Midterm 2: Version B Answer Key

Terrance Berg

Midterm 2: Version B Answer Key

$x+y=5$

$x$ $y$

5 0

0 5

3 2

$2x-y=1$

$x$ $y$

1 1

0 −1

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

License

Icon for the Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License

$x+y=5$
$x$	$y$
5	0
0	5
3	2

$2x-y=1$
$x$	$y$
1	1
0	−1
−1	−3

License

Share This Book