Question

Найдите корень уравнения:
а) (y+4)-(y-1)=6y
б) 3p-1-(p+3)=1
в) 6x-(7x-12)=101
г) 20x=19-(3+12x)

Answer 1

a)

$(y+4)-(y-1)=6y\\y+4-y+1-6y=0\\-6y+5=0\\-6y=-5\\\\\underline{y=\dfrac56\qquad\qquad\quad\;\qquad\qquad\ }\\\\\bigg(\dfrac56+4\bigg)-\bigg(\dfrac56-1\bigg)=6\cdot\dfrac56\\\\{}\qquad\qquad\qquad\qquad\ \,\,5=5$

б)

$3p-1-(p+3)=1\\3p-1-p-3=1\\2p=1+4\\p=5:2\\\underline{p=2.5\qquad\qquad\qquad\qquad}\\3\cdot2.5-1-(2.5+3)=1\\{}\qquad\qquad\qquad\qquad\ \,\,1=1$

в)

$6x-(7x-12)=101\\6x-7x+12=101\\-1x=101-12\\\underline{x=-89\qquad\qquad\qquad\qquad\quad\; }\\6(-89)-(7(-89)-12)=101\\{}\qquad\qquad\qquad\qquad\ \,\,\, 101=101$

г)

$20x=19-(3+12x)\\20x=19-3-12x\\20x+12x=16\\32x=16\\x=16:32\\\underline{x=0.5\qquad\qquad\qquad\qquad\quad}\\20\cdot0.5=19-(3+12\cdot0.5)\\{}\qquad10=10$