[br][br][color=#0000ff]Example 1. Solve [math]\large 2x+5=7.[/math][/color][br][br][math]\large \begin{array}{lrcll}[br]\text{Add -5 to both sides: } &2x+5&=&7&|-5\\[br] \text{ }&2x+\cancel 5-\cancel 5&=&7-5 \\[br] \text{Divide with multiplier of the variable: }&2x&=&2&|:2\\[br]\text{Give the answer:} &x&=&1&[br]\end{array}[br][/math] [br][br][br][color=#0000ff]Example 2. Solve [math]\large 3x-5=4x+1.[/math] [/color][br][br][math]\large \begin{array}{lrcll}[br]\text{Add 5 to both sides: } & 3x-5&=&4x+1&|+5\\[br]\text{Simplify: }&3x&=&4x+6&|-4x \\[br] \text{Add -4x to both sides: }&3x-4x&=&4x+6-4x& \\[br]\text{Simplify: }&-x&=&6&|:(-1) \\[br] \text{Divide with multiplier of the variable: }&\dfrac{\cancel{-}x}{\cancel{-1}}&=&\dfrac{6}{-1}&\\[br]\text{Give the answer:} &x&=&-6&[br]\end{array}[br][/math] [br][br][br][color=#0000ff]Example 3. [/color][br][br][math]\large \begin{array}{lrcll}[br]\text{Remove brackets:} &x-7&=&3x-(6x-8)&\\[br] \text{Change the signs! }&x-7&=&3x\textcolor{blue}{-}6x\textcolor{blue}{+}8 &|+7\\[br] \text{Add 7 to both sides: }&x&=&-3x+8+7&|+3x\\[br]\text{Add 3x} &x+3x&=&15&\\[br]\text{Simplify: } &4x&=&15&|:4\\[br]\text{Divide with the multiplier of a variable} &x&=&\frac{15}{4}&[br]\end{array}[br][/math] [br][br][br][color=#0000ff]Example 4. [/color][br][br][math]\large \begin{array}{lrcll}[br]\text{} & \frac{4x-2(x-4)}{3}&=&8&|\cdot 3\\[br]\text{Multiply with the denominator the both sides: } & \cancel 3 \cdot \frac{4x-2(x-4)}{\cancel 3}&=&3\cdot 8& \\[br]\text{Simplify: } & 4x-2(x-4)&=&24\\[br]\text{Remove brackets: } & 4x-2x\textcolor{blue}{+}8&=&24&|-8\\[br]\text{Subtract 8 from both sides: } & 2x&=&16&|:2\\[br]\text{Divide with the multiplier of a variable:} & x&=&8[br]\end{array}[/math][br] [br] [br][color=#0000ff]Example 5. [/color][br][br][math]\large \begin{array}{rcll}[br]\frac{2}{x}&=&\frac{3}{x+3}&|\cdot (x+3)\\[br]\frac{2\cdot (x+3)}{x}&=&3&|\cdot x\\[br]2\cdot (x+3)&=&3x&\\[br]2x+6&=&3x&|-6-3x\\[br]2x-3x&=&-6\\[br]-x&=&-6\\[br]x&=&6[br]\end{array}[/math][br][br]In this example, we could have used cross-multiplications [br][br][math]\large\frac a b =\frac c d \Leftrightarrow ad=bc[/math][br][br]to get [br][br][math]\large \begin{array}{rcll}[br]\frac{2}{x}&=&\frac{3}{x+3}&\\[br]2\cdot (x+3)&=&3x&\\[br]\end{array}[/math][br][br]