Opposite angles are equal, so \(-6x+9y=117\). Adjacent angles sum to \(180^\circ\), so \(2x+3y=180-117=63\). Multiply the second equation by 3: \(6x+9y=189\). Add: \(18y=306\), hence \(y=17\). Then \(2x+51=63\), so \(x=6\).

\(3(4x-3)=5(7-2x)\), so \(12x-9=35-10x\). Thus \(22x=44\), \(x=2\).

(a) \(p^2=(-15)^2-8(7)=169\), so \(p=\pm13\).

(b) \(8t=r^2-p^2\), giving \(t=\frac{r^2-p^2}{8}\).

\(7x=25-4=21\), hence \(x=\boxed{3}\).

Add the equations: \(3x=15\), so \(\boxed{x=5}\). Then \(y=12-5\), giving \(\boxed{y=7}\).

(a) \(E=5(\sqrt{18})^2-2^2=90-4=\boxed{86}\).

(b) \(5c^2=E+d^2\), so \(c^2=\frac{E+d^2}{5}\). Therefore \(\boxed{c=\pm\sqrt{\frac{E+d^2}{5}}}\).

(a) \(c=4(-2)^2+2(3)=16+6=\boxed{22}\).

(b) Multiply by \(x-5\): \(yx-5y=2x-6\). Collect the x-terms: \(x(y-2)=5y-6\). Hence \(\boxed{x=\frac{5y-6}{y-2}}\), for \(y\ne2\).

Multiply by 12: \(3(w+8)=4(2w-3)\). Thus \(3w+24=8w-12\), \(5w=36\), and \(\boxed{w=7.2}\).

(a) \(8x=32\), so \(\boxed{x=4}\).

(b) \(10y-2=24\), hence \(10y=26\) and \(\boxed{y=13/5=2.6}\).

Multiply the equations by 2 and 5: \(8x-10y=26\) and \(15x-10y=40\). Subtract: \(7x=14\), so \(x=2\). Then \(6-2y=8\), giving \(\boxed{x=2,y=-1}\).

\(4c=22\), so \(\boxed{c=5.5}\).

\(12n+9(n-10)=277.50\), so \(21n-90=277.50\). Thus \(n=367.50/21=\boxed{\$17.50}\).

Multiply the second equation by 3: \(9w+3y=33\). Add the first: \(11w=44\), so \(w=4\). Then \(12+y=11\), giving \(\boxed{w=4,y=-1}\).

\(3t+w=51\) and \(6t+5w=129\). Double the first: \(6t+2w=102\). Subtract to get \(3w=27\), so \(w=9\). Then \(3t=51-9=42\), giving \(\boxed{t=14,w=9}\).

\(2x=11-4=7\), so \(\boxed{x=3.5}\).

\(8x=44\), so \(\boxed{x=5.5}\).

Add the equations: 5x=50, giving \(\boxed{x=10}\). Then 30+5y=5, so \(\boxed{y=-5}\).

Equate the first two: \(5x-2=3(10-x)=30-3x\), hence 8x=32 and x=4. The common value is 10−4=6. Thus y+11=6, giving \(\boxed{y=-5}\).

Multiply the first equation by 2: 3x+10y=10. Multiply this by 3 and the second equation by 10: \(9x+30y=30\), \(40x-30y=460\). Add to get 49x=490, so \(\boxed{x=10}\). Then 40−3y=46, giving \(\boxed{y=-2}\).

Substitute: \(4(x^2-18)+3x=13\), hence \(4x^2+3x-85=0=(4x-17)(x+5)\). Thus x=17/4 or −5. Corresponding y-values from \(y=x^2-18\) are 1/16 and 7. The pairs are \(\boxed{(x,y)=(17/4,1/16)\text{ or }(-5,7)}\).

Add the equations: 8t=24, so \(\boxed{t=3}\). Then 9+2w=5, giving \(\boxed{w=-2}\).

Multiply the second equation by 3: \(9x-6y=-51\). Add to the first: \(14x=-42\), so \(\boxed{x=-3}\). Then \(-15+6y=9\), giving \(\boxed{y=4}\).

Add to obtain 4p=8, so \(\boxed{p=2}\). Then 2+2q=1, giving \(\boxed{q=-\frac12}\).

Add 5h and subtract 7: \(4h=-4\), so \(\boxed{h=-1}\).

\(16t=-4\ \Rightarrow\ t=-\frac14\).

\(25-2u=6\ \Rightarrow\ 2u=19\ \Rightarrow\ u=9.5\).

Multiply the second equation by 2 and add: \(5x=25\Rightarrow x=5\). Then \(y=3-5=-2\).

\(c(2x-5)=3x\Rightarrow2cx-5c=3x\Rightarrow x(2c-3)=5c\).

\(x=\frac{5c}{2c-3}\), with \(c\ne3/2\).

\(Rm=2m-2k\Rightarrow m(2-R)=2k\). Hence \(m=\frac{2k}{2-R}\) for \(R\ne2\).

\(v=4-9.8(-7)=4+68.6=72.6\).

(a)(i) \(3^3x^{2(3)}y^{4(3)}=27x^6y^{12}\). (ii) For \(x,y\ne0\), reverse the fraction: \(\left(\frac{x^{16}y^8}{16}\right)^{3/2}=\frac{x^{24}y^{12}}{64}\).

(b)(i) \((x-3)(x+3)\). (ii) Denominator \(=2y(x-3)+5(x-3)=(x-3)(2y+5)\). Cancel to get \(\frac{x+3}{2y+5}\), with original restrictions \(x\ne3, y\ne-5/2\).

(c) Substitute \(y=7-2x\): \(7-2x=5x^2+2x-13\Rightarrow5x^2+4x-20=0\).

\(x=\frac{-4\pm\sqrt{16+400}}{10}=\frac{-4\pm\sqrt{416}}{10}\). Use \(y=7-2x\) with the unrounded roots. The pairs are \((x,y)=(1.64,3.72)\) and \((-2.44,11.88)\).

(a) \(4x+3(x+27)=194.75\Rightarrow7x=113.75\Rightarrow x=16.25\). One shirt costs \$16.25.

(b) \(y=-8-5x\). Substitute: \(x^2+4(-8-5x)=37\Rightarrow x^2-20x-69=0\).

\((x+3)(x-23)=0\). For \(x=-3\), \(y=7\); for \(x=23\), \(y=-123\). Thus \((-3,7),(23,-123)\).

(c) Cylinder area \(=2\pi x(6x)+2\pi x^2=14\pi x^2\). Equate \(4\pi r^2=14\pi x^2\) and divide by \(4\pi\) to obtain \(r^2=\frac72x^2\).

(a) \(x=8-3=5\). (b) \(y=63/7=9\), so \(10y=90\).

\(2w^2=y+x\Rightarrow w^2=\frac{y+x}2\Rightarrow w=\pm\sqrt{\frac{y+x}2}\). For real values, \(y+x\ge0\).

\(x+1=x^2+x-3\Rightarrow x^2=4\Rightarrow x=\pm2\). With \(y=x+1\), points \((2,3),(-2,-1)\).

\(v=30+(-2)(7)=16\).

\(T^2=3d-e\Rightarrow3d=T^2+e\Rightarrow d=\frac{T^2+e}3\). The original square-root formula has \(T\ge0\).

\(108=\frac{16y^2}3\Rightarrow y^2=\frac{324}{16}=20.25\). Positive \(y=4.5\).

Substitute \(y=9-4x\): \(9-4x+x^2=5\Rightarrow x^2-4x+4=0\).

\((x-2)^2=0\Rightarrow x=2\). Then \(y=9-4(2)=1\). The single intersection point is \((2,1)\).

(a) \(s=0.5(9.8)(20)^2=1960\). (b) \(4y-3=3\Rightarrow4y=6\Rightarrow y=1.5\).

(c) \(15x-24-6x+14=9x-10\). (d) \(3c^3=2b^2-A\Rightarrow c=\sqrt[3]{\frac{2b^2-A}3}\).

(e) \(2q(3p-2)-(3p-2)=(2q-1)(3p-2)\).

\(x+2(x-24)+(x-16)=100\Rightarrow4x-64=100\Rightarrow x=41\). The jacket costs \$41.

(a) \(s=0.9(16)/2=7.2\). (b) \(t^2=2s/a\Rightarrow t=\pm\sqrt{2s/a}\). If \(t\) denotes elapsed time, use the nonnegative root.

(a)(i) \(x+x+7+2x-1=30\), hence \(x=6\). (ii) There are 13 oranges and 11 bananas. Banana cost \((555-6(15)-13(18))/11=21\) cents.

(b)(i) \(3w/16=3/2\) gives \(w=8\). (ii) \(2^{-y}=8=2^3\) gives \(y=-3\).

(c)(i) Add the equations: \(3p=3/2\) gives \(p=1/2,q=1\). (ii) \(\sin u=1/2,\cos v=1\), hence \(u=30,150\)° and \(v=0,360\)°.

(a) \(7=14p\) gives \(p=1/2\). (b) \(m(c^2-g)=2k\) gives \(m=\frac{2k}{c^2-g}\).

(c) \(2x+3+4(x-3)=(x-3)(2x+3)\). Hence \(6x-9=2x^2-3x-9\), so \(x(2x-9)=0\). Thus \(x=0\) or \(x=4.5\); neither is an excluded denominator value.

(d) \(x=12-2y\). Substitute: \(5(12-2y)+y^2=39\), giving \(y^2-10y+21=(y-3)(y-7)=0\). Thus \((x,y)=(6,3),(-2,7)\). (e) \((x+6)(x-4)=x^2+2x-24\). Multiply by \(2x-3\) to obtain \(2x^3+x^2-54x+72\).

Subtract the first equation from the second: \(x=4\). Substitute \(4-3y=7\) to obtain \(y=-1\).

Add equations: \(8x=72\), so \(x=9\). \(27-2y=21\) gives \(y=3\).

\(2mh=g-gh\quad\Rightarrow\quad h(2m+g)=g\quad\Rightarrow\quad h=\frac{g}{2m+g}\), provided \(2m+g\ne0\).

Multiply the first equation by 5: \(10x+5y=15\). Add the second equation: \(11x=55\), so \(x=5\). Substitute: \(y=3-10=-7\).

\(b^2=5ca=5(2)(5.625)=56.25\). Thus \(b=\pm7.5\).

(a) Subtract to obtain 3y = 9, so y = 3 and x = 7. (b) Equate: \(x^2+2x+2=2-x\), hence \(x(x+3)=0\). Solutions (x, y) = (0, 2) and (−3, 5).

\(y(1-x)=3x-2\quad\Rightarrow\quad y+2=x(y+3)\quad\Rightarrow\quad x=\frac{y+2}{y+3}\), with \(y\ne-3\).

Double the first equation: \(8x-4y=-26\). Add the second: \(5x=-15\), hence \(x=-3\). Then \(-12-2y=-13\) gives \(y=.5\).

Either \(5x-3=0\) or \(2x+7=0\), giving \(x=3/5\) or \(x=-7/2\).

(a) \(P=100(3^2+4.5^2)=2925\). (b) \(g^2=P/M-h^2\) gives \(g=\pm\sqrt{P/M-h^2}\), for \(M\ne0\) and nonnegative radicand. If g denotes a nonnegative quantity, use the positive root.

(a)(i) \(42-12x=3x-8\), so \(15x=50\) and \(x=10/3\). (ii) \(6x=2(x-5)\), so \(x=-2.5\). (b)(i) \(2(x^2-144y^2)=2(x-12y)(x+12y)\). (ii) \((5x-8)(x+5)\). (c) Cancel \(x^3\): \(4x^2-17x+9=0\). \(x=\frac{17\pm\sqrt{17^2-4(4)(9)}}8=\frac{17\pm\sqrt{145}}8\). Thus \(x=.62\) or \(x=3.63\).

(a)(i) \(8x-12=24\) gives \(x=4.5\). (ii) \(6x>-8\) gives \(x>-4/3\). (b) \(3y^3=2x^3-V\quad\Rightarrow\quad y=\sqrt[3]{\frac{2x^3-V}3}\). (c) \((2n-5)^2-13=4n^2-20n+12=4(n^2-5n+3)\). The bracket is an integer for every integer n, proving divisibility by 4. (d)(i) \(5+12x-2x^2=23-2(x-3)^2\), so \(p=-3,q=23\). (ii) Maximum at (3, 23). (e) Multiplier \(1.3^2=1.69\), hence 69% increase.

(a) Double the first equation and add the second: \(7p=21\), so \(p=3\) and \(q=2p-7=-1\). (b) Multiply by 12: \(3x+8x=12\), giving \(x=12/11\). (c)(i) Add 2, divide by 3: \(-2<x\le3\). (ii) −1, 0, 1, 2, 3. (d) \(4a(4-a)\). (e)(i) \(\frac1{2a}\times\frac{4b}3=\frac{2b}{3a}\). (ii) \(\frac{2(x-1)-x}{x-1}=\frac{x-2}{x-1}\), \(x\ne1\).

\(1-x=15\) gives x = −14.

\(2y^2=h^2-x^2\), so \(y=\pm\sqrt{\frac{h^2-x^2}{2}}\). If y denotes a non-negative length, use the positive root.

(a) \(17\le5m\Rightarrow m\ge17/5=3.4\). (b) Cross-multiply: \(15(2x+5)=14(3-x)\), so \(44x=-33\Rightarrow x=-3/4\).

(c) Substitute y=4−x: \(x^2+2(4-x)^2=67\). Expand to \(3x^2-16x-35=0\) and factor \((3x+5)(x-7)=0\). The pairs are \((x,y)=(7,-3),\quad(-5/3,17/3)\).

Leeks cost \(2.8(2.65)=7.42\). Potatoes cost \(13.72-7.42=6.30\). Per kg: \(6.30/3.6=1.75\), so \$1.75.

\(5x=2y+7\Rightarrow x=\frac{2y+7}5\).

\(6=5x\Rightarrow x=6/5=1.2\).

Add the equations: \(5x=15\), so x=3. Substitute into2x+y=7: y=1.

Double the second equation: \(2x+8y=8\). Add to the first: \(5x=30\), giving x=6. Then \(6+4y=4\) gives y=−1/2.

\(m-2p=\sqrt{x/y}\). Square both sides: \((m-2p)^2=x/y\). Thus \(x=y(m-2p)^2\), retaining the original domain conditions.

(a) \(3x+2(2x+1)=12.36\Rightarrow7x=10.36\Rightarrow x=1.48\), so \$1.48. (b) \(18x=7(2x+1)\Rightarrow4x=7\Rightarrow x=1.75\), so \$1.75. (c) \(\frac4x=\frac{13}{2x+1}\Rightarrow8x+4=13x\Rightarrow x=.8\), so \$0.80.

(d) \(\frac{20}x+\frac{10}{2x+1}=45\). Multiplying by \(x(2x+1)\) gives \(20(2x+1)+10x=45x(2x+1)\), so \(90x^2-5x-20=0\Rightarrow18x^2-x-4=(9x+4)(2x-1)=0\). Roots −4/9 and1/2. A price must be positive, so x=\$0.50.

(a) Multiply the first equation by3 and the second by5: \(18x+15y=81,\quad25x-15y=220\). Add:43x=301, so x=7; then42+5y=27 gives y=−3.

(b) \(y=k/(x+3)^2\), and k=8(5²)=200. At x=7, \(y=200/10^2=2\). (c) \(3x-6<7x+14\Rightarrow-20<4x\Rightarrow x>-5\).

Multiply the second equation by 10: \(5x+30y=70\). Subtract the first: \(22y=66\Rightarrow y=3\). Then \(5x+24=4\Rightarrow x=-4\).

\(Pm=k+m\Rightarrow(P-1)m=k\Rightarrow m=k/(P-1)\), P≠1, with m≠0 in the original formula.

\(6f=12\Rightarrow f=2\).

\(w+h=P/2\Rightarrow w=P/2-h\).

\(P=r(2+\pi)\Rightarrow r=\frac P{2+\pi}\).

\(x-2=9\Rightarrow x=11\).

(a) \(2.45x+3.15y\). (b) \(2.45(8)+3.15y=60.55\Rightarrow y=\frac{40.95}{3.15}=13\) cards.

\(5n=\frac{391+n+n-1}3\Rightarrow15n=390+2n\Rightarrow13n=390\Rightarrow n=30\).

(a) \(21x+21(x+2)=420\Rightarrow42x=378\Rightarrow x=9\). (b) \(5r+p=245,\quad2r+3p=215\). Multiply the first by 3 and subtract the second: \(13r=520\), giving r=40 and p=45 cents.

(c)(i) \(\frac{12}x+\frac6{x-1}=5\Rightarrow12(x-1)+6x=5x(x-1)\). Expanding: \(18x-12=5x^2-5x\Rightarrow5x^2-23x+12=0\). (ii) \((5x-3)(x-4)\). (iii) \(x=3/5\text{ or }4\). (iv) Since x−1 must be positive, x=4. Final speed is 3 km/h.

(a)(i) Triangle sum \((a+2b)+2a+(3b+10)=180\Rightarrow3a+5b=170\). (ii) Quadrilateral sum \(8a+(3a+2b)+(b+50)+(4b-2a)=360\Rightarrow9a+7b=310\). (iii) Triple the first: \(9a+15b=510\). Subtract the second: \(8b=200\), so b=25. Then \(3a=170-125=45\), so a=15. (iv) Angles 65°,30°,85°; smallest=30°.

(b) \(6x=-9\Rightarrow x=-1.5\). (c) \(8x-2y=5x-3\Rightarrow2y=3x+3\Rightarrow y=\frac{3x+3}2\). (d) \(27^{2/3}x^{9(2/3)}=9x^6\). (e) \(\frac{x(x+5)}{(x-5)(x+5)}=\frac{x}{x-5}\), x≠±5.

Double the first equation: \(4x+y=26\). Double again: \(8x+2y=52\). Subtract the second: \(5x=35\), so x=7. Then y=26−28=−2.

(a) w=70/10=7. (b) x=3, so 12x=36.

\(1-p=12\Rightarrow-p=11\Rightarrow p=-11\).

(a)(i) \((m-3)(2n+m)\). (ii) \((2y-9)(2y+9)\). (iii) \((t-4)(t-2)\). (b) \(kx=2m-x\Rightarrow x(k+1)=2m\Rightarrow x=\frac{2m}{k+1}\). (c) Double the first equation: x−6y=18. Substitute y=28−5x: 31x−168=18, so x=6, y=−2. (d)(i) Multiply by m(m+4), m≠0,−4: \(3m-4(m+4)=6m(m+4)\Rightarrow6m^2+25m+16=0\). (ii) \(m=\frac{-25\pm\sqrt{625-384}}{12}=\frac{-25\pm\sqrt{241}}{12}\), giving m=−0.79 or −3.38.

Multiply the first equation by 2 and the second by 3:4 x+6 y=−24,15 x+6 y=42. Subtract:11 x=66, so x=6. Then 12+3 y=−12 gives y=−8.

\(3w=39\Rightarrow w=13\).

\(s=2(3)+16(3^2)=6+144=150\).

\(1=6x\Rightarrow x=\frac16\).

\(q^2=p/2\Rightarrow q=\pm\sqrt{p/2}\).

\(5(7)-3(-2)=35+6=41\).

\(\sqrt a=x-y\Rightarrow a=(x-y)^2\) (the original equation requires x≥y).

(a)(i) \(8x+20-15x+35=-7x+55\). (ii) \(x^2-14x+49\).

(b)(i) \(2x/3=-12\Rightarrow x=-18\). (ii) \(4x+9=6x-21\Rightarrow30=2x\Rightarrow x=15\). (iii) \(3x^2=75\Rightarrow x^2=25\Rightarrow x=\pm5\).

(a) Multiply the first equation by 3 and the second by 2: \(6x+9y=33,\quad6x-10y=-100\). Subtract: 19y=133, so y=7; then 2x+21=11 gives x=−5.

(b) Comparing \(x^2+2bx+b^2\) gives b=−6, a=36. (c) \(\frac{x(x-1)+(3x+2)(2x-5)}{(2x-5)(x-1)}=\frac{7x^2-12x-10}{(2x-5)(x-1)}\), x≠5/2,1.

\(2x-x/2=1\Rightarrow3x/2=1\Rightarrow x=2/3\); \(y=1/3\).

\(y^2=x^2+1\Rightarrow x^2=y^2-1\Rightarrow x=\pm\sqrt{y^2-1}\). Real solutions require y≥1.

(a) \(5=14n\Rightarrow n=5/14\). (b) \(p-3=15\Rightarrow p=18\).

\(12m+4xy=xp\Rightarrow12m=x(p-4y)\Rightarrow x=\frac{12m}{p-4y}\), p≠4y.

(a) \(8x=-22\Rightarrow x=-11/4=-2.75\). (b)(i) \((x+11)(x-2)\). (ii) x=−11 or 2.

(c) \(xy=2x-2a\Rightarrow x(2-y)=2a\Rightarrow x=2a/(2-y)\). (d) \(\frac{x(x-6)}{(x-6)(x+6)}=\frac{x}{x+6}\), original domain x≠±6.

\(y=(-2)(-7)-3=11\).

\(py=qx\Rightarrow x=\frac{py}{q}\).

\(rp-3p=8r-5\Rightarrow p(r-3)=8r-5\Rightarrow p=\frac{8r-5}{r-3}\).

Double the second equation: \(10x+4y=42\). Subtract the first: \(7x=28\), so \(x=4\). Substitute in \(5x+2y=21\) to obtain \(20+2y=21\), giving \(y=0.5\).

\(y+1=\frac96=\frac32\Rightarrow y=\frac12\).

Double the second equation: \(2x+4y=18\). Subtract the first: \(y=5\). Then \(x=9-2(5)=-1\).

(a) \(y=(-5)^2+3(-7)=25-21=4\). (b) \(p^2=y-qr\Rightarrow p=\pm\sqrt{y-qr}\). The principal square root gives \(p=\sqrt{y-qr}\) when p is non-negative.

\(k-8=78/6=13\), so \(k=21\).

Double the first equation to give \(x+2y=16\). Add the second: \(2x=18\), so \(x=9\). Then \(y=8-9/2=3.5\).

\(V=4(3)^2=36\).

(a) \(12x=27\Rightarrow x=2.25\). (b) \(4x\ge14\Rightarrow x\ge3.5\). (c) \((x-7)(x+3)\), because −7 + 3 = −4 and (−7)(3) = −21. (d) \(12x^2+9xy-8xy-6y^2=12x^2+xy-6y^2\).

(a) \(x+(x+4)+(x+4)=26\). (b) \(3x+8=26\Rightarrow x=6\), so \$6.

Double the first equation: \(x-16y=2\). Subtract it from the second: \(18y=4.5\Rightarrow y=1/4\). Then \(x=6.5-2(1/4)=6\).

Multiply the first by 5 and the second by 2: \(25x+10y=-10,\quad6x-10y=34.8\). Add: \(31x=24.8\Rightarrow x=0.8\). Substitute: \(4+2y=-2\Rightarrow y=-3\).

\(w+4000=60000/5=12000\Rightarrow w=8000\).

\(3x+12=8x-2\Rightarrow14=5x\Rightarrow x=2.8\).

(a) Multiply by t: \(At-tx=xr\). Collect x terms: \(At=x(t+r)\), hence \(x=At/(t+r)\). (b) Comparing coefficients gives \(2b=-16\Rightarrow b=-8\) and \(a=b^2=64\). (c) \(\frac{6(3x-2)-5(x-4)}{(x-4)(3x-2)}=\frac{13x+8}{(x-4)(3x-2)}\).

(a) \(x+(x+12)+3(x+12)=123\Rightarrow5x+48=123\Rightarrow x=15\). (b) \(t<8\) and \(t\ge37/7\), so integers 6, 7. (c)(i) \(21-x=4x+12\Rightarrow9=5x\Rightarrow x=1.8\). (ii) \(x=\frac{-7\pm\sqrt{49+60}}6=\frac{-7\pm\sqrt{109}}6\), giving \(x=-2.91\) or \(x=0.57\).

(a)(i) \(A=\pi(6.2)(10.8)+\pi(6.2)^2=105.4\pi\approx331\) cm² (3 significant figures). (ii) \(A-\pi r^2=\pi rl\Rightarrow l=(A-\pi r^2)/(\pi r)\). (b)(i) Time \(10/4+8/5=4.1\) h, distance 18 km. Average \(18/4.1=4.390\ldots\approx4.39\) km/h. (ii) \(\frac{2x+4}{x/5+(x+4)/10}=7\), so \(20x+40=21x+28\Rightarrow x=12\). (c)(i) \(19.80/1.2=16.50\) dollars. (ii) If original price is C, \(C(1+y/100)=x\), hence \(C=100x/(100+y)\) dollars.

\(x^2=(y-b)/a\Rightarrow x=\pm\sqrt{(y-b)/a}\).

\(s-ut=\tfrac12at^2\Rightarrow a=2(s-ut)/t^2\).

(a) Work in cents: \(6x+11(x-5)=1356\Rightarrow17x=1411\Rightarrow x=83\). (b) \(y(y+3)=\tfrac12(2y+1)(y+1)\). Expanding gives \(2y^2+6y=2y^2+3y+1\Rightarrow y=1/3\). (c) Equate bottle counts: \(480/(w-1)=780/(2w-11)\). Cross multiply: \(960w-5280=780w-780\Rightarrow w=25\). (d)(i) \(\tfrac12u(3u-2)=2.5\) gives the required quadratic. (ii) \((3u-5)(u+1)\). (iii) Positive lengths require \(u=5/3\); base \(3u-2=3\). Therefore \(t=\tan^{-1}[(5/3)/3]\approx29.1^\circ\).

\(n-8=22\Rightarrow n=30\).

\((x-4)^2=y-6\Rightarrow x=4\pm\sqrt{y-6}\) (real values require y ≥ 6).

(a) \(V=15(7)/3=35\). (b) Multiply by 3 and divide by A: \(h=3V/A\).

Add to give \(5x=10\Rightarrow x=2\). Then \(y=2x-7=-3\).

Let n and p be their dollar prices: \(3n+4p=10.80,\quad5n+2p=14.50\). Double the second and subtract the first: \(7n=18.20\Rightarrow n=2.60\). Then \(p=(14.50-13)/2=0.75\).

Multiply the equations by 5: \(2x-25y=135,\ 10x+y=45\). Five times the first minus the second gives \(-126y=630\Rightarrow y=-5\). Then \(10x=50\Rightarrow x=5\).

\(v^3=p+r\Rightarrow r=v^3-p\).

\(2x+5=24\Rightarrow2x=19\Rightarrow x=9.5\).

\(y-2=\sqrt{x-8}\Rightarrow(y-2)^2=x-8\Rightarrow x=8+(y-2)^2\) (with y ≥ 2).

(a)(i) \((3x-5)+(2x-3)+(2x+7)+(15-2x)=5x+14\) cm. (ii) \(5x+14=32\Rightarrow x=3.6\). Side lengths 5.8, 4.2, 14.2, 7.8 cm; longest 14.2 cm. (b) \(14+(7a-6b)+(3b+a)=32.5\Rightarrow8a-3b=18.5\). \((5a-2b)+(6b-a)+13.5+a=39.75\Rightarrow5a+4b=26.25\). Multiply the first equation by 4 and second by 3: \(32a-12b=74,\quad15a+12b=78.75\). Add: \(47a=152.75\Rightarrow a=3.25\). Substitute: \(26-3b=18.5\Rightarrow b=2.5\).

\(2y-17=12\Rightarrow2y=29\Rightarrow y=14.5\).

\(y=2x-19\). Substitute: \(3x+5(2x-19)=48\Rightarrow13x=143\Rightarrow x=11\). Then \(y=3\) and \(2x+y=25\).

(a) \(y=\sqrt{10}=3.16227766\ldots\approx3.1623\). (b) \(y^2=8+4/x\Rightarrow x(y^2-8)=4\Rightarrow x=4/(y^2-8)\).

(a) \(\frac{x+2}{2x+3}=\frac9{16}\Rightarrow16x+32=18x+27\Rightarrow x=2.5\). Thus AB = 4.5 cm and AC = 8 cm. \(BC=\sqrt{8^2-4.5^2}=6.6144\ldots\approx6.61\) cm. (b)(i) Let masses be w and r: \(r=w+2.5,\quad7w+5r=114.5\). Therefore \(12w+12.5=114.5\Rightarrow w=8.5,\ r=11\) g. (ii)(a) \(\tfrac7{12}\tfrac6{11}=7/22\). (b) \(\tfrac7{12}\tfrac5{11}+\tfrac5{12}\tfrac7{11}=35/66\).

\(c^2=a^2+b^2\Rightarrow b^2=c^2-a^2\Rightarrow b=\pm\sqrt{c^2-a^2}\). If b is a length, take the nonnegative root.

\(24=5x\Rightarrow x=4.8\).

Double the first equation: \(4x-14y=4\). Subtract it from the second to obtain \(19y=38\Rightarrow y=2\). Then \(x=8\). Intersection (8, 2).

\(x^2=y-4\Rightarrow x=\pm\sqrt{y-4}\).

\(2x=-16\Rightarrow x=-8\).

Multiply the second equation by 3 and subtract the first: \(16y=144\Rightarrow y=9\). Then \(x=56-7(9)=-7\).

\(cw+3c=4+w\Rightarrow(c-1)w=4-3c\Rightarrow w=(4-3c)/(c-1)\), with \(c\ne1\).

\(8+4.5d=50-10.50\Rightarrow4.5d=31.5\Rightarrow d=7\) days.

\(3w/a=2-t\Rightarrow w=a(2-t)/3\).

(a) \(10x+4y=10.70,\quad8x+6y=10.10\). Multiply by 3 and 2 respectively: \(30x+12y=32.10,\ 16x+12y=20.20\Rightarrow14x=11.90\Rightarrow x=0.85\). Substitute to obtain y = 0.55. (b) \(x=(5\pm\sqrt{25+64})/4=(5\pm\sqrt{89})/4\approx3.61,\ -1.11\).

(a) \(12x+7(x+1)=31.70\Rightarrow19x=24.70\Rightarrow x=1.30\) dollars. (b)(i) \(36/y-36/(y+1)=25\Rightarrow36(y+1)-36y=25y(y+1)\Rightarrow25y^2+25y-36=0\). (ii) \((5y+9)(5y-4)\). (iii) y = −1.8, 0.8. (iv) Reject the negative price; total \(0.8+1.8=2.60\) dollars.

\(yx-4y=x+2\Rightarrow x(y-1)=4y+2\Rightarrow x=\frac{4y+2}{y-1}\), with \(y\ne1\).

\(4x-12=22-6x\Rightarrow10x=34\Rightarrow x=3.4\).

\(\pi y^2=\pi x^2-A\Rightarrow y=\sqrt{\frac{\pi x^2-A}{\pi}}\) (positive root when y represents a length; algebraically both signs are possible).

Multiply the first equation by 3 and add: \(9x+3y=90;\ 11x=143\Rightarrow x=13\). Then \(y=30-39=-9\).

\(c(a-b)=4\Rightarrow a-b=4/c\Rightarrow a=b+4/c\), \(c\ne0\).

\(x=5y;\ 75y+10y=17\Rightarrow y=1/5=0.2;\ x=1\).

\(7t+11(t+5)=2215\Rightarrow18t=2160\Rightarrow t=120\) cents.

\(y-5=x/3\Rightarrow x=3(y-5)\).

Double the first equation and subtract the second: \(7y=6-13=-7\Rightarrow y=-1\). Then \(x=3-2(-1)=5\).

Subtract to get \(2y=10\Rightarrow y=5\). Substitute: \(x=12-15=-3\).

Multiply by 2 to give \(2x+y=14,\ 2x-y=34\). Add: \(4x=48\Rightarrow x=12\). Substitute: \(y=14-24=-10\).

\(6y-24+y=18\Rightarrow7y=42\Rightarrow y=6\).

(a) \(y=(p+q)^2-5\). (b) \(6x+9(x-3)=51\Rightarrow15x=78\Rightarrow x=5.20\), so \$5.20. (c) \(a+c=52,\ 3a+2c=139\). Subtract twice the first equation from the second: \(a=139-104=35\), then \(c=17\).

Substitute \(x=2y+10\): \(2(2y+10)+y=5\Rightarrow5y=-15\). Thus y = −3, x = 4.

Twice the first plus the second: \(11x=-11\Rightarrow x=-1\). Then \(-1+2y=9\Rightarrow y=5\).