In exercises 1 through 14, solve the equations.
-
\[
5(x – 3) = 3(x + 7) + x
\]
-
\[
y – (6 – 2y) = 8(y – 2)
\]
-
\[
\frac{1}{2} (2x-1) = 3\left(x+\frac{1}{4} \right)
\]
-
\[
\frac{x}{4} – \frac{x}{3} = \frac{7}{6} – \frac{4x}{3}
\]
-
\[
\frac{2x-1}{5} = \frac{2+x}{3}
\]
-
\[
\frac{7z+1}{6} + \frac{3}{2} = \frac{3z}{4}
\]
-
\[
\frac{x-1}{3} – \frac{2-3x}{14} = \frac{4x-3}{7}
\]
-
\[
\frac{x-3}{6} – \frac{2x-1}{5} =-1
\]
-
\[
\frac{x+1}{5} + \frac{x+2}{6} = \frac{x-1}{4} + \frac{x+7}{10}
\]
\[
\begin{aligned}
&\frac{x+1}{5} + \frac{x+2}{6}
\\[1em]
&\hspace{3em}= \frac{x-1}{4} + \frac{x+7}{10}
\end{aligned}
\]
-
\[
\frac{5x-2}{3} – \frac{1}{2} (3x-1) = \frac{9x+7}{6} – \frac{2}{9} (5x-1)
\]
\[
\begin{aligned}
&\frac{5x-2}{3} – \frac{1}{2} (3x-1)
\\[1em]
&\hspace{3em}= \frac{9x+7}{6} – \frac{2}{9} (5x-1)
\end{aligned}
\]
-
\[
(x-3)^2 = (x-1)^2
\]
-
\[
(x-5)(x+1) = (x+2)(x-3) + 13
\]
\[
\begin{aligned}
&(x-5)(x+1)
\\[1em]
&\hspace{4em} = (x+2)(x-3) + 13
\end{aligned}
\]
-
\[
(2x-5) (x-1) + x^2 = (3x-1) (x+2) +1
\]
\[
\begin{aligned}
&(2x-5) (x-1) + x^2
\\[1em]
&\hspace{3em} = (3x-1) (x+2) +1
\end{aligned}
\]
-
\[
8x(x+2) (x-1) = (2x+1)^3 – (2x+3)^2
\]
\[
\begin{aligned}
&8x(x+2) (x-1)
\\[1em]
&\hspace{3em} = (2x+1)^3 – (2x+3)^2
\end{aligned}
\]
In exercises 15 through 22, solve the equations for the variable \(\boldsymbol{x}\).
-
\[
5(5x-a) = a^2 (x-1)
\]
-
\[
a(x+b) + x(b-a) = 2b (2a-x)
\]
-
\[
x^2 + b^2 + b(b-1) = (x+b)^2
\]
-
\[
(x+a)^3 -2x^3 = 12a^3 – (x-a)
\]
-
\[
\frac{x-a}{b} + \frac{x-b}{a} = 2
\]
-
\[
\frac{x-3m}{m^2} + \frac{x-2m}{mn} = -\frac{1}{m}
\]
-
\[
\frac{a-x}{a} – \frac{b-x}{b} = \frac{2(a-b)}{ab}
\]
-
\[
\frac{x-a}{a+b} + \frac{a+b}{a-b} = \frac{x+b}{a+b} + \frac{x-b}{a-b}
\]
In exercises 23 through 26, solve each equation for the indicated variable in terms of the remaining variables.
-
\[
A = \pi (r^2+rs) ,\; s
\]
-
\[
S = a\frac{1-r^n}{1-r} ,\; a
\]
-
\[
S = \frac{f}{H-h} ,\; h
\]
-
\[
\frac{1}{x} + \frac{1}{y} = \frac{1}{a} ,\; x
\]
Solve the equations from 27 to 40 using factorization.
-
\[
x^2-4x-12=0
\]
-
\[
x^2-6x+9=0
\]
-
\[
x^2+24=-11x
\]
-
\[
2x^2-3x+1=0
\]
-
\[
9x^2-17x-2=0
\]
-
\[
(2x-1)^2 – (x+5)^2 =-19
\]
-
\[
(x-5)^2 – (x-4)^2 = (2x+3)^2 +12
\]
\[
\begin{aligned}
&(x-5)^2 – (x-4)^2
\\[1em]
&\hspace{5em}= (2x+3)^2 +12
\end{aligned}
\]
-
\[
(x-2)^3 – (x+1)^3 = -x(3x+4) -24
\]
\[
\begin{aligned}
&(x-2)^3 – (x+1)^3
\\[1em]
&\hspace{4em}= -x(3x+4) -24
\end{aligned}
\]
-
\[
6x^2 -\frac{5x}{2} = -\frac{1}{4}
\]
-
\[
\frac{2(x+5)}{5} + \frac{x-4}{4} = \frac{x^2-53}{5}
\]
-
\[
x^4-17x^2+16=0
\]
-
\[
6y^4 = \frac{y^2}{2} + \frac{1}{4}
\]
-
\[
x^{2/3}+x^{1/3}-6=0
\]
-
\[
2x^{2/3}+3x^{1/3}-2=0
\]
Solve the equations from 41 to 46 using the quadratic formula.
-
\[
9(x-1)^2=5
\]
-
\[
\sqrt{3}x-3=4x^2
\]
-
\[
2x(2x-3)=-1
\]
-
\[
(x+15)^2=6x(x+5)
\]
-
\[
x^2-2x-(a^2+2a)=0
\]
-
\[
\frac{x^2}{2a} – \frac{a+2}{2a} x+1=0
\]
In exercises 47 through 60, solve the given equations.
-
\[
\frac{x-6}{x} = \frac{x+6}{x-6} + \frac{6}{x}
\]
-
\[
\frac{x}{x+2} – \frac{x}{x-2} = \frac{x-15}{x^2-4}
\]
-
\[
\frac{1}{3x-3} + \frac{1}{4x+4} = \frac{1}{12x-12}
\]
-
\[
\frac{4x+1}{4x-1} = \frac{4x-1}{4x+1} + \frac{6}{16x^2-1}
\]
-
\[
\frac{1}{x} + \frac{1}{4-x} = 1
\]
-
\[
\frac{x}{1+x} + \frac{1}{1-x} = 0
\]
-
\[
\frac{3y-2}{3y+2} = \frac{2y+3}{4y-1}
\]
-
\[
\frac{ x+5 }{ (x-1)(x+2) } = \frac{2x}{x+2}
\]
-
\[
\frac{1}{x-1} – \frac{1}{x-2} = \frac{1}{x-3}
\]
-
\[
\frac{3x}{x-2} – \frac{1}{x^2-4} = 2
\]
-
\[
\frac{1}{x^2} + \frac{2}{x} -15 = 0
\]
-
\[
\frac{12}{x-1} + \frac{12}{x} = 10
\]
-
\[
\frac{2x}{x-1} = \frac{8}{x-1} – \frac{5}{x}
\]
-
\[
\frac{1}{x^2-4} + \frac{2x+3}{x+2} + \frac{x+3}{x-2} = 0
\]
In exercises 61 through 76, solve the equations and eliminate extraneous solutions.
-
\[
5-\sqrt{2x+3}=
\]
-
\[
\sqrt{ \frac{x}{18} + 1 } = \frac{2}{3}
\]
-
\[
(5x-1)^{1/2}=7
\]
-
\[
(y+9)^{3/2} = 4^3
\]
-
\[
\sqrt{x^2-5} = 5-x
\]
-
\[
\sqrt{z+7} – \sqrt{z} = 1
\]
-
\[
\sqrt{ 9x^2 – 10x } = 3x-2
\]
-
\[
\sqrt{ \frac{1}{x} } – \sqrt{ \frac{8}{4x+1} } = 0
\]
-
\[
\sqrt{4x+1} + 1 = 2x
\]
-
\[
\sqrt{x^2+5} = 2x-1
\]
-
\[
\sqrt{x+5} = 2\sqrt{x} – 1
\]
-
\[
\sqrt{x} + \sqrt{x-3} = \sqrt{x+5}
\]
-
\[
\sqrt{ x + \sqrt{x+8} } = 2\sqrt{x}
\]
-
\[
\sqrt{ x + \sqrt{x+8} } = 2\sqrt{x}
\]
-
\[
\sqrt{ x + \sqrt{x+8} } = 2\sqrt{x}
\]
-
\[
\frac{x}{2} = \frac{ \sqrt{x+2} – \sqrt{x-2} }{ \sqrt{x+2} + \sqrt{x-2} }
\]
In the exercises 77 and 78, solve the equations by using a change of variable.
-
\[
\left( \frac{3x}{x+1} \right)^2 – \frac{6x}{x+1} = 8
\]
-
\[
\sqrt[3]{ \frac{5x+4}{x-1} } + \sqrt[3]{ \frac{x-1}{5x+4} } = \frac{5}{2}
\]
In the exercises 79 and 80, use the remainder theorem to find the remainders.
-
\[
3x^4 – 5x^3 – 4x^2 + 3x – 2, \; \text{ entre } (x-2)
\]
\[
\begin{aligned}
&3x^4 – 5x^3 – 4x^2 + 3x – 2,
\\[1em]
&\hspace{4em}\text{entre } (x-2)
\end{aligned}
\]
-
\[
x^3 – 6x^2 +11x – 6, \; \text{ entre } (x+2)
\]
\[
\begin{aligned}
&x^3 – 6x^2 +11x – 6,
\\[1em]
&\hspace{4em} \text{entre } (x+2)
\end{aligned}
\]
In exercises 81 through 88, find the roots of the equations and factorize the polynomials.
-
\[
x^3 + 2x^2 – x – 2 = 0
\]
-
\[
x^3 – 3x^2 + 2 = 0
\]
-
\[
4x^3 – 7x^2 + 3 = 0
\]
-
\[
2x^3 – 2x^2 – 11x + 2 = 0
\]
-
\[
x^4 – x^3 – 5x^2 + 3x + 6 = 0
\]
-
\[
3x^4 + 5x^3 – 5x^2 – 5x + 2 = 0
\]
-
\[
x^5 -3x^4 -5x^3 +15x^2+4x -12 = 0
\]
\[
\begin{aligned}
x^5 -3x^4 -5x^3 &+15x^2
\\[1em]
&+ 4x -12 = 0
\end{aligned}
\]
-
\[
x^5 + 4x^4 – 4x^3 -34x^2 -45x-18 = 0
\]
\[
\begin{aligned}
x^5 + 4x^4 – 4x^3 &-34x^2
\\[1em]
&-45x-18 = 0
\end{aligned}
\]
In the exercises 89, 90 and 91, use the factor theorem to prove:
-
\(x – a\) is a factor of \(x^n – a^n\), where \(n\) is positive integer.
-
\(x + a\) is a factor of \(x^n – a^n\), where \(n\) is even integer.
-
\(x + a\) is a factor of \(x^n + a^n\), where \(n\) is odd integer.