Midterm 2: Version C Answer Key

Terrance Berg

Midterm 2: Version C Answer Key

$2x-y-2=0$

$x$ $y$

0 −2

1 0

2 2

$2x+3y+6=0$

$x$ $y$

0 −2

−3 0

−6 2
$\phantom{1}$
$x+y=2\Rightarrow x=2-y \\$
$\begin{array}{rrrrrrl} 3(2&-&y)&-&4y&=&13 \\ 6&-&3y&-&4y&=&13 \\ -6&&&&&&-6 \\ \midrule &&&&-7y&=&\phantom{-}7 \\ &&&&y&=&-1 \\ \\ &&&&x&=&2--1 \\ &&&&x&=&3 \end{array}$
$(3,-1)$
$\begin{array}{rrrrrr} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ &4x&-&3y&=&6 \\ +&4x&+&3y&=&2 \\ \midrule &&&8x&=&8 \\ &&&x&=&1 \\ \\ &3y&+&4(1)&=&2 \\ &&-&4&&-4 \\ \midrule &&&3y&=&-2 \\ \\ &&&y&=&-\dfrac{2}{3} \end{array}$
$\left(1,-\dfrac{2}{3}\right)$
$\begin{array}{ll} \begin{array}{rrrrrrrl} \\ \\ \\ \\ \\ \\ \\ \\ \left[1\right]&(x&+&y&+&z&=&\phantom{-}6)(-2) \\ \left[2\right]&&&(-2x&+&z&=&-3)(-1) \\ \\ \left[1\right]&-2x&-&2y&-&2z&=&-12 \\ +&&&2y&+&4z&=&\phantom{-}10 \\ \midrule &&&-2x&+&2z&=&-2 \\ +&&\left[2\right]&2x&-&z&=&\phantom{-}3 \\ \midrule &&&&&z&=&1 \\ \end{array} & \hspace{0.25in} \begin{array}{rrrrr} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ 2y&+&4z&=&10 \\ 2y&+&4(1)&=&10 \\ &-&4&&-4 \\ \midrule &&2y&=&6 \\ &&y&=&3 \\ \\ -2x&+&z&=&-3 \\ -2x&+&1&=&-3 \\ &-&1&&-1 \\ \midrule &&-2x&=&-4 \\ &&x&=&2 \\ \end{array} \end{array}$
$(2,3,1)$
$36+\{\cancel{-2x-\left[6x-3(5-2x)\right]\}^0}1+6x^3$
$36+1+6x^3\Rightarrow 6x^3+37$
$6a^2b(a^2-9)$
$6a^4b-54a^2b$
$\begin{array}{rrrrrlrrrr} \\ \\ \\ \\ \\ \\ &x^2&+&3x&+&5&&&& \\ \times &x^2&+&3x&+&5&&&& \\ \midrule &x^4&+&3x^3&+&5x^2&&&& \\ &&&3x^3&+&9x^2&+&15x&& \\ +&&&&&5x^2&+&15x&+&25 \\ \midrule &x^4&+&6x^3&+&19x^2&+&30x&+&25 \end{array}$
$\polylongdiv{2x^4+x^3+4x^2-4x-5}{2x+1}$
$x^2-x+18x-18$
$x(x-1)+18(x-1)$
$(x-1)(x+18)$
$2(a^2-2ab-15b^2)$
$2(a^2-5ab+3ab-15b^2)$
$2\left[a(a-5b)+3b(a-5b)\right]$
$2(a-5b)(a+3b)$
$(2x)^3-y^3$
$(2x-y)(4x^2+2xy+y^2)$
$(4y^2-x^2)(4y^2+x^2)$
$(2y-x)(2y+x)(4y^2+x^2)$
$\phantom{1}$
$B+S=30\Rightarrow B=30-S \\$
$\begin{array}{rrrrrrr} B&-&10&=&4(S&-&10) \\ 30-S&-&10&=&4S&-&40 \\ +S&+&40&&+S&+&40 \\ \midrule &&60&=&5S&& \\ \\ &&S&=&\dfrac{60}{5}&=&12 \\ \\ &&\therefore B&=&30&-&S \\ &&B&=&30&-&12 \\ &&B&=&18&& \end{array}$
$\begin{array}{rrrrrl} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ &(D&+&N&=&\phantom{1}18)(-1) \\ &(10D&+&5N&=&120)(\div 5) \\ \\ &-D&-&N&=&-18 \\ +&2D&+&N&=&\phantom{-}24 \\ \midrule &&&D&=&6 \\ \\ \therefore &D&+&N&=&18 \\ &6&+&N&=&18 \\ &-6&&&&-6 \\ \midrule &&&N&=&12 \end{array}$
$\phantom{1}$
$\text{if }x=5\%, \text{ then }10-x=30\% \\$
$\begin{array}{rrrrcrl} 5x&+&30(10&-&x)&=&25(10) \\ 5x&+&300&-&30x&=&\phantom{-}250 \\ &-&300&&&&-300 \\ \midrule &&&&-25x&=&-50 \\ \\ &&&&x&=&\dfrac{-50}{-25}\text{ or 2 L of 5\%} \\ \\ &&10&-&x&=&\text{8 L of 30\%} \end{array}$

License

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

$2x-y-2=0$
$x$	$y$
0	−2
1	0
2	2

$2x+3y+6=0$
$x$	$y$
0	−2
−3	0
−6	2

License

Share This Book