Question

Soluciona las ecuaciones lineales y cuadráticas: a. 1/9X -3 = X + 8 b. 2/11 X + 2/3 = - ½ c. X² -11 = 6 d. X² + 2/3 = 12 e. X² - 1/9 = -1/10

Answer 1

a)

$\frac{1}{9}x-3=x+8$

$\frac{1}{9}x-3+3=x+8+3$

$\frac{1}{9}x=x+11$

$\frac{1}{9}x-x=x+11-x$

$-\frac{8}{9}x=11$

$9\left(-\frac{8}{9}x\right)=11\cdot \:9$

$-8x=99$

$-x=\frac{99}{-8}$

$x=-\frac{99}{8}$

b)

$\frac{2}{11}x+\frac{2}{3}=-\frac{1}{2}$

$\frac{2}{11}x+\frac{2}{3}-\frac{2}{3}=-\frac{1}{2}-\frac{2}{3}$

$\frac{2}{11}x=-\frac{7}{6}$

$11\cdot \frac{2}{11}x=11\left(-\frac{7}{6}\right)$

$2x=-\frac{77}{6}$

$=-\frac{77}{6\cdot \:2}$

$x=-\frac{77}{12}$

c)

$x^2-11=6\\x^2-11+11=6+11\\x^2=17\\x=\sqrt{17}$

d)

$x^2+\frac{2}{3}=12$

$x^2+\frac{2}{3}-\frac{2}{3}=12-\frac{2}{3}$

$x^2=12-\frac{2}{3}$

$x^2=\frac{12\cdot \:3}{3}-\frac{2}{3}$

$x^2=\frac{12\cdot \:3-2}{3}$

$12\cdot \:3-2=34$

$x^2=\frac{34}{3}$

$x=\sqrt{\frac{34}{3}}$

e)

$x^2-\frac{1}{9}=-\frac{1}{10}$

$x^2-\frac{1}{9}+\frac{1}{9}=-\frac{1}{10}+\frac{1}{9}$

-------------------------------

MCM de 10 y 9:

10-9 l 10

01-9 l 9

01-1  l

MCM = 10x9 = 90

--------------------------------

$\frac{1}{10}=\frac{1\cdot \:9}{10\cdot \:9}=\frac{9}{90}$

$\frac{1}{9}=\frac{1\cdot \:10}{9\cdot \:10}=\frac{10}{90}$

--------------------------------

$x^2=-\frac{9}{90}+\frac{10}{90}$

$x^2=\frac{1}{90}$

$=\sqrt{\frac{1}{90}}$

$=\frac{\sqrt{1}}{\sqrt{90}}$

$\sqrt{90}=3\sqrt{10}$

$=\frac{\sqrt{1}}{3\sqrt{10}}$

$=\frac{1\cdot \sqrt{10}}{3\sqrt{10}\sqrt{10}}$

$3\sqrt{10}\sqrt{10}=30$

$x=\frac{\sqrt{10}}{30}$