\ \\ \\ \\ \\ \\
&x^2&+&2x&+&3&&&& \\
\times&x^2&+&2x&+&3&&&& \\
\midrule
&x^4&+&2x^3&+&3x^2&&&& \\
&&&2x^3&+&4x^2&+&6x&& \\
+&&&&&3x^2&+&6x&+&9 \\
\midrule
&x^4&+&4x^3&+&10x^2&+&12x&+&9
\end{array}\) - \(\polylongdiv{2x^3-7x^2+15}{x-2}\)
- \(a(2b+3c)-2(2b+3c)\)
\((2b+3c)(a-2)\) - \(a^2-5ab+3ab-15b^2\)
\(a(a-5b)+3b(a-5b)\)
\((a-5b)(a+3b)\) - \(x^2(x+1)-9(x+1)\)
\((x^2-9)(x+1)\)
\((x+3)(x-3)(x+1)\) - \((x)^3-(4y)^3\)
\((x-4y)(x^2+4xy+16y^2)\) - \(\phantom{1}\)
\(B+S=35\Rightarrow B=35-S \\ \)
\(\begin{array}{ll}
\begin{array}{rrrrrrrr}
&B&-&10&=&2(S&-&10) \\
&35-S&-&10&=&2S&-&20 \\
&25&-&S&=&2S&-&20 \\
+&20&+&S&&S&+&20 \\
\midrule
&&&45&=&3S&&
\end{array}
& \hspace{0.25in}
\begin{array}{rrl}
S&=&\dfrac{45}{3}\text{ or }15 \\ \\
\therefore B&=&35-S \\
B&=&35-15 \\
B&=&20 \\
\end{array}
\end{array}\) - \(\phantom{1}\)
\(D+Q=20\Rightarrow Q=20-D \\ \)
\(\begin{array}{ll}
\begin{array}{rrlrr}
10D&+&25Q&=&275 \\
10D&+&25(20-D)&=&275 \\
10D&+&500-25D&=&275 \\
&-&500&&-500 \\
\midrule
&&\phantom{500}-15D&=&-225
\end{array}
& \hspace{0.25in}
\begin{array}{rrl}
D&=&\dfrac{-225}{-15}\text{ or }15 \\ \\
\therefore Q&=&20-D \\
Q&=&20-15 \\
Q&=&5 \\
\end{array}
\end{array}\) - \(\phantom{1}\)
\(A+B=50\Rightarrow A=50-B \\ \)
\(\begin{array}{rrrrrrl}
(3.95A&+&3.70(50&-&A)&=&191.25)(100) \\ \\
395A&+&370(50&-&A)&=&19125 \\
395A&+&18500&-&370A&=&19125 \\
&-&18500&&&&-18500 \\
\midrule
&&&&25A&=&625 \\ \\
&&&&A&=&\dfrac{625}{25}\text{ or 25 kg} \\ \\
&&&&B&=&50-25 \\
&&&&B&=&25 \text{ kg} \\
\end{array}\)
