How do you solve $y - 3 < 5 y + 1$ $y-3<5y+1$ ?

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Answer 1 · 2018-03-05T19:10:48

$y > - 1$ $y> -1$

Move like terms to one side:

$y - 5 y < 1 + 3$ $y-5y<1+3$

$- 4 y < 4$ $-4y<4$

When multiplying or dividing by the negative, make sure to flip the inequality sign:

$y > - 1$ $y> -1$

Answer 2 · 2018-03-05T19:12:40

$y < - 1$ $y<-1$

$y - 3 < 5 y + 1$ $y-3<5y + 1$

$y - y - 3 - 1 < 5 y + 1 - 1 - y$ $y-y-3-1<5y + 1-1-y$

$- \frac{4}{4} > \frac{4}{4} y$ $-4/4>4/4y$

$- 1 > y$ $-1>y$

$y < - 1$ $y<-1$

How do you solve y-3<5y+1y−3<5y+1y-3<5y+1?