Solving Equations by using Addition/ Subtraction: Solving an equation using all techniques

To open the parentheses on the left side, we'll use the formula:

$-a\left(b+c\right)=-ab-ac$

$-12a-24=27a$

We'll arrange the equation so that the terms with 'a' are on the right side, and maintain the plus and minus signs during the transfer:

$-24=27a+12a$

Let's group the terms on the right side:

$-24=39a$

Let's divide both sides by 39:

$-\frac{24}{39}=\frac{39a}{39}$

$-\frac{24}{39}=a$

Note that we can reduce the fraction since both numerator and denominator are divisible by 3:

$-\frac{8}{13}=a$

First, let's isolate a from the parentheses in the equation on the right side. We'll remember that minus times minus becomes plus, so we get the equation:

$a^4+7a-5=2a+a^4+3a+a$

Let's continue solving the equation on the right side by adding $2a+3a+a=5a+a=6a$

Now the equation we got is:

$a^4+7a-5=6a+a^4$

Let's divide both sides by $a^4$ and we get:

$7a-5=6a$

Now let's move 6a to the left side and the number 5 to the right side, remembering to change the plus and minus signs accordingly.

The equation we got now is:

$7a-6a=5$

Let's solve the subtraction and we get:

$1a=5$

Let's divide both sides by 1 and we find that $a=5$

First, we'll open the parentheses by multiplying each term by 4:

$4\times\frac{b}{2}+4\times b-\frac{1}{3}=6b$

Let's solve the multiplication exercise $4\times\frac{b}{2}=\frac{4b}{2}=2b$

Now the equation is:

$2b+4b-\frac{1}{3}=6b$

We'll combine the left side between the two b terms and get:

$6b-\frac{1}{3}=6b$

We'll reduce both sides by 6b and get:

$-\frac{1}{3}=0$

Since the result obtained is impossible, the exercise has no solution.