Rational Expressions and Rational Equations: Problems with Solutions

Solution:
Every denominator must be different from zero:
[tex]x^2-4=(x-2)(x+2)\ne 0\ \Rightarrow\ x\ne 2,\ x\ne -2[/tex]
[tex]x^2+3x+2=(x+1)(x+2)\ne 0\ \Rightarrow\ x\ne -1,\ x\ne -2[/tex]
[tex]x+4\ne 0\ \Rightarrow\ x\ne -4[/tex]
Moreover, we divide by the fraction [tex]\frac{x-4}{x+4}[/tex], so this divisor must also be non-zero (after turning the division into multiplication, [tex]x-4[/tex] becomes a denominator):
[tex]x-4\ne 0\ \Rightarrow\ x\ne 4[/tex]
Hence the admissible values are all real [tex]x[/tex] except [tex]-4[/tex], [tex]-2[/tex], [tex]-1[/tex], [tex]2[/tex] and [tex]4[/tex]:
[tex]x\in\mathbb{R}\setminus\{-4, \,-2, \,-1, \,2, \,4\}[/tex]

Solution:
Factor the numerators and the denominators:
[tex]a^2-b^2=(a-b)(a+b),\qquad a^3-b^3=(a-b)(a^2+ab+b^2)[/tex]
Since [tex]a\ne b[/tex], we can cancel [tex]a-b[/tex] in both fractions:
[tex]\frac{(a-b)(a+b)}{a-b}=a+b,\qquad \frac{(a-b)(a^2+ab+b^2)}{(a-b)(a+b)}=\frac{a^2+ab+b^2}{a+b}[/tex]
Bring to the common denominator [tex]a+b[/tex], which is non-zero because [tex]a\ne -b[/tex]:
[tex]a+b-\frac{a^2+ab+b^2}{a+b}=\frac{(a+b)^2-(a^2+ab+b^2)}{a+b}=\frac{a^2+2ab+b^2-a^2-ab-b^2}{a+b}=\frac{ab}{a+b}[/tex]

Solution:
Domain: [tex]x\ne\pm 2[/tex]. Change the signs in the right-hand fraction:
[tex]\frac{24-x^2}{4-x^2}=\frac{x^2-24}{x^2-4}=\frac{x^2-24}{(x-2)(x+2)}[/tex]
Multiply both sides by [tex](x-2)(x+2)\ne 0[/tex]:
[tex]5(x-2)-3(x+2)=x^2-24\ \Rightarrow\ 2x-16=x^2-24\ \Rightarrow\ x^2-2x-8=0\ \Rightarrow\ (x-4)(x+2)=0[/tex]
The value [tex]x=-2[/tex] makes the denominators zero, so it is an extraneous root and is rejected. The value [tex]x=4[/tex] is admissible. Check:
[tex]\frac{5}{6}-\frac{3}{2}=-\frac{2}{3},\qquad \frac{24-16}{4-16}=\frac{8}{-12}=-\frac{2}{3}[/tex]
The only root is [tex]x=4[/tex].

Solution:
Factor the denominator: [tex]x^3+1=(x+1)(x^2-x+1)[/tex]. The second factor is positive for every [tex]x[/tex], because
[tex]x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\gt 0[/tex]
so the only restriction is [tex]x\ne -1[/tex]. Multiplying both sides by the common denominator [tex](x+1)(x^2-x+1)[/tex], we get
[tex]2x-1+(x^2-x+1)=2(x+1)\ \Rightarrow\ x^2+x=2x+2\ \Rightarrow\ x^2-x-2=0[/tex]
Its roots are [tex]x=2[/tex] and [tex]x=-1[/tex]. The value [tex]x=-1[/tex] makes the denominators [tex]x+1[/tex] and [tex]x^3+1[/tex] zero, so it is an extraneous root and is rejected. Check for [tex]x=2[/tex]:
[tex]\frac{3}{9}+\frac{1}{3}=\frac{2}{3},\qquad \frac{2}{4-2+1}=\frac{2}{3}[/tex]
The equation has the single root [tex]x=2[/tex].

Solution:
Domain: [tex]y^2-9\ne 0[/tex], i.e. [tex]y\ne\pm 3[/tex]. Multiply both sides by [tex](y-3)(y+3)\ne 0[/tex]:
[tex](y+1)(y+3)-(y-1)(y-3)=8y[/tex]
[tex](y^2+4y+3)-(y^2-4y+3)=8y\ \Rightarrow\ 8y=8y[/tex]
This holds for every [tex]y[/tex], so the equation is an identity on its domain: every admissible value of [tex]y[/tex] is a solution, while [tex]y=\pm 3[/tex] are not solutions because the fractions are not defined there. Hence
[tex]y\in\mathbb{R}\setminus\{-3, \,3\}[/tex]

Solution:
Let [tex]n[/tex] be the number of days the student actually needed; then he read [tex]\frac{240}{n}[/tex] pages a day. With [tex]4[/tex] more pages a day he would need [tex]n-2[/tex] days (so [tex]n\gt 2[/tex]), i.e. he would read [tex]\frac{240}{n-2}[/tex] pages a day. Hence
[tex]\frac{240}{n-2}=\frac{240}{n}+4[/tex]
Multiply by [tex]n(n-2)\ne 0[/tex]:
[tex]240n=240(n-2)+4n(n-2)\ \Rightarrow\ 4n^2-8n-480=0\ \Rightarrow\ n^2-2n-120=0[/tex]
The roots are [tex]n=12[/tex] and [tex]n=-10[/tex]; the number of days cannot be negative, so [tex]n=12[/tex]. Check: [tex]\frac{240}{12}=20[/tex] pages a day, and with [tex]24[/tex] pages a day the book takes [tex]\frac{240}{24}=10=12-2[/tex] days. The student read the book in [tex]12[/tex] days.

Solution:
Domain: the denominators must be non-zero, so [tex]x\ne 0[/tex], [tex]y\ne x[/tex], [tex]y\ne -x[/tex] and [tex]y\ne 2x[/tex]. First bracket:
[tex]1+\frac{x^2-xy+y^2}{x^2-y^2}=\frac{x^2-y^2+x^2-xy+y^2}{x^2-y^2}=\frac{2x^2-xy}{x^2-y^2}=\frac{x(2x-y)}{x^2-y^2}[/tex]
Second bracket:
[tex]\frac{3x}{2x-y}-\frac{2x+y}{x}=\frac{3x^2-(2x+y)(2x-y)}{x(2x-y)}=\frac{3x^2-4x^2+y^2}{x(2x-y)}=-\frac{x^2-y^2}{x(2x-y)}[/tex]
Multiplying, we get
[tex]A=\frac{x(2x-y)}{x^2-y^2}\times \left(-\frac{x^2-y^2}{x(2x-y)}\right)=-1[/tex]
So for all admissible values the expression equals [tex]-1[/tex].

Solution:
From [tex]\frac{b}{a}=-\frac{1}{2}[/tex] we get [tex]a=-2b[/tex] with [tex]b\ne 0[/tex]; then [tex]a-b=-3b\ne 0[/tex] and [tex]a+b=-b\ne 0[/tex], so the expression is defined. Factor:
[tex]a^4-b^4=(a-b)(a+b)(a^2+b^2),\qquad ab^4-b^5=b^4(a-b)[/tex]
[tex]a^3-b^3=(a-b)(a^2+ab+b^2),\qquad a^2-b^2=(a-b)(a+b)[/tex]
Hence the expression equals
[tex]\frac{(a+b)(a^2+b^2)}{b^4}\times \frac{a^2+ab+b^2}{a+b}=\frac{(a^2+b^2)(a^2+ab+b^2)}{b^4}=\left(\frac{a^2}{b^2}+1\right)\left(\frac{a^2}{b^2}+\frac{a}{b}+1\right)[/tex]
Here [tex]\frac{a}{b}=-2[/tex] (not [tex]-\frac{1}{2}[/tex]), so the value is
[tex](4+1)(4-2+1)=5\times 3=15[/tex]