## Precalculus (6th Edition)

(a) identity; solution set = $\emptyset$ (b) contradiction; solution set = $\left\{x|x \text{ is a real number}\right\}$ (c) conditional; solution set = $\left\{\frac{11}{4}\right\}$
RECALL: (1) An equation in one variable is an identity if any real number satisfies the equation. (2) An equation in on variable is conditional if only one real number satisfies the equation. (3) An equation is a contradiction if no real number satisfies the equation. Solve each equation to obtain: (a) $4x-5=-2(3-2x)+3 \\4x-5=-6-(-4x)+3 \\4x-5=-6+4x+3 \\4x-5=4x-3 \\4x-4x=-3+5 \\0=2$ The statement above is false therefore the given equation has no solution. Thus, the equation is a contradiction and its solution set is $\emptyset$. (b) $5x-9=5(-2+x)+1 \\5x-9=-10+5x+1 \\5x-9=5x-9$ The statement above is true for all real numbers therefore the equation is an identity. The solution set is $\left\{x|x \text{ is a real number}\right\}$. (c) $5x-4=3(6-x) \\5x-4=18-3x \\5x-4+3x=18-3x+3x \\8x-4=18 \\8x-4+4=18+4 \\8x=22 \\\frac{8x}{8}=\frac{22}{8} \\x=\frac{11}{4}$ Thus, the equation is conditional and its solution set is $\left\{\dfrac{11}{4}\right\}$