How do you solve $3 x + 3 < x + 6$ $3x + 3< x + 6$ ?

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Answer 1 · 2018-04-26T14:55:38

$x < \frac{3}{2}$ $x<3/2$ or $x < 1.5$ $x<1.5$

Given: $3 x + 3 < x + 6$ $3x+3<x+6$ .

Subtract $x$ $x$ from both sides.

$3x+3-x<color(red)cancelcolor(black)x+6-color(red)cancelcolor(black)x$

$2x+3<6$

Subtract $3$ from both sides.

$2x+color(red)cancelcolor(black)3-color(red)cancelcolor(black)3<6-3$

$2x<3$

Divide by $2$ to get $x$ .

$(color(red)cancelcolor(black)2x)/(color(red)cancelcolor(black)2)<3/2$

$x<3/2=1.5$

How do you solve 3x + 3< x + 63x+3<x+63x + 3< x + 6?