Question

In the inequality 6a+4b>10, what could be the possible value of a if b=2?

Answer 1

We are given the following inequality:

$6a+4b>10$

If we replace b = 2, we get:

$\begin{gathered} 6a+4(2)>10 \\ 6a+8>10 \end{gathered}$

Now we solve for "a" first by subtracting 8 on both sides:

$\begin{gathered} 6a+8-8>10-8 \\ 6a>2 \end{gathered}$

Now we divide both sides by 6

$\frac{6a}{6}>\frac{2}{6}$

Simplifying:

$a>\frac{1}{3}$

Therefore, for b = 2, the possible values of "a" are those that are greater than 1/3