Question

$2a \leqslant - 4 + a$solving inequalities by multiplying and dividing

Answer 1

$2a\leqslant-4+a$

Solve the inequality for a

step 1

subtract a both sides

$\begin{gathered} 2a-a\leqslant-4+a-a \\ a\leq-4 \end{gathered}$

the solution is the interval for a (-infinite, -4}

Part 2

we have

$\frac{3}{4}t\ge9$

step 1

Multiply by 4 both sides

$3t\ge36$

step 2

Divide by 3 both sides

$t\ge12$

the solution is the interval {12, infinite)

If you have

2a > 8

In this case, to isolate the variable a, divide both sides by 2

so

a > 8/2

a>4

If you have

2a +5 > 10

subtract 5 both sides, to isolate teh variable

2a+5-5 > 10-5

2a > 5

then

divide by 2 both sides

a > 5/2

If you have

2a > a+3

group the terms with the variable

subtract a both sides

2a-a>a+3-a

a>3

If you have

3a>a+3

group the terms with the variable

subtract a both sides

3a-a>a+3-a

2a>3

divide by 2 both sides

a>3/2