Identify the LCD of the expressions in the following rational equations:
1) \(\Large{\frac{2x-1}{x-6}=\frac{x+3}{x}+\frac{4x+3}{6}}\)
2) \(\Large{\frac{x+6}{2x}+\frac{x}{x-2}=\frac{3x}{x^2-4}}\)
Solve the following rational equations.
3) \(\Large{\frac{3x-1}{x+1}=\frac{5x+4}{x+1}}\)
4) \(\Large{\frac{x^2-20}{x-5}=\frac{-2x+15}{x-5}}\)
5) \(\Large{\frac{5}{x+3}=\frac{2}{x}}\)
6) \(\Large{\frac{6}{x^2-9}=\frac{x+8}{x-3}}\)
7) \(\Large{\frac{x+7}{x-4}+\frac{2}{x+1}=\frac{2x-1}{x-4}}\)
8) \(\Large{\frac{3x+4}{x+2}+\frac{3}{x-5}=\frac{6x-9}{x^2-3x-10}}\)
9) \(\Large{\frac{x}{x+2}\normalsize-3=\Large\frac{x-7}{x^2+5x+6}}\)
10) Meghan found the solutions of \(\Large{\frac{3x-7}{x-2}=\frac{-5}{x^2+x-6}}\) to be \(x=2\) and \(\large{x =-\frac{8}{3}}\).
a) Without solving, explain why Meghan cannot be correct.
b) Solve the equation.
Solve the following rational equations.
11) \(\Large{\frac{x-2}{x+4}=\frac{x+6}{x}}\)
12) \(\Large{\frac{5}{x^2-16}-1=\frac{x+3}{x-4}}\)
13) \(\Large{\frac{x+10}{2x^2-6x}-\frac{1}{x-3}=-2}\)
14) \(\Large{\frac{-2x}{x-3}+\frac{x}{2x-8}=\frac{2}{x-4}}\)
15) \(\Large{\frac{x+5}{x+2}-\frac{3}{2}=\frac{-2}{x^2-5x-14}}\)
16) \(\Large{\frac{x^2-5}{2x^2+5x-3}+\frac{5-x}{x+3}=\frac{x-1}{x+3}}\)
Review
17) Solve \(2x^3-21x^2+28x-9=0\), given that \((2x-1)\) is a factor.
18) Rewrite as an equivalent expression: \((3+2i)(-6-i)\).
19) Add the rational expressions: \(\large{\frac{2x}{x^2-9}+\frac{x-5}{x-3}}\).
Solution Bank
1) \(\Large{\frac{2x-1}{x-6}=\frac{x+3}{x}+\frac{4x+3}{6}}\)
2) \(\Large{\frac{x+6}{2x}+\frac{x}{x-2}=\frac{3x}{x^2-4}}\)
Solve the following rational equations.
3) \(\Large{\frac{3x-1}{x+1}=\frac{5x+4}{x+1}}\)
4) \(\Large{\frac{x^2-20}{x-5}=\frac{-2x+15}{x-5}}\)
5) \(\Large{\frac{5}{x+3}=\frac{2}{x}}\)
6) \(\Large{\frac{6}{x^2-9}=\frac{x+8}{x-3}}\)
7) \(\Large{\frac{x+7}{x-4}+\frac{2}{x+1}=\frac{2x-1}{x-4}}\)
8) \(\Large{\frac{3x+4}{x+2}+\frac{3}{x-5}=\frac{6x-9}{x^2-3x-10}}\)
9) \(\Large{\frac{x}{x+2}\normalsize-3=\Large\frac{x-7}{x^2+5x+6}}\)
10) Meghan found the solutions of \(\Large{\frac{3x-7}{x-2}=\frac{-5}{x^2+x-6}}\) to be \(x=2\) and \(\large{x =-\frac{8}{3}}\).
a) Without solving, explain why Meghan cannot be correct.
b) Solve the equation.
Solve the following rational equations.
11) \(\Large{\frac{x-2}{x+4}=\frac{x+6}{x}}\)
12) \(\Large{\frac{5}{x^2-16}-1=\frac{x+3}{x-4}}\)
13) \(\Large{\frac{x+10}{2x^2-6x}-\frac{1}{x-3}=-2}\)
14) \(\Large{\frac{-2x}{x-3}+\frac{x}{2x-8}=\frac{2}{x-4}}\)
15) \(\Large{\frac{x+5}{x+2}-\frac{3}{2}=\frac{-2}{x^2-5x-14}}\)
16) \(\Large{\frac{x^2-5}{2x^2+5x-3}+\frac{5-x}{x+3}=\frac{x-1}{x+3}}\)
Review
17) Solve \(2x^3-21x^2+28x-9=0\), given that \((2x-1)\) is a factor.
18) Rewrite as an equivalent expression: \((3+2i)(-6-i)\).
19) Add the rational expressions: \(\large{\frac{2x}{x^2-9}+\frac{x-5}{x-3}}\).
Solution Bank