Definition and how to solve linear equations

We have to follow the steps, mainly isolate the x, and pass the rest of the terms to the other side of the equality.

1. In the case of the first equation: $$2x+1=3 \Rightarrow 2x=3-1 \Rightarrow 2x=2 \Rightarrow x=\dfrac{2}{2}=1$$

2. In the second case, it is necessary to use least common multiple: $$6x+\dfrac{1}{2}=\dfrac{4}{3} \Rightarrow 6x=\dfrac{4}{3}-\dfrac{1}{2} \Rightarrow 6x=\dfrac{8}{6}-\dfrac{3}{6} \Rightarrow$$ $$\Rightarrow 6x=\dfrac{5}{6} \Rightarrow x=\dfrac{5}{6 \cdot 6}=\dfrac{5}{36}$$

3. In the third case: $$3x+5=-5x+3 \Rightarrow 3x+5x=3-5 \Rightarrow 8x=-2 \Rightarrow x=-\dfrac{2}{8}=-\dfrac{1}{4}$$

4. Finally: $$-8(6+3x)=-7(-6-3x)$$ It is necessary to solve first the products and then continue as in the previous cases: $$-48-24x=42+21x \Rightarrow -24x-21x=42+48 \Rightarrow$$ $$\Rightarrow -45x=90 \Rightarrow x=\dfrac{90}{-45}=-2$$

We raise an equation that corresponds to the statement of the problem.

If $x$ is the number of candies that I bought yesterday, then, $2\cdot x$ is the number of candies that I have bought today.

If I give $3$ candies to my friend I have to subtract $3$ from the quantity of candies that I have today: $$2x-3$$

Since I am left with just with $1$ candy, the statement is translated into the following equation: $$2x-3=1$$

We solve this equation: $$2x=1+3 \Rightarrow 2x=4 \Rightarrow x=2 $$