Question

Решите уравнение:
а) (14+2х)+(3х-11)=-2
б) (9-4х)-(2х-3)=6
в) 4х(х-5)-2х(2х-5)=20
г)3-2y/5+y/4=0

Answer 1

1. $2x+3x+3=-2$
$5x+3=-2$
$5x=-3-2$
$5x=-5$
$frac{5x}{5} =- frac{5}{5}$
$x=- frac{5}{5}$
$x=-1*1$
$x=-1$

2. $9-4x+(-2x+3)=6$
$9-4x-2x+3=6$
$-4x-2x+12=6$
$-6x+12=6$
$-6x=-12+6$
$-6x=-6$
$- frac{6x}{-6} =- frac{6}{-6}$
$x=- frac{6}{-6}$
$x=1$

3. $4x^{2} -20x+(-4x^2+10x)=20$
$4x^2-20x-4x^2+10x=20$
$-20x+0+10x=20$
$-20x+10x=20$
$-10x=20$
$- frac{10x}{-10} = frac{20}{-10}$
$x= frac{20}{-10}$
$x=-2$

4. $3- frac{2y}{5} + frac{y}{4} =0$
$3+(- frac{2y}{5} * frac{4}{4} + frac{y}{4} )=0$
$3+(- frac{2y}{5} * frac{4}{4} + frac{y}{4} * frac{5}{5} )=0$
$3+(- frac{2y*4}{20} + frac{y*5}{20} )=0$
$3+ frac{-(2y*4)+y*5}{20} =0$
$3- frac{3y}{20} =0$
$- frac{3y}{20} =-3$
$-3y=-3*20$
$-3y=-60$
$- frac{3y}{-3}=- frac{60}{-3}$
$y=- frac{60}{-3}$
$y=20$