Answers to the Sample Test on Review Material (Chapters P,1 and 2)

1. a) $\ \frac{-4}{7}$ b) $\ \sqrt{9}=3$, f) $(\sqrt[3]{-5})^{6}=25$, g) $.123456789$ and h) $2.123333..$ are rational.

2. a) $x=1.035353535...=\frac{1025}{990}= \frac{205}{198}$

b) $x=\frac{2}{7}= .\, 285\,714285714285714...=.\overline{285714}$

3. a) The distributive law: $5x-3x=(5-3)x=2x\qquad$

b) The associative law: $\ \ (3x)y=3(xy)\qquad$

c) $\$The existence of an additive identity element: $3x+0=3x$

4. a) $48\div 6-2\times \left( 5-1\right) = 8-2\cdot 4=0$

b) $(6+2^{3}\times 5\div 4)\div 2^{2}+4= \lbrack \left( 6+10\right) /4\rbrack +4=4+4=8$

c) $\ \left\vert 3-2\left( 9-5\right) \right\vert = \left\vert 3-8\right\vert =\left\vert -5\right\vert =5$

d) $\frac{48}{-12}= -4$

e) $\frac{-2(5-7)}{3(6-12)}= \frac{4}{-18}=-\frac{2}{9}$

f) $\frac{3^{-2}}{9^{\frac{-1}{2}}}= \frac{9^{\frac{1}{2}}}{3^{2}}=\frac{3}{9}=\frac{1}{3}$
g) $\frac{\sqrt{25\cdot 64}}{\sqrt{16\cdot 100}}= \frac{5\cdot 8}{4\cdot 10}=\frac{40}{40}=1$

5. There are many possible answers. For example, $\left\vert 2+(-3)\right\vert =\left\vert -1\right\vert =1\neq 5=\left\vert 2\right\vert +\left\vert -3\right\vert$

6. a) $\left\vert x-3\right\vert =9\qquad$ $x-3=\pm 9$ There are two solutions: $x=12,x=-6$.

b) $\frac{4}{12}=\frac{9}{x}$
$4x=9\cdot 12$.
$x=9\cdot 3=27$

7. a) $\frac{-9xy}{6y}= -\frac{3x}{2}$

b) $\frac{4x-12}{9-3x}= -\frac{4}{3}$

c) $\frac{x-5}{3}+\frac{4x-3}{5}-x= \frac{2x-34}{15}$

d) $\frac{3}{2x-7}-\frac{5}{x+1}= -\frac{7x-38}{\left(2x-7\right) \left( x+1\right) }$

e) $\frac{2x+3}{3x-9}\cdot \frac{x-3}{4x+6}= \frac{1}{6}$

f) $\frac{\frac{5x-7}{4x+2}}{\frac{6x}{2x+1}}= \frac{5x-7}{12x}$

g) $\frac{\frac{4}{x-2}-\frac{3}{x-3}}{\frac{1}{x-3}+\frac{2}{x-2}}= \frac{x-6}{3x-8}$

h) $\frac{(a^{2}b)((4ab^{-1})^{3}}{(2ab)^{5}}= \frac{2}{b^{7}}$

i) $\left( \frac{x^{-1}y^{2}}{xy^{3}}\right) ^{-2}= x^{4}y^{2}$

8. a) $(5x-4)+(2x^{2}-2x+1)-(x^{2}-3)= x^{2}+3x$

9. a) $(x^{2}+5y^{2})(2x-3y)= 2x^{3}-3x^{2}y+10y^{2}x-15y^{3}$

b) $(a+2b)(a+2b)= a^{2}+4ab+4b^{2}$

c) $2x(x-3)((5x+1)(x^{2}-2x+5)= 10x^{5}-48x^{4}+100x^{3}-128x^{2}-30x$

10. a) $(x^{2}+3x-1)\overline{)x^{5}-6x^{4}+2x^{2}+12x-7}$ The quotient is $x^{3}-9x^{2}+28x-91$ and the remainder is $313x-98$

b) $(x-5)\overline{)x^{4}-9x^{3}-16x-10}$ The quotient is $x^{3}-4x^{2}-20x-116$, and the remainder is $-590$

11. $\frac{x^{5}-6x^{4}+2x^{2}+12x-7}{x^{2}+3x-1}= x^{3}-9x^{2}+28x-91+\frac{313x-98}{x^{2}+3x-1}$

12. $P(x)= 5x^{3}-8x^{2}-7x+6$.
a) $P(0)=6\qquad$

b) $P(1)=5-8-7+6= -4$        

c) $P(-1)=-5-8+7+6= 0$

d) $P(2)=40-32-14+6= 0$

e) $P(-2)=-40-32+14+6= -52$

f) $P(-1+\sqrt{2})=5\left( -1+\sqrt{2}\right) ^{3} -8\left( -1+\sqrt{2}\right) ^{2}-7\left( -1+\sqrt{2}\right) +6$ $= 5\left( (-1)^{3}+3\left( -1\right) ^{2}\sqrt{2}+3\left( -1\right) (\sqrt{2})^{2}+\left( \sqrt{2}\right) ^{3}\right)$ $-8\left( \left( -1\right) ^{2}+2\left( -1\right) \sqrt{2} +\left( \sqrt{2}\right) ^{2}\right) +7-7\sqrt{2}+6$ $=5\left( -1+3\sqrt{2}-6+2\sqrt{2}\right) -8\left( 1-2\sqrt{2}+2\right) +13-7\sqrt{2}= 34\sqrt{2}-46$

b) From parts c) and d) above, we see that $x+1$ and $x-2$ are factors of $P(x).$ $P(x)= 5x^{3}-8x^{2}-7x+6=\left( x+1\right) \left( 5x^{2}-13x+6\right)$ $=\left( x+1\right) \left( x-2\right) \left( 5x-3\right)$.
Thus the three roots of $P(x)$ are $x=-1$, $x=2$ and $x=\frac{3}{5}$.

13. a) $4x^{2}-9= \left( 2x-3\right) \left( 2x+3\right)$

b) $16x^{2}+8x+1= \left( 4x+1\right) ^{2}$

c) $25x^{2}-20xy+4y^{2}= \left( 5x-2y\right) ^{2}$

d) $4y^{2}-12y+5xy-15x= \left( y-3\right) \left( 4y+5x\right)$

e) $10x^{2}+x-3= \left( 5x+3\right) \left( 2x-1\right)$

f) $\ 6t^{2}+7t-20= \left( 2t+5\right) \left( 3t-4\right)$

g) $8x^{3}-27= \left( 2x-3\right) \left( 4x^{2}+6x+9\right)$

h) $x^{3}-x^{2}-5x-3= \left( x-3\right) \left( x+1\right) ^{2}$

14. a) $\frac{2x+5}{x^{2}-7x+12}-\frac{x-1}{x^{2}-3x}+5 =\frac{2x+5}{\left( x-3\right) \left( x-4\right) } -\frac{x-1}{x\left( x-3\right) }+5$ $=\frac{x\left( 2x+5\right) -\left( x-1\right) \left( x-4\right) +5x\left( x-3... ...right) } = \frac{5x^{3}-34x^{2}+70x-4}{\left( x-3\right) x\left( x-4\right) }$

b) $\frac{5x}{x^{2}+1}-\frac{3}{x+1}=\frac{5x\left( x+1\right) -3\left( x^{2}+1\ri... ...t( x+1\right) }= \frac{2x^{2}+5x-3}{\left( x^{2}+1\right) \left( x+1\right) }$

c) $\frac{x^{2}+4x}{x^{2}+4x+4}\div \frac{x^{2}-x-20}{x^{2}+2x}= \frac{x\left( x+4... ...5\right) \left( x+4\right) }= \frac{x^{2}}{\left(x-5\right) \left( x+2\right) }$

d) $\frac{\frac{x-1}{x^{2}-3x-10}-\frac{x-2}{x^{2}+2x}}{\frac{x+1}{x^{2}-5x}+ \fra... ... x+2\right) ^{2}\left( x-5\right) }{x\left( x+2\right) ^{2}\left( x-5\right) }$ $=\frac{\left( x-1\right) x\left( x+2\right) -\left( x-2\right) \left( x+2\righ... ...-\left( x^{3}-5x^{2}-4x+20\right) } {\left(x^{3}+5x^{2}+8x+4\right) +x^{2}-5x}$ $= \frac{6x^{2}+2x-20}{ x^{3}+6x^{2}+3x+4}=\frac{ 2\left( 3x-5\right) (x+2)}{x^{3}+6x^{2}+3x+4}$

15. Simplify each of the following: a) $\left( \frac{-64}{125}\right) ^{-\frac{2}{3}}=\frac{\left( 125\right) ^{\frac{2}{3}}}{\left( -64\right) ^{\frac{2}{3}}}=\frac{5^{2}}{(-4)^{2}}=%% \frac{25}{16}$
b) $\left( \frac{(x^{3}y^{-1})(x^{-\frac{3}{2}}y^{\frac{1}{2}})}{%% (xy^{-1})^{3}}... ...{3}y^{-3}}\right) ^{2}=\frac{x^{3}y^{-1}}{x^{6}y^{-6}}= \frac{%% y^{5}}{x^{3}}$

16. $(xy^{5})^{\frac{3}{5}}+x^{\frac{8}{5}}=x^{\frac{3}{5}}y^{3}+x^{\frac{8}{5}}=x^{\frac{3}{5}}\left( y^{3}+x\right)$

17. a) $3\sqrt{x-2}+x^{2}\sqrt{x-2}-(4x-1)\sqrt{x-2} =\left(3+x^{2}-4x+1\right) \sqrt{x-2}$
                      $=\left( x^{2}-4x+4\right) \sqrt{x-2} =\left( x-2\right) ^{2}\sqrt{x-2}=\left( x-2\right) ^{\frac{5}{2}}$
b) $\sqrt{x^{2}+3}-\frac{x^{2}}{\sqrt{x^{2}+3}}=\frac{x^{2}+3}{\sqrt{%% x^{2}+3}}-... ...}}{\sqrt{x^{2}+3}}=\frac{3}{\sqrt{x^{2}+3}}=\frac{3\sqrt{%% x^{2}+3}}{x^{2}+3}$

18. a) $\frac{2}{\sqrt{7}}=\frac{2\sqrt{7}}{7}\qquad$
b) $\frac{2x+1}{\sqrt{x-1}}=\frac{\left( 2x+1\right) \sqrt{x-1}}{x-1}$
c) $\frac{2}{3+\sqrt{5}}=\frac{2}{3+\sqrt{5}}\cdot \frac{3-\sqrt{5}}{3-%% \sqrt{5}}=\frac{6-2\sqrt{5}}{9-5}=\frac{6-2\sqrt{5}}{4}=\frac{3-\sqrt{5}}{2}$
d) $\frac{x}{\sqrt[3]{(x-1)^{2}}}=\frac{x}{\left( x-1\right) ^{\frac{2}{3}}%% }\cd... ...t) ^{\frac{1}{3}}}{\left( x-1\right) ^{\frac{1}{3}}}=\frac{x\sqrt[3]{x-1}}{x-1}$

19. a) $_{8}C_{5}=\frac{8!}{5!3!}= 56$    b ) $_{37}C_{37}=\frac{37!}{0!37!}=1$    c) $_{37}C_{36}=\frac{37!}{36!0!}=37$

20. $(2x-y)^{8}=_{8}C_{0}\left( 2x\right) ^{8}-_{8}C_{1}\left( 2x\right) ^{7}y+_{8}C_{2}\left( 2x\right) ^{6}y^{2}-_{8}C_{3}\left( 2x\right) ^{5}y^{3}$
                      $+_{8}C_{4}\left( 2x\right) ^{4}y^{4}-_{8}C_{5}\left( 2x\right) ^{3}y^{5}+_{8}C_{6}\left( 2x\right) ^{2}y^{6}-_{8}C_{7}\left( 2x\right) y^{7}+_{8}C_{8}y^{8}$ $= \left( 2x\right) ^{8}-8\left( 2x\right) ^{7}y+28\left( 2x\right) ^{6}y^{2}-56\left( 2x\right) ^{5}y^{3}$ $+70\left( 2x\right) ^{4}y^{4}-56\left( 2x\right) ^{3}y^{5}+28\left( 2x\right) ^{2}y^{6}-8\left( 2x\right) y^{7}+y^{8}$ $= 256x^{8}-1024x^{7}y+1792x^{6}y^{2}-1792x^{5}y^{3}+ 1120x^{4}y^{4}-448x^{3}y^{5}$ $+112x^{2}y^{6}-16xy^{7}+ y^{8}$
a) $-16\qquad$
b) $-448$
c) $-1792$
d) $256$

21. $(x-3)^{5}= x^{5}-_{5}C_{1}x^{4}3+_{5}C_{2}x^{3}3^{2}-_{5}C_{3}x^{2}3^{3}+_{5}C_{4}x3^{4}-_{5}C_{5}3^{5}$ $= x^{5}-5\cdot 3x^{4}+10\cdot 9x^{3}-10\cdot 27x^{2}+5\cdot 81x-243$ $=x^{5}-15x^{4}+90x^{3}-270x^{2}+405x-243$

22. a) $\frac{3}{4}x-5(x-\frac{2}{3})=2x+5\qquad$(Multiply by 12)
$9x-5\left( 12x-8\right) =24x+60$
$75x=-20$
$x=-\frac{20}{75}=-\frac{4}{15}$

b) $\ 6(x-3)+3(2x-1)=9(x-7)+2(3x+6)$
$6x-18+6x-3=9x-63+6x+12$
$x=10$

c) $\frac{2}{5x-1}=\frac{3}{4+7x}\qquad \ \ \ \ \ \ \ \ \ \ \$(Multiply by $\left( 5x-1\right) \left( 7x+4\right)$)
$2(7x+4)=3\left( 5x-1\right)$
$x=11$

d) $\frac{3x+1}{5x-3}=\frac{6x-7}{10x+9}\qquad$(Multiply by $\left( 5x-3\right) \left( 10x+9\right)$)
$\left( 3x+1\right) \left( 10x+9\right) =\left( 6x-7\right) \left( 5x-3\right) \ \ \ \$
$30x^{2}+37x+9= 30x^{2}-53x+21$
$90x=12$
$x=\frac{2}{15}$

e) $\frac{5}{x+1}-\frac{2}{x+3}=\frac{3x+8}{x^{2}+4x+3}\qquad$(Multiply by $x^{2}+4x+3=\left( x+1\right) \left( x+3\right)$)
$5\left( x+3\right) -2\left( x+1\right) =3x+8$
$3x+13=3x+8$ There are no solutions.

f) $\frac{5}{x+1}-\frac{2}{x+3}=\frac{x-7}{x^{2}+4x+3}\qquad$(Multiply by $x^{2}+4x+3=\left( x+1\right) \left( x+3\right)$)
$5\left( x+3\right) -2\left( x+1\right) =x-7$
$3x+13=x-7$
$x=-10$

g) $\frac{5}{x+2}-\frac{2}{x-2}=\frac{x-10}{x^{2}-4}$ (Multiply by $\ x^{2}-4=\left( x+2\right) \left( x-2\right)$)
$5\left( x-2\right) -2\left( x+2\right) =x-10\qquad$
$3x-14=x-10$
$x=2$
However, this solution is extraneous, since if $x$ were $2$ then $x-2$ would be $0$, and we can't have $0$ in a denominator. Thus, there is no solution.

h) $\sqrt{5x-4}+14=0$
$\sqrt{5x-4}=-14\qquad$(Square both sides) (Actually, we can tell here that there is no solution, since the left hand side of the equation is positive and the right hand side is negative.)
$5x-4=196$
$5x=200$
$x=40$. But this solution is extraneous since if we substitute $x=40$ in the original equation we get
$\sqrt{196}+14=14+14=28\neq 0$. Thus there is no solution.

i) $\sqrt{5x-4}-14=0$
$\sqrt{5x-4}=14\qquad \ \$(Square both sides)
$5x-4=196$
$5x=200$
$x=40\qquad$(This time the solution checks.)

j) $\sqrt{4x-1}+3=4x+10$
$\sqrt{4x-1}=4x+7$
$4x-1=\left( 4x+7\right) ^{2}$
$4x-1= 16x^{2}+56x+49$
$16x^{2}+52x+50=0$
$8x^{2}+26x+25=0$
This problem was my mistake. It should not be here. We will not learn how to do this until Chapter 3.

k) $\sqrt{4x-1}+5=\sqrt{4x+10}\qquad$(Square both sides. Don't forget the middle term)
$4x-1+10\sqrt{4x-1}+25=4x+10$
$10\sqrt{4x-1}=-14\qquad$ Here again we can see that there is no solution as the left hand side is positive but the right hand side is negative.

l) $\left\vert x+5\right\vert -8=0$
$x+5=\pm 8$
$x=3,$ or $\ x=-13$

m) $\left\vert x+5\right\vert +8=0$
$\left\vert x+5\right\vert =-8$
Since the left hand side is positive and the right hand side is negative, there is no solution.

n) $\left\vert \frac{4x-3}{x+5}\right\vert =3$
$\frac{4x-3}{x+5}=\pm 3$
If $\frac{4x-3}{x+5}=3$
$4x-3=3\left( x+5\right) =3x+15$
$x=18$
If $\frac{4x-3}{x+5}=-3$
$4x-3=-3\left( x+5\right) =-3x-15$
$7x=-12$
$x=-\frac{12}{7}$
Thus, we have two solutions.

23. Solve each of the following: a) $2x+7<5x-8$
$15<3x$
The solution is $x>5$ (or $\left( 5,\infty \right)$)

b) $3(x+4)\geq 5(x-7)$
$3x+12\geq 5x-35$
$47\geq 2x$
The solution is : $x\leq \frac{47}{2}$ or $(-\infty ,47\rbrack$

c) $\frac{x+1}{2x-1}\geq 0$
The key points on a number line are $-1$ and $\frac{1}{2}$.
Testing values, we find that the solution is $(-\infty ,-1\rbrack \cup (\frac{1}{2},\infty )$

d) $\frac{3x+10}{x+2}\leq 2$
$\frac{3x+10}{x+2}-2\leq 0$
$\frac{3x+10}{x+2}-\frac{2\left( x+2\right) }{x+2}\leq 0$
$\frac{x+6}{x+2}\leq 0$
The solution is : $(-\infty ,-2)\cup \lbrack -6,\infty )$

e) $\frac{3x+5}{x+2}\leq 2$
$\frac{3x+5}{x+2}-2\leq 0$
$\frac{3x+5}{x+2}-\frac{2(x+2)}{x+2}\leq 0$
$\frac{x+1}{x+2}\leq 0$
The solution is : $\lbrack -1,-2)$

f) $\left\vert 4x+8\right\vert <15$
$-15<4x+8<15$
$-23<4x<7$
$-\frac{23}{4}<x<\frac{7}{4}$
The solution is $\left( -\frac{23}{4},\frac{7}{4}\right)$

g) $\ \left\vert 5x+6\right\vert \leq 10$
$-10\leq 5x+6\leq 10$
$-16\leq 5x\leq 4$
$-\frac{16}{5}\leq x\leq \frac{4}{5}$
The solution is $\left[ -\frac{16}{5},\frac{4}{5}\right]$

h) $\left\vert 3x-9\right\vert \geq 6$
$3x-9\leq -6$ or $3x-9\geq 6$
$x-3\leq -2$ or $x-3\geq 2$
$x\leq 1$ or $x\geq 5$
The solution is $(-\infty ,1\rbrack \cup \lbrack 5,\infty )$

24. Let $P_{1}(0,0)$, $P_{2}(3,4)$, and $P_{3}(7,1)$ be three points in the plane.
i) $d(P_{1},P_{2})=\sqrt{\left( 3-0\right) ^{2}+\left( 4-0\right) ^{2}}=%% \sqrt{9+16}=\sqrt{25}=5$
ii) $d(P_{2},P_{3})=\sqrt{\left( 7-3\right) ^{2}+\left( 1-4\right) ^{2}}=%% \sqrt{16+9}=\sqrt{25}=5$
iii) $d(P_{1},P_{3})=\sqrt{\left( 7-0\right) ^{2}+\left( 1-0\right) ^{2}}=%% \sqrt{49+1}=\sqrt{50}=5\sqrt{2}$

b) $d(P_{1},P_{2})^{2}+d(P_{2},P_{3})^{2}=25+25=50=d(P_{1},P_{2})^{2}$.
Therefore, by the Pythagorean Theorem, $P_{1}P_{2}P_{3}$ is a right triangle.

c) I showed you in class how to do this using distances, but it is easier to do it using slopes:
The slope, $m_{12}$ of the side $P_{1}P_{2}$ is $\frac{4-0}{3-0}= \frac{4}{3}$, so the slope $m_{34}$ of side $P_{3}P_{4}$ must also be $\frac{4}{%% 3}$, as opposite sides of a square are parallel (and hence have the same slope). Thus the equation of the line through $P_{3}$ and $P_{4}$ is $y-1=\frac{4}{3}(x-7)$. Similarly, $m_{23}=\frac{1-4}{7-3}=-\frac{3}{4}$, and so $m_{14}=-\frac{3}{4}$, and the equation of the line through $P_{1}$ and $P_{4}$ is $y-0 =-\frac{3}{4}(x-0)$. These two sides meet in point $P_{4}$, so $y=\frac{4}{3}(x-7)+1$ $y=-\frac{3}{4}x$.
$\frac{4}{3}(x-7)+1=-\frac{3}{4}x$
$16x-112+12=-9x$
$25x=100$
$x=4$ and so $y=-3$
$P_{4}=(4,-3)$.

25. a) i) $M_{12}=\left( \frac{0+3}{2},\frac{0+4}{2}\right) =\left( \frac{3}{2},2\right)$
ii) $M_{13}=\left( \frac{0+7}{2},\frac{0+1}{2}\right) =\left( \frac{%% 7}{2},\frac{1}{2}\right)$
iii) $M_{23}=\left( \frac{3+7}{2},\frac{4+1}{2}\right) =\left( 5,\frac{5}{2}\right)$
b) $d(M_{12},M_{23})=\sqrt{\left( \frac{3}{2}-5\right) ^{2}+\left( 2-\frac{5}{2}\r... ...) ^{2}}=\sqrt{\left( -\frac{7}{2}\right) ^{2}+\left( -\frac{1}{2}\right) ^{2}}$ $=\sqrt{\frac{49}{4}+\frac{1}{4}} =\sqrt{\frac{50}{4}}= \frac{5}{2}\sqrt{2}$
From problem 24, we found that $d(P_{1},P_{3})=5\sqrt{2}$, so $d(M_{12},M_{23})=\frac{1}{2}\cdot 5\sqrt{2}=\frac{1}{2}d(P_{1},P_{3})$

c) One could do this by the Pythagorean Theorem. I will do it using slopes.
The slope of $M_{12}M_{13}$ is $\frac{\frac{1}{2}-2}{\frac{7}{2}-\frac{3}{2}} =\frac{-\frac{3}{2}}{\frac{4}{2}}=-\frac{3}{4}$. [Note that $M_{12}M_{13}$ is parallel to $P_{2}P_{3}$. ]
The slope of $M_{23}M_{13}$ is $\frac{\frac{1}{2}-\frac{5}{2}}{\frac{7}{2}-5}=\frac{-\frac{4}{2}}{-\frac{3}{2}}=\frac{4}{3}$. [Note that $M_{23}M_{13}$ is parallel to $P_{2}P_{1}$. ]
Since the slope of $M_{23}M_{13}$ is the negative reciprocal of the slope of $M_{12}M_{13}$, these two lines are perpendicular and so angle $%% M_{12}M_{13}M_{23}$ is a right angle and triangle $M_{12}M_{13}M_{23}$ is a right triangle.

26. I actually did all parts of this problem in my solutions to problems 24 and 25.

27. a) The slope is $\frac{16-7}{5-3}=\frac{9}{2}$, so the equation is $y-7=\frac{9}{2}\left( x-3\right)$, or, equivalently, $y=\frac{9}{2}x-%% \frac{13}{2}$.

b) The slope is $\frac{3-7}{5-3}=-2$, so the equation is $y-7=-2\left( x-3\right)$, or, equivalently, $y=-2x+13$.

c) This is a horizontal line. The slope is $0$ and the equation is $y=7$.

d) This is a vertical line. It has no slope but its equation is $x=3$.

e) Use the point-slope form: $y+1=5\left( x-2\right)$, or, equivalently, $y=5x-11$

f) Use the point-slope form: $y+1=-2\left( x-2\right)$, or, equivalently, $y=-2x+3$

g) $y=-1.$

h) $x=2.$

i) The slope is $\frac{8-\left( -1\right) }{0-2}=-\frac{9}{2}$, so the equation is $y+1=-\frac{9}{2}\left( x-2\right)$, or, equivalently, $y=-%% \frac{9}{2}x+8$.

j) The slope is $\frac{0-\left( -1\right) }{8-2}=\frac{1}{6}$, so the equation is $y+1=\frac{1}{6}\left( x-2\right)$, or, equivalently, $y=%% \frac{1}{6}x-\frac{4}{3}$.

k) The slope of $4x+2y=11$ is $-2$ so the equation of the line through $P(2,-1)$ which is parallel is $y+1=-2\left( x-2\right)$, or, equivalently, $y=-2x+3$.

l) The line parallel to the line $4x+2y=11$ which has $y$-intercept $8$ is y=-2x+8.

m) The line $4x+2y=11$ has slope $-2$, so the slope of a perpendicular line is $\frac{1}{2}$. Thus, the equation of the desired line is $y=%% \frac{1}{2}x+8$.

n) The line through $P(3,7)$ and $Q(5,7)$ is horizontal, so any perpendicular line is vertical. The midpoint of $P$ and $\ Q$ is $%% \left( 4,7\right)$, so the equation of the desired line is $x=4$.

28. a) $x+y=7$ and $4x+5y=33$.
Solving for $y$, $y=7-x$, so, by substitution, $4x+5\left( 7-x\right) =33$. Thus, $x=2$, and so $y=5$.
b) $2x-5y=4$ and $3x+2y=6$.
Multiply the first equation by $2$ and the second equation by $5$, and then add.
$4x-10y=8$
$15x+10y=30$
$19x=38$
$x=2$; $y=0$.

c) $3x-4y=18$ and $8y-6x=4$.
Multiply the first equation by $-2$ and add.
$6x-8y=36$
$-6x+8y=4$
$0=40$, which is impossible. Therefore, there are no solutions.

d) $3x-4y=18$ and $y=-5x+7$.
Substitute for $y$.
$3x-4\left( -5x+7\right) =18$
$23x-28=18$
$23x=46$
$x=2$; $y=-3$

e) $3x-6y=21$ and $y=\frac{1}{2}x-\frac{7}{2}$.
Substitute for $y$.
$3x-6\left( \frac{1}{2}x-\frac{7}{2}\right) =21$
$3x-3x+21=21$.
This holds for all values of $x.$ Therefore, there are infinitely many solutions: $x=t$; $y=\frac{1}{2}t-\frac{7}{2}$ for any value of $t$.

29. a) Let $b$ be the price of the blouse, $k$ the price of the skirt, and $w$ the price of the sweater. Then $b-\frac{1}{3}b=\frac{2}{3}b=\$12$; so $b=\$18$. $k-\frac{1}{4}k=\frac{3}{4}k=\$21$; so $k=\$28$. $w-\frac{1}{2}w=\frac{1}{2}w=\$18$; so $w=\$36$. In all, the original price of the three items was $\$18+\$28+\$36= \$82$, but Sally only paid $\$12+\$21+\$18= \$51$ - a $\ \$31$ discount. Thus the percentage discount was $\frac{31}{82}\times 100\approx 37.8\%$.

b) If Mary has driven 120 miles and averaged 40 miles per hour, then she has been driving for 3 hours. [time=distance/rate]. If she has sixty miles to go, then the entire trip would be 180 miles. To average 45 miles per hour, she would have to drive 180 miles in 4 hours [rate=distance/time[. She therefore has 1 hour to complete her trip. Since she still has 60 miles to go, she must drive the remainder of her trip at 60 miles per hour in order to arrive in time.