# Midterm 2: Version C Answer Key

1. $2x-y-2=0$
$x$ $y$
0 −2
1 0
2 2
$2x+3y+6=0$
$x$ $y$
0 −2
−3 0
−6 2

2. $x+y=2\Rightarrow x=2-y$
$\begin{array}{rrrrrrl} 3(2&-&y)&-&4y&=&13 \\ 6&-&3y&-&4y&=&13 \\ -6&&&&&&-6 \\ \hline &&&&-7y&=&\phantom{-}7 \\ &&&&y&=&-1 \\ \\ &&&&x&=&2--1 \\ &&&&x&=&3 \end{array}$
$(3,-1)$
3. $\phantom{a}$
$\begin{array}[t]{rrrrrr} &4x&-&3y&=&6 \\ +&4x&+&3y&=&2 \\ \hline &&&8x&=&8 \\ &&&x&=&1 \\ \\ &3y&+&4(1)&=&2 \\ &&-&4&&-4 \\ \hline &&&3y&=&-2 \\ \\ &&&y&=&-\dfrac{2}{3} \end{array}$
$\left(1,-\dfrac{2}{3}\right)$
4. $\phantom{a}$
$\begin{array}[t]{ll} \begin{array}[t]{rrrrrrrl} \left[1\right]&(x&+&y&+&z&=&\phantom{-}6)(-2) \\ \left[2\right]&&&(-2x&+&z&=&-3)(-1) \\ \\ \left[1\right]&-2x&-&2y&-&2z&=&-12 \\ +&&&2y&+&4z&=&\phantom{-}10 \\ \hline &&&-2x&+&2z&=&-2 \\ +&&\left[2\right]&2x&-&z&=&\phantom{-}3 \\ \hline &&&&&z&=&1 \\ \end{array} & \hspace{0.25in} \begin{array}[t]{rrrrr} 2y&+&4z&=&10 \\ 2y&+&4(1)&=&10 \\ &-&4&&-4 \\ \hline &&2y&=&6 \\ &&y&=&3 \\ \\ -2x&+&z&=&-3 \\ -2x&+&1&=&-3 \\ &-&1&&-1 \\ \hline &&-2x&=&-4 \\ &&x&=&2 \\ \end{array} \end{array}$
$(2,3,1)$
5. $36+\{\cancel{-2x-\left[6x-3(5-2x)\right]\}^0}1+6x^3$
$36+1+6x^3\Rightarrow 6x^3+37$
6. $6a^2b(a^2-9)$
$6a^4b-54a^2b$
7. $\phantom{a}$
$\begin{array}[t]{rrrrrlrrrr} &x^2&+&3x&+&5&&&& \\ \times &x^2&+&3x&+&5&&&& \\ \hline &x^4&+&3x^3&+&5x^2&&&& \\ &&&3x^3&+&9x^2&+&15x&& \\ +&&&&&5x^2&+&15x&+&25 \\ \hline &x^4&+&6x^3&+&19x^2&+&30x&+&25 \end{array}$
8. $\phantom{a}$
9. $x^2-x+18x-18$
$x(x-1)+18(x-1)$
$(x-1)(x+18)$
10. $2(a^2-2ab-15b^2)$
$2(a^2-5ab+3ab-15b^2)$
$2\left[a(a-5b)+3b(a-5b)\right]$
$2(a-5b)(a+3b)$
11. $(2x)^3-y^3$
$(2x-y)(4x^2+2xy+y^2)$
12. $(4y^2-x^2)(4y^2+x^2)$
$(2y-x)(2y+x)(4y^2+x^2)$
13. $B+S=30\Rightarrow B=30-S$
$\begin{array}{rrrrrrr} B&-&10&=&4(S&-&10) \\ 30-S&-&10&=&4S&-&40 \\ +S&+&40&&+S&+&40 \\ \hline &&60&=&5S&& \\ \\ &&S&=&\dfrac{60}{5}&=&12 \\ \\ &&\therefore B&=&30&-&S \\ &&B&=&30&-&12 \\ &&B&=&18&& \end{array}$
14. $\phantom{a}$
$\begin{array}[t]{rrrrrl} &(D&+&N&=&\phantom{1}18)(-1) \\ &(10D&+&5N&=&120)(\div 5) \\ \\ &-D&-&N&=&-18 \\ +&2D&+&N&=&\phantom{-}24 \\ \hline &&&D&=&6 \\ \\ \therefore &D&+&N&=&18 \\ &6&+&N&=&18 \\ &-6&&&&-6 \\ \hline &&&N&=&12 \end{array}$
15. $\text{if }x=5\%, \text{ then }10-x=30\%$
$\begin{array}{rrrrcrl} 5x&+&30(10&-&x)&=&25(10) \\ 5x&+&300&-&30x&=&\phantom{-}250 \\ &-&300&&&&-300 \\ \hline &&&&-25x&=&-50 \\ \\ &&&&x&=&\dfrac{-50}{-25}\text{ or 2 L of 5%} \\ \\ &&10&-&x&=&\text{8 L of 30%} \end{array}$