Midterm 2: Version D Answer Key

Terrance Berg

Midterm 2: Version D Answer Key

$x-2y=-6$

$x$ $y$

0 3

−6 0

$x+y=6$

$x$ $y$

0 6

6 0
$\begin{array}{rrrrrl} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ &(3x&-&2y&=&0)(5) \\ &(2x&+&5y&=&0)(2) \\ \\ &15x&-&10y&=&0 \\ +&4x&+&10y&=&0 \\ \midrule &&&19x&=&0 \\ &&&x&=&0 \\ \\ &\therefore \cancel{2x}0&+&5y&=&0 \\ &&&5y&=&0 \\ &&&y&=&0 \end{array}$
$(0,0)$
$\begin{array}{rrrrrr} \\ \\ &2x&-&3y&=&8 \\ +&-2x&+&3y&=&4 \\ \midrule &&&0&=&12 \\ \end{array}$
$\phantom{1}$
∴ No solution. Parallel lines.
$\begin{array}{ll} \begin{array}{rrrrrrrl} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ &(2x&+&y&-&3z&=&-7)(2) \\ \\ &4x&+&2y&-&6z&=&-14 \\ +&&-&2y&+&3z&=&\phantom{-1}9 \\ \midrule &&&4x&-&3z&=&-5 \\ \\ &&&(3x&+&z&=&6)(3) \\ \\ &&&4x&-&3z&=&-5 \\ +&&&9x&+&3z&=&18 \\ \midrule &&&&&13x&=&13 \\ &&&&&x&=&1 \end{array} & \hspace{0.25in} \begin{array}{rrrrr} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ 3x&+&z&=&6 \\ 3(1)&+&z&=&6 \\ -3&&&&-3 \\ \midrule &&z&=&3 \\ \\ -2y&+&3z&=&9 \\ -2y&+&3(3)&=&9 \\ &&-9&&-9 \\ \midrule &&-2y&=&0 \\ &&y&=&0 \end{array} \end{array}$
$(1,0,3)$
$36-\cancel{\{-2x-\left[6x-3(5-2x)\right]\}^0}1+3x^2$
$36-1+3x^2$
$35+3x^2$
$3a^2(a^2-4a+4)$
$3a^4-12a^3+12a^2$
$\begin{array}{rrrrrlrrrr} \\ \\ \\ \\ \\ \\ &x^2&+&2x&-&4&&&& \\ \times &x^2&+&2x&-&4&&&& \\ \midrule &x^4&+&2x^3&-&4x^2&&&& \\ &&&2x^3&+&4x^2&-&8x&& \\ +&&&&-&4x^2&-&8x&+&16 \\ \midrule &x^4&+&4x^3&-&4x^2&-&16x&+&16 \end{array}$
$\polylongdiv{x^4+4x^3+4x^2+10x+20}{x+2}$
$x^2-3x+6x-18$
$x(x-3)+6(x-3)$
$(x-3)(x+6)$
$3x^2+xy+24xy+8y^2$
$x(3x+y)+8y(3x+y)$
$(3x+y)(x+8y)$
$(5x)^3-y^3$
$(5x-y)(25x^2+5xy+y^2)$
$(9y^2-4x^2)(9y^2+4x^2)$
$(3y-2x)(3y+2x)(9y^2+4x^2)$
$\phantom{1}$
$B+G=18\Rightarrow G=18-B \\$
$\begin{array}{rrrcrrrr} &G&-&4&=&4(B&-&4) \\ &18-B&-&4&=&4B&-&16 \\ +&16+B&&&&+B&+&16 \\ \midrule &&&30&=&5B&& \\ \\ &&&B&=&\dfrac{30}{5}&=&6 \\ \\ &&&\therefore G&=&18&-&B \\ &&&\phantom{\therefore}G&=&18&-&6 \\ &&&\phantom{\therefore}G&=&12&& \end{array}$
$\begin{array}{rrrrrl} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ &(D&+&Q&=&20)(-10) \\ &10D&+&25Q&=&350 \\ \\ &-10D&-&10Q&=&-200 \\ +&10D&+&25Q&=&\phantom{-}350 \\ \midrule &&&15Q&=&150 \\ \\ &&&Q&=&\dfrac{150}{15}\text{ or }10 \\ \\ \therefore &D&+&Q&=&\phantom{-}20 \\ &D&+&10&=&\phantom{-}20 \\ &&-&10&&-10 \\ \midrule &&&D&=&10 \end{array}$
$\phantom{1}$
$A+B=60\Rightarrow B=60-A \\$
$\begin{array}{llclrll} \\ \\ \\ \\ \\ \\ \\ &&3.80A&+&3.55B&=&\phantom{-}218.50 \\ 3.80A&+&3.55(60&-&A)&=&\phantom{-}218.50 \\ 3.80A&+&213&-&3.55A&=&\phantom{-}218.50 \\ &-&213&&&=&-213 \\ \midrule &&&&0.25A&=&5.50 \\ \\ &&&&A&=&\dfrac{5.50}{0.25}\text{ or 22 kg} \\ \\ &&&&B&=&60-A \\ &&&&B&=&60-22 \\ &&&&B&=&38\text{ kg} \end{array}$

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$x-2y=-6$
$x$	$y$
0	3
−6	0

$x+y=6$
$x$	$y$
0	6
6	0

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