Question

Help asap please and thanks

Answer 1

 $-\frac{2}{3}(2x - \frac{1}{2}) \leq \frac{1}{5}x - 1$

$-\frac{4x}{3} + \frac{1}{3} \leq \frac{1}{5}x - 1$   distributed =2/3 on left side

$-\frac{4x}{3}(15) + \frac{1}{3}(15) \leq \frac{1}{5}x (15) - 1(15)$  

-20x + 5 ≤ 3x - 15

-3x           -3x        

-23x + 5 ≤     -15

         -5        -5

-23x      ≤     -20

    x      ≥     $\frac{20}{23}$     divided by a negative so the symbol reversed

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2x² + 2 = 3x

2x² - 3x + 2 = 0

2x² - 3x       = -2

2(x² - $\frac{3}{2}x}$ + __) = -2 + (2)(___)

2(x² - $\frac{3}{2}x}+(-\frac{3}{4})^{2}$) = -2 + $2(-\frac{3}{4})^{2}$

2(x - $\frac{3}{4}$)² = -2 + $2(\frac{9}{16})$

2(x - $\frac{3}{4}$)² = -2 + $\frac{9}{8}$

2(x - $\frac{3}{4}$)² = -2 + $\frac{9}{8}$

2(x - $\frac{3}{4}$)² = $-\frac{16}{8}$ + $\frac{9}{8}$

2(x - $\frac{3}{4}$)² = $-\frac{7}{8}$

(x - $\frac{3}{4}$)² = $-\frac{7}{16}$

x - $\frac{3}{4}$ = +/- $\sqrt{\frac{-7}{16} }$

x - $\frac{3}{4}$ = +/- $\frac{i\sqrt{7} }{4}$

x = $\frac{3}{4}$ + $\frac{i\sqrt{7} }{4}$, x = $\frac{3}{4}$ - $\frac{i\sqrt{7} }{4}$

x = $\frac{3 + i\sqrt{7} }{4}$, x = $\frac{3 - i\sqrt{7} }{4}$