Question

solve $8(t + 2) + 7(t + 2) - 3(t - 2) < 0$.
$t < \square$

Explanation:

Step1: Expand the expressions

First, we expand each term using the distributive property (also known as the distributive law: \(a(b + c)=ab+ac\) and \(a(b - c)=ab - ac\)).

For \(8(l + 2)\), we get \(8\times l+8\times2 = 8l+16\).

For \(7(l + 2)\), we get \(7\times l+7\times2=7l + 14\).

For \(-3(l - 2)\), we get \(-3\times l-3\times(- 2)=-3l + 6\).

Now, substitute these expanded forms back into the original inequality:

\(8l + 16+7l + 14-3l + 6<0\)

Step2: Combine like terms

Next, we combine the \(l\)-terms and the constant terms separately.

For the \(l\)-terms: \(8l+7l-3l=(8 + 7-3)l=12l\)

For the constant terms: \(16 + 14+6=36\)

So the inequality simplifies to:

\(12l+36<0\)

Step3: Isolate the variable \(l\)

Subtract 36 from both sides of the inequality:

\(12l+36-36<0 - 36\)

Which simplifies to:

\(12l<-36\)

Then, divide both sides by 12 (since 12 is positive, the direction of the inequality sign remains the same):

\(\frac{12l}{12}<\frac{- 36}{12}\)

\(l<-3\)

Answer:

\(-3\)