How do you solve the inequality $3( d + 2 )> 6$ ?

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Answer 1 · 2018-07-16T10:22:48

$d > 0$

As explained below:

$3(d+2)>6$

Open the brackets:

$3d + 6 > 6$

Subtract $6$ both sides:

$3d+6-6 > 6 -6$

$3d+cancel(6-6) > 0$

$3d > 0$

Divide by $3$ both sides:

$(3d)/3 > 0/3$

$d > 0$

Check the answer:

Let $d = 1$

$3(1+2) > 6$

$3(3) > 6$

$9 > 6$ ---> Right!

How do you solve the inequality 3( d + 2 )> 63( d + 2 )> 6?