Midterm 2: Version B Answer Key
[latexpage]
-
\(x+y=5\) \(x\) \(y\) 5 0 0 5 3 2 \(2x-y=1\) \(x\) \(y\) 1 1 0 −1 −1 −3 
- \(\begin{array}{ll}
\begin{array}{rrrrrrr}
\\ \\ \\ \\
4(4y&+&2)&+&3y&=&8 \\
16y&+&8&+&3y&=&8 \\
&-&8&&&&-8 \\
\midrule
&&&&19y&=&0 \\
&&&&y&=&0
\end{array}
& \hspace{0.25in}
\begin{array}{rrrrr}
\\ \\
x&=&4y&+&2 \\
x&=&\cancel{4y}0&+&2 \\
x&=&2&&
\end{array}
\end{array}\)
\((2,0)\) - \(\begin{array}{rrrrrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\
&5x&-&3y&=&2 \\
&(3x&+&y&=&4)(3) \\ \\
&5x&-&3y&=&\phantom{1}2 \\
+&9x&+&3y&=&12 \\
\midrule
&&&14x&=&14 \\
&&&x&=&1 \\ \\
&\therefore 3(1)&+&y&=&4 \\
&3&+&y&=&4 \\
&&&y&=&1
\end{array}\) - \(\begin{array}{ll}
\begin{array}{rrrrrrrl}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
&&&(x&-&2z&=&-7)(-1) \\ \\
&x&+&y&+&z&=&3 \\
+&-x&&&+&2z&=&7 \\
\midrule
&&&y&+&3z&=&10 \\
&&&(-2y&+&4z&=&20)(\div 2) \\
&&&&&\Downarrow&& \\
&&&y&+&3z&=&10 \\
&&+&-y&+&2z&=&10 \\
\midrule
&&&&&5z&=&20 \\
&&&&&z&=&4 \\
\end{array}
& \hspace{0.25in}
\begin{array}{rrrrr}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
x&-&2z&=&-7 \\
x&-&2(4)&=&-7 \\
x&-&8&=&-7 \\
&+&8&&+8 \\
\midrule
&&x&=&1 \\ \\
y&+&3z&=&10 \\
y&+&3(4)&=&10 \\
y&+&12&=&10 \\
&-&12&&-12 \\
\midrule
&&y&=&-2 \\
\end{array}
\end{array}\)
\((1,-2,4)\) - \(5-3\left[4x-2\cancel{(6x-5)^0}1-(7-2x)\right]\)
\(5-3\left[4x-2(1)-(7-2x)\right]\)
\(5-3\left[6x-9\right]\)
\(5-18x+27\Rightarrow -18x+32\) - \(\begin{array}{rrrlrl}
\\ \\ \\ \\ \\ \\ \\
&a&+&3&& \\
\times &a&+&3&& \\
\midrule
&a^2&+&3a&& \\
+&&&3a&+&9 \\
\midrule
&a^2&+&6a&+&9 \\
\times&&&&&3a^2 \\
\midrule
&3a^4&+&18a^3&+&27a^2
\end{array}\) - \(\begin{array}{rrrlrrrrrr}
\\ \\ \\ \\ \\ \\
&x^2&+&x&+&5\phantom{x^2}&&&& \\
\times&x^2&+&x&-&5\phantom{x^2}&&&& \\
\midrule
&x^4&+&x^3&+&5x^2&&&& \\
&&&x^3&+&x^2&+&5x&& \\
+&&&&&-5x^2&-&5x&-&25 \\
\midrule
&x^4&+&2x^3&+&x^2&-&25&&
\end{array}\) - \((x^{4n-3n}x^{-6})^{-1}\)
\((x^nx^{-6})^{-1}\)
\(x^{-n}x^6\text{ or }\dfrac{x^6}{x^n}\) - \(2a(7xy-3z)-1(7xy-3z)\)
\((7xy-3z)(2a-1)\) - \(a^2-3ab+5ab-15b^2\)
\(a(a-3b)+5b(a-3b)\)
\((a-3b)(a+5b)\) - \(2x^2(x+4)-1(x+4)\)
\((x+4)(2x^2-1)\) - \((3x)^3+(2y)^3\)
\((3x+2y)(9x^2-6xy+4y^2)\) - \(\phantom{1}\)
\(F+D=38\Rightarrow F=38-D \\ \)
\(\begin{array}{rrrrrrr}
(F&+&6)&=&4(D&+&6) \\
38-D&+&6\phantom{)}&=&4D&+&24 \\
-24+D&&&&+D&-&24 \\
\midrule
&&20&=&5D&& \\ \\
&&D&=&\dfrac{20}{5}&=&4 \\ \\
&&\therefore F&=&38&-&D \\
&&F&=&38&-&4 \\
&&F&=&34&&
\end{array}\) - \(\phantom{1}\)
\(A+B=90\Rightarrow B=90-A \\ \)
\(\begin{array}{rrrrrrl}
3A&+&5(90&-&A)&=&\phantom{-}370 \\
3A&+&450&-&5A&=&\phantom{-}370 \\
&-&450&&&&-450 \\
\midrule
&&&&-2A&=&-80 \\ \\
&&&&A&=&\dfrac{-80}{-2}\text{ or }40\text{ kg} \\ \\
&&&&\therefore B&=&90-A \\
&&&&B&=&90-40 \\
&&&&B&=&50\text{ kg}
\end{array}\) - \(\begin{array}{rrrrrrr}
\\ \\ \\ \\ \\ \\ \\
10x&+&25(40)&=&15(x&+&40) \\ \\
10x&+&1000&=&15x&+&600 \\
-10x&-&600&&-10x&-&600 \\
\midrule
&&400&=&5x&& \\ \\
&&x&=&\dfrac{400}{5}&=&80
\end{array}\)
