# Equivalent Equations

Equations, that have same roots are called equivalent. In particular, equations that don't have roots are equivalent.

Example 1. Equations ${x}+{2}={7}$ and ${2}{x}+{1}={11}$ are equivalent, because each of them has unique root 5.

Example 2. Equations ${{x}}^{{2}}+{1}={0}$ and ${2}{{x}}^{{2}}+{3}={1}$ are equivalent, because they don't have roots.

Example 3. Equations ${{x}}^{{2}}+{2}{x}={3}$ and ${{x}}^{{3}}={8}$ are not equivalent, because first of them has two roots: 1 and -3, and another has one root: 2.

Warning. Equations are equivalent only when they have exactly same roots.

Example 4. Equations ${x}-{6}={0}$ and ${{x}}^{{2}}={36}$ are not equivalent, because, although 6 is root of both of them, but second equation also has another root: -6.

Rules for getting equivalent equation:

• Adding (subtracting) same number (or expression) to both sides of the equation will give equivalent equation.
• Multiplying (dividing) both sides of the equation by the same non-zero number (or expression) will give equivalent equation.
• Raising both sides of equation to the same odd power (or taking odd root) will give equivalent equation.
• If both sides of equation are non-negative, then raising both sides of equation to the same even power (or taking even root) will give equivalent equation.

These rules give us an ability to "simplify" equations and even solve them.

Example 1. Solve the equation ${x}+{1}={0}$.

To solve it, let's try to isolate variable (leave it on one side) and everything else move to another.

To obtain equivalent equation, we can subtract 1 from both sides of the equation:

${x}+{1}-{1}={0}-{1}$

${x}=-{1}$

We isolated variable. The only root of the equation is $-{1}$.

Example 2. Solve the equation ${2}{x}-{8}={0}$.

To solve it, let's try to isolate variable (leave it on one side) and everything else move to another.

To obtain equivalent equation, we can add 8 to both sides of the equation:

${2}{x}-{8}+{8}={0}+{8}$

${2}{x}={8}$

Now, we can divide both sides by 2:

$\frac{{{2}{x}}}{{2}}=\frac{{8}}{{2}}$

${x}={4}$

We isolated variable. The only root of the equation is ${4}$.

Example 3. Simplify the equation: ${2}{{x}}^{{3}}+{2}{{x}}^{{2}}+{5}={4}{{x}}^{{2}}-{1}$.

Usually, "simplify" means rewriting equation in such way that right side equals 0.

So, let's subtract ${4}{{x}}^{{2}}$ from both sides:

${2}{{x}}^{{3}}+{2}{{x}}^{{2}}+{5}-{4}{{x}}^{{2}}={4}{{x}}^{{2}}-{1}-{4}{{x}}^{{2}}$

${2}{{x}}^{{3}}+{2}{{x}}^{{2}}+{5}-{4}{{x}}^{{2}}=-{1}$

When we deal with expressions, involving variables, we also can use all properties, that are true for number expressions (we used commutative property of addition)

Now, add 1 to both sides of the equation:

${2}{{x}}^{{3}}+{2}{{x}}^{{2}}+{5}-{4}{{x}}^{{2}}+{1}=-{1}+{1}$

${2}{{x}}^{{3}}+{2}{{x}}^{{2}}+{5}-{4}{{x}}^{{2}}+{1}={0}$

We can simplify even more, using properties of expressions.

${2}{{x}}^{{3}}+{2}{{x}}^{{2}}-{4}{{x}}^{{2}}+{5}+{1}={0}$ (commutative property of addition)

${2}{{x}}^{{3}}+{2}{{x}}^{{2}}-{4}{{x}}^{{2}}+{6}={0}$

${2}{{x}}^{{3}}+{\left({2}-{4}\right)}{{x}}^{{2}}+{6}={0}$ (distributive property of multiplication)

${2}{{x}}^{{3}}-{2}{{x}}^{{2}}+{6}={0}$

Finally, multiply both sides of the equation by $\frac{{1}}{{2}}$:

$\frac{{1}}{{2}}{\left({2}{{x}}^{{3}}-{2}{{x}}^{{2}}+{6}\right)}=\frac{{1}}{{2}}\cdot{0}$

$\frac{{1}}{{2}}{\left({2}{{x}}^{{3}}-{2}{{x}}^{{2}}+{6}\right)}={0}$

$\frac{{1}}{{2}}\cdot{2}{{x}}^{{3}}-\frac{{1}}{{2}}\cdot{2}{{x}}^{{2}}+\frac{{1}}{{2}}\cdot{6}={0}$ (distributive property of multiplication)

${{x}}^{{3}}-{{x}}^{{2}}+{3}={0}$

Done.

So, ${2}{{x}}^{{3}}+{2}{{x}}^{{2}}+{5}={4}{{x}}^{{2}}-{1}$ is equivalent to ${{x}}^{{3}}-{{x}}^{{2}}+{3}={0}$, but last one looks like much simpler.

Now, it is time to exercise.

Exercise 1. Find roots of the equation ${x}-{2}={0}$.

Exercise 2. Solve the equation $-\frac{{1}}{{2}}{x}+{4}={0}$.
Exercise 3. "Simplify" the equation ${3}{{x}}^{{2}}+{{x}}^{{3}}-{5}-{2}{x}={7}+{2}{{x}}^{{3}}+{x}-{3}{x}-{{x}}^{{2}}$.
Answer: ${{x}}^{{3}}-{4}{{x}}^{{2}}+{12}={0}$.