Question

Determine a equaçao...A) -5x+9=2x-12 B) -2x-5=-9 C)-x-(-20)=-30
me ajudemn preciso disso hj aindaaa

Answer 1

A ---      -5x+9 =2x -12    -----      -5x -2x = -12-9 ----  -7x=-21 .(-1) ---- x= 3



B --      -2x -5 = -9          -----   -2x = -4.(-1) ---- 2x= 4 ------  x=2 



c-------          -x-(-20) =-30 ------   -x= -50 ----------------- x=50

Answer 2

a) -5x + 9 - 2x + 12 = 0   => -7x + 21 = 0  (-1)    => 7x - 24 = 0


Δ = $b^{2} - 4 * a * c$
Δ = (-21)^{2} - 4 * 7* 0
Δ = 441 + 0
Δ = 441

$\frac{-(-21)+- \sqrt{441} }{2*7 }} \\ \\ x' = \frac{21+21}{14x1} = \frac{42}{14} => x' = 3$

$x'' = \frac{21-21}{14} => x'' = \frac{0}{14} => x'' = 0$
S = {3 , 0}

b) -2x - 5 =-9   => -2x -5 + 9 (-1)  => 2x +4 = 0

Δ = (-4)^{2} - 4 * 2 * 0
Δ = 16 + 0
Δ = 16 

$\frac{-(-4)+ \sqrt{16}}{2*2} => \frac{4+- 4}{4} => \\ \\ x' = \frac{4+4}{4} => x' = \frac{8}{4} => x' = 2 \\ \\ x'' = \frac{4-4}{4} => \frac{0}{4} => x'' = 0$
S = {2, 0 }

c) -x-(-20) = - 30   => -x + 20 + 30    => -x + 50 *(-1)  => x - 50

Δ = (-20)^{2} - 4 * 1 * 0 
Δ = 2500 + 0
Δ = 2500

$\frac{-(-50)+- \sqrt{2500} }{2} => \frac{50+-50}{2} => \\ \\ x' = \frac{50+50}{2} => x' = \frac{100}{2} x' = 50 \\ \\ x'' = \frac{50-50}{2} => x'' = \frac{0}{2} => x'' = 0$

S= {50, 0}