Factor the numerator: \(2x^2+3x-20=(2x-5)(x+4)\). Group the denominator: \(a(x+4)-3b(x+4)=(a-3b)(x+4)\). Cancel \(x+4\) to obtain \(\frac{2x-5}{a-3b}\), with the original restrictions \(x\ne-4\), \(a\ne3b\).

Use denominator \((x+3)(x-2)\): \(\frac{(x-2)^2+(x+3)^2}{(x+3)(x-2)}=\frac{x^2-4x+4+x^2+6x+9}{x^2+x-6}=\boxed{\frac{2x^2+2x+13}{x^2+x-6}}\), for \(x\ne-3,2\). No common factor cancels.

Factorise: \(\frac{6(x-5)(x+5)(x+2)}{3(x-5)(x+5)}=\boxed{2(x+2)}\), with the original restrictions \(x\ne-5,5\).

\(\frac8{x+2}+\frac3{x-4}=\frac{8(x-4)+3(x+2)}{(x+2)(x-4)}=\boxed{\frac{11x-26}{(x+2)(x-4)}}\), with \(x\ne-2,4\).

Factorise: \(\frac{8x(x+3)}{2(x^2-9)}=\frac{8x(x+3)}{2(x-3)(x+3)}=\boxed{\frac{4x}{x-3}}\), with the original restrictions \(x\ne-3,3\).

(a) With denominator 24, \(\frac{6x+9x-2(x+2)}{24}=\boxed{\frac{13x-4}{24}}\).

(b) \(3x(a+4y)-y(a+4y)=\boxed{(3x-y)(a+4y)}\).

Factorise: \(\frac{2x(x+5)}{(x-5)(x+5)}=\boxed{\frac{2x}{x-5}}\), with \(x\ne\pm5\) as in the original expression.

(a) \(\frac{15ab}{6a}=\boxed{5b/2}\), for \(a\ne0\).

(b) \(\frac{2p+3t}{4}\) is the required fraction.

(c) \(\frac{2(x+1)-3(x-2)}{(x-2)(x+1)}=\boxed{\frac{8-x}{(x-2)(x+1)}}\).

\(\frac{x(7-x)}{(7-x)(7+x)}=\boxed{\frac x{7+x}}\), with \(x\ne\pm7\).

\(\frac{h(h+4)}{(h-4)(h+4)}=\boxed{\frac h{h-4}}\), with \(h\ne\pm4\).

Factor the numerator as \(5a(2x-5)+3b(2x-5)=(5a+3b)(2x-5)\). The denominator is \((2x-5)(2x+5)\). Cancel the common factor to give \(\boxed{(5a+3b)/(2x+5)}\), for \(x\ne\pm5/2\).

(a) Multiply by the reciprocal: \(\frac pt\frac t2=\boxed{p/2}\), for \(t\ne0\).

(b) \(\frac{3x-2(x-1)}4=\boxed{(x+2)/4}\).

(a) Cancel m and reduce 15/25: \(\boxed{3p/(5y)}\), for m,y nonzero.

(b) \(\frac{3(x-3)+4(2x-5)}{(2x-5)(x-3)}=\boxed{(11x-29)/[(2x-5)(x-3)]}\), for \(x\ne5/2,3\).

Numerator \(=(2x+3)(x-7)\); denominator \(=(x-7)(x+7)\). Cancel the common factor to get \(\boxed{(2x+3)/(x+7)}\), for \(x\ne\pm7\).

\(\frac7{2m}+\frac3{8m}=\frac{28+3}{8m}=\boxed{\frac{31}{8m}}\), for \(m\ne0\).

(a)(i) \(\boxed{(x-4)(x+3)}\). (ii) \(\frac{(x-4)(x+4)}{(x-4)(x+3)}=\boxed{\frac{x+4}{x+3}}\), with x≠4,−3.

(b) \((4x^2-12x+9)-(x^2+2x+1)=\boxed{3x^2-14x+8}\). (c) Common denominator \((x+1)(x-3)\). Numerator \((2x+4)(x-3)-x(x+1)=x^2-3x-12\). Hence \(\boxed{\frac{x^2-3x-12}{(x+1)(x-3)}}\), for x≠−1,3.

(d) \((x^2-8x+15)(2x+1)=\boxed{2x^3-15x^2+22x+15}\).

(e) From the first equation, \(y=(x-13)/3\). Substitute into the second: \(2x^2-3(x-13)=116\), hence \(2x^2-3x-77=0\). Factor: \((2x+11)(x-7)=0\). Substituting into y gives \(\boxed{(x,y)=(7,-2)\text{ or }(-11/2,-37/6)}\).

Common denominator y(y+1): \(\frac{2y-3(y+1)}{y(y+1)}=\boxed{\frac{-y-3}{y(y+1)}}\).

Numerator=\(p(t-1)-(t-1)=(p-1)(t-1)\). Denominator=\((1-t)(1+t)=-(t-1)(t+1)\). Cancel t−1 to obtain \(\boxed{\frac{1-p}{1+t}}\), with original restrictions t≠±1.

(a)(i) C=¼×5×64=\(\boxed{80}\). (ii) y²=4×15/2.4=25, so \(\boxed{y=5}\). (b) \(\frac{4(2x+5)-3(x-1)}{(x-1)(2x+5)}=\boxed{\frac{5x+23}{(x-1)(2x+5)}}\). (c) \((2x+3)(x^2-8x+16)=\boxed{2x^3-13x^2+8x+48}\). (d) \(16^{3/4}x^{12}y^{-6}=\boxed{\frac{8x^{12}}{y^6}}\).

Numerator=\((3x-2)-y(3x-2)=(3x-2)(1-y)\). Denominator=\((1-y)(1+y)\). Cancel 1−y: \(\boxed{\frac{3x-2}{1+y}}\), with original restrictions y≠±1.

\(\frac{5x^2-19x+12}{x^2-9}=\frac{(5x-4)(x-3)}{(x-3)(x+3)}=\frac{5x-4}{x+3}\), with the original restrictions \(x\ne\pm3\).

\(\frac{5(2x-1)+4(3x+2)}{(3x+2)(2x-1)}=\frac{22x+3}{(3x+2)(2x-1)}\), \(x\ne-2/3,1/2\).

\(\frac{(2x-3)(x+4)}{(2x-3)(2x+3)}=\frac{x+4}{2x+3}\), with original restrictions \(x\ne\pm3/2\).

(a)(i) \(3(9y^2-1)=3(3y-1)(3y+1)\). (ii) \((2-p)m+(2-p)k=(2-p)(m+k)\).

(b) With \(x\ne\pm1\), multiply by \((x+1)(x-1)\): \((x-1)^2-6(x+1)=x^2-1\).

\(x^2-8x-5=x^2-1\Rightarrow-8x=4\Rightarrow x=-1/2\). This is allowed by the original restrictions.

(c) \(x=\frac{3\pm\sqrt{9+32}}8=\frac{3\pm\sqrt{41}}8\), so \(x=-0.43\text{ or }1.18\).

(d) \(k=4m+kpm\Rightarrow k(1-pm)=4m\). Hence \(k=\frac{4m}{1-pm}\), provided \(1-pm\ne0\).

(a) \(\frac{10x(x-6)}{(x-6)(x+5)}=\frac{10x}{x+5}\), with original \(x\ne6,-5\).

(b) \(\frac{7(8x-1)+5(x+3)}{(x+3)(8x-1)}=\frac{61x+8}{(x+3)(8x-1)}\), for \(x\ne-3,1/8\).

Numerator \(a(x-2)-(x-2)=(a-1)(x-2)\). Denominator \((a-1)(a+1)\). Cancel: \(\frac{x-2}{a+1}\), with original \(a\ne\pm1\).

\(\frac{x(5-x)}{(5-x)(5+x)}=\frac{x}{5+x},\quad x\ne\pm5\).

For \(x\ne-1,2.5\), multiply by \((x+1)(2x-5)\): \(4(2x-5)+2(x+1)=3(x+1)(2x-5)\). Expand: \(10x-18=6x^2-9x-15\). Thus \(6x^2-19x+3=(6x-1)(x-3)=0\). Solutions \(x=1/6,3\) are both allowed.

\(\frac{(x-4)(2x+3)}{3x(x-4)}=\frac{2x+3}{3x}\), with original exclusions \(x\ne0,4\).

Exclude \(x=-1,-9\). Multiply by \((x+1)(x+9)\): \(x+9+9(x+1)=(x+1)(x+9)\). This reduces to \(x^2=9\), so \(x=3,-3\). Both satisfy the original equation.

\(\frac{14-(x+2)(x+3)}{7(x+3)}=\frac{8-5x-x^2}{7(x+3)}\), \(x\ne-3\).

\(\frac{3y(x+12)-5(x+12)}{2(x-12)(x+12)}=\frac{3y-5}{2(x-12)}\), with \(x\ne12,-12\).

\(\frac{3x(x-6)}{a(x-6)+2c(x-6)}=\frac{3x}{a+2c}\), with \(x\ne6,\ a+2c\ne0\).

(a) \(\frac{(x+3)(x+2)-(x-2)(x-3)}{(x-3)(x+2)}=\frac{10x}{(x-3)(x+2)}\). (b) \(12-k/2=5\), so k = 14. (c) \((y^2-y-12)(2y-1)=2y^3-3y^2-23y+12\). (d) \(xy=3+x\), \(x(y-1)=3\), so \(x=\frac3{y-1}\).

\(\frac{4p^2q}{2qt}=\frac{2p^2}{t}\).

Numerator \((2x-5)(x+3)\); denominator \(a(x+3)-2b(x+3)=(a-2b)(x+3)\). Cancel x + 3 to obtain \(\frac{2x-5}{a-2b}\), retaining the original restrictions x ≠ −3 and a ≠ 2b.

\(\frac{x(x-5)}{2(x-5)(x+5)}=\frac{x}{2(x+5)}\), retaining the original restrictions x≠±5.

\(\frac{2(x+1)-(2x-1)}{x+1}=\frac3{x+1}\), with x≠−1.

\(\frac{u(x-2)-(x-2)}{(u-1)(u+1)}=\frac{(u-1)(x-2)}{(u-1)(u+1)}=\frac{x-2}{u+1}\), retaining u≠±1.

\(\frac{(x-5)(x+5)}{(x-5)(x-12)}=\frac{x+5}{x-12}\), retaining x≠5,12.

\(\frac{b(a-b)}{(a-b)(a+b)}=\frac b{a+b}\), retaining a≠±b.

\(\frac{3x-1-2(x+2)}{(x+2)(3x-1)}=\frac{x-5}{(x+2)(3x-1)}\), for x≠−2, 1/3.

\(\frac{2x(x-5)+(x+3)^2}{(x+3)(x-5)}=\frac{3x^2-4x+9}{(x+3)(x-5)}\), x≠−3,5.

\(\frac{x^2(x+5)}{(x-5)(x+5)}=\frac{x^2}{x-5}\), retaining x≠±5.

\(\frac{x(x+1)-2(2x+4)}{2(x+1)}=\frac{x^2-3x-8}{2(x+1)}\), x≠−1.

Multiply by x(x+1), with x≠0,−1: \(x+1-2x=3x(x+1)\Rightarrow1-x=3x^2+3x\Rightarrow3x^2+4x-1=0\).

\(x=\frac{-4\pm\sqrt{4^2-4(3)(-1)}}6=\frac{-4\pm\sqrt{28}}6\), giving \(x=-1.55\text{ or } .22\). Neither root is an excluded value.

(a) \(t=\pm\sqrt{k-s}\). (b)(i) \((x-5)(x+5)\). (ii) \(\frac{(x-5)(x+5)}{(x-7)(x+5)}=\frac{x-5}{x-7}\), x≠−5,7. (c) \(\frac{(x-8)(x+1)+3x^2}{x(x+1)}=\frac{4x^2-7x-8}{x(x+1)}\), x≠0,−1. (d) \(18<8n\Rightarrow n>2.25,\quad5n\le30\Rightarrow n\le6\). Thus n=3,4,5,6.

\(\frac{x+3}{(3-x)(3+x)}=\frac1{3-x}\), for x≠±3.

\(\frac{y-(y-1)}{y(y-1)}=\frac1{y(y-1)}\), y≠0,1.

\(\frac{(x-5)(x+2)+18}{3(x+2)}=\frac{x^2-3x+8}{3(x+2)}\), x≠−2.

\(\frac{x+1-x}{x(x+1)}=\frac1{x(x+1)}\), x≠0,−1.

\(\frac{(2x+1)(x-1)}{x(2x+1)}=\frac{x-1}x\), x≠0,−1/2.

(a) \(\frac{x(x-3)}{(x-3)(x+3)}=\frac{x}{x+3}\), with x≠±3. (b) \(\frac{3(2x+5)+2(x-4)}{(x-4)(2x+5)}=\frac{8x+7}{(x-4)(2x+5)}\), with x≠4,−5/2.

\(\frac{(2x-1)(x+1)-6}{3(x+1)}=\frac{2x^2+x-7}{3(x+1)}\), x≠−1.

\(\frac{10(x+7)+6(x-3)+(x-3)(x+7)}{2(x-3)(x+7)}=\frac{x^2+20x+31}{2(x-3)(x+7)}\), \(x\ne3,-7\).

\(\frac{y(x+1)-x(y-1)}{xy}=\frac{y+x}{xy}\), x,y≠0.

(a) \(y=k\sqrt{x+2}\). \(9=k\sqrt9\Rightarrow k=3\), so \(y=3\sqrt{25}=15\). (b) \(\frac{(x+6)^2}{(x+6)(x-2)}=\frac{x+6}{x-2}\), with the original restrictions \(x\ne-6,2\). (c) Square both sides: \(W^2=\frac{X-a}{a}\). Then \(aW^2=X-a\Rightarrow a(W^2+1)=X\Rightarrow a=\frac{X}{W^2+1}\). (d) \(\frac{(x-2)(x-1)-(x+3)(x+1)}{(x+1)(x-1)}=\frac{(x^2-3x+2)-(x^2+4x+3)}{x^2-1}=\frac{-7x-1}{x^2-1}\).

\(\frac{x^3y+2xy^3}{x^2y^2}=\frac{xy(x^2+2y^2)}{x^2y^2}=\frac{x^2+2y^2}{xy}\), with \(x,y\ne0\).

\(1-\frac2p-\frac3t=\frac{pt}{pt}-\frac{2t}{pt}-\frac{3p}{pt}=\frac{pt-2t-3p}{pt}\).

Factor the numerator and denominator: \(\frac{7n(6p-1)}{2t(6p-1)+3m(6p-1)}=\frac{7n(6p-1)}{(6p-1)(2t+3m)}=\frac{7n}{2t+3m}\), wherever the original denominator is nonzero.

Exclude \(x=-3,\ 1/2\). Multiply by \(12(x+3)(2x-1)\): \(24(2x-1)+(x+3)(2x-1)=36(x+3)\). Expand and rearrange: \(48x-24+2x^2+5x-3=36x+108\), hence \(2x^2+17x-135=0\). Factorise: \((x-5)(2x+27)=0\). The solutions are \(x=5\) and \(x=-27/2\), and neither is excluded.

\(\frac{3(2x-5)-4(x+2)}{(x+2)(2x-5)}=\frac{2x-23}{(x+2)(2x-5)}\).

\(\frac{(x-4)(x+4)}{(x-4)(x+1)}=\frac{x+4}{x+1}\), with original exclusions \(x\ne4,-1\).

(a) \(3x/2\). (b) \((x^2+2)/x\).

\(\frac{2(2+5w)}{2(4-25w^2)}=\frac{2+5w}{(2-5w)(2+5w)}=\frac1{2-5w}\), with original exclusions \(w\ne\pm2/5\).

(a) \((x-5)(x+2)\). (b)(i) Multiply by \(x(x+1)\): \(x(x+2)+3(x+1)=3x(x+1)\). Expanding and rearranging gives \(2x^2-2x-3=0\). (ii) \(x=(2\pm\sqrt{4+24})/4=(1\pm\sqrt7)/2\), so \(x=-0.823\) or \(x=1.823\). (c) \(\frac{(2x+3)(x+1)-x(x+2)}{(x+2)(x+1)}=\frac{x^2+3x+3}{(x+2)(x+1)}\).

\(\frac{2(x+1)-2x}{x(x+1)}=\frac2{x(x+1)}\).

For x ≠ 0, −1, multiply by \(2x(x+1)\): \(3(x+1)+2x=0\Rightarrow5x=-3\Rightarrow x=-3/5\).

\(\frac{(x+7)(x-1)}{3(x+7)}=\frac{x-1}3\), for x ≠ −7.

(a)(i) \(600/x\) km/h. (ii) \(600/(x+1)\) km/h. (b)(i) \(600/x-600/(x+1)=20\). Multiply by \(x(x+1)\): \(600=20x^2+20x\), giving the required equation. (ii) \((x+6)(x-5)=0\). Time must be positive, so x = 5. Speed \(600/6=100\) km/h.

(a)(i) Multiply by \(2(x+4)\): \(14+(2x-3)(x+4)=2(x+4)\). Expand and collect to obtain \(2x^2+3x-6=0\). (ii) \(x=\frac{-3\pm\sqrt{3^2-4(2)(-6)}}4=\frac{-3\pm\sqrt{57}}4\), giving 1.14 and −2.64. (b) Total area \(\pi x^2+\pi x(3x)=4\pi x^2=\pi r^2\). Hence \(r^2=4x^2\) and radii are positive, so r = 2x.

\(3x+15=7x\Rightarrow4x=15\Rightarrow x=15/4=3.75\).

(a) \(\frac{3(x+2)-(2x-1)}{(2x-1)(x+2)}=\frac{x+7}{(2x-1)(x+2)}\). (b) \(\frac{4x(x-4)}{2(x+7)(x-4)}=\frac{2x}{x+7}\), retaining the original exclusions x ≠ 4, −7.

Common denominator 6x gives \(\frac{9+4x^2+18x+12x^2}{6x}=\frac{16x^2+18x+9}{6x}\), x ≠ 0.

(a) \(x^2+x-30=(x+6)(x-5)\). (b) Cancel the common factor to obtain \((x+4)/(x+6)\), for \(x\ne5,-6\).

\(\frac{2(x+2)+3(x+3)}{(x+3)(x+2)}=\frac{5x+13}{(x+3)(x+2)}\).

\(\frac{(x+3)(x+1)-(x-1)(x-3)}{(x-3)(x+1)}=\frac{8x}{(x-3)(x+1)}\).

(a)(i) \((25-8x)/20\). (ii) \(\frac{12+(2x-1)(x+3)}{3(x+3)}=\frac{2x^2+5x+9}{3(x+3)}\). (b) Double the first equation and add the second: \(21x=14\Rightarrow x=2/3,\ y=-3\). (c) \(\frac{7(x+3)}{(2x+3)(x+3)}=\frac7{2x+3}\), retaining the original exclusions \(x\ne-3,-3/2\).

\(\frac{3(t-1)-(t+2)}{t-1}=\frac{2t-5}{t-1}\).

(a) \([5(2x-1)-2(3x+1)]/10=(4x-7)/10\). (b) \(4x^2-12x+9-3x^2+12x=x^2+9\). (c)(i) \((2x-1)(x+3)\). (ii) \(\frac{(2x-1)(x+3)}{2(x-3)(x+3)}=\frac{2x-1}{2(x-3)}\), with \(x\ne\pm3\).

(a) \(\frac{x(x-3)}{(x-3)(x+3)}=\frac{x}{x+3}\), where \(x\ne\pm3\). (b) \(x\ne0,-1\). Multiply by \(x(x+1)\): \(15(x+1)-20x=2x(x+1)\Rightarrow2x^2+7x-15=0\Rightarrow(2x-3)(x+5)=0\). Thus \(x=3/2\ \text{or}\ x=-5\). Both satisfy the original equation.

\(\frac{(x-5)(x+4)}{x(x-5)^2}=\frac{x+4}{x(x-5)}\), with \(x\ne0,5\).

\(\frac{4(x+2)-3(2x-1)+12}{12}=\frac{23-2x}{12}\).

\(\frac{(h-5)(h+4)}{(h-5)(h+5)}=\frac{h+4}{h+5}\), with \(h\ne\pm5\).

\(\frac{(x+1)^2-x(x+5)}{(x+1)(x+5)}=\frac{1-3x}{(x+1)(x+5)}\), \(x\ne-1,-5\).

\(\frac{2(x+2)+3(x-2)}{(x-2)(x+2)}=\frac{5x-2}{(x-2)(x+2)}\), \(x\ne\pm2\).

\(\frac{3(x+4)-(x+10)}{(x+10)(x+4)}=\frac{2x+2}{(x+10)(x+4)}\), with \(x\ne-10,-4\).

\(\frac{(1-x)(1-2x)-x(2+x)}{x(1-2x)}=\frac{1-3x+2x^2-2x-x^2}{x(1-2x)}=\frac{x^2-5x+1}{x(1-2x)}\), for \(x\ne0,1/2\).

\(15a/40+32/40=(15a+32)/40\).

\((2x+3x-3)/6=(5x-3)/6\).

\(\frac{(x-2)(x-3)}{(x-2)(x+2)}=\frac{x-3}{x+2}\), retaining the original restrictions x ≠ 2,−2.

\(\frac4{2x}+\frac1{2x}+\frac{x}{2x}=\frac{x+5}{2x}\), with x ≠ 0.

\(\frac{3(x-1)-2(x+2)}{(x+2)(x-1)}=\frac{x-7}{(x+2)(x-1)}\), x ≠ −2,1.

(a) \((x-2y)/(xy)\), x,y ≠ 0. (b) \(x(x+1)/[3(x+1)]=x/3\), preserving x ≠ −1.