eolas/zk/Equivalent_equations.md

---
tags: [algebra]
---

# Equivalent equations

> Two equations are equivalent if they have the same
> [solution](Algebra%20key%20terms.md#678811) set.

We know from the distributive property of multiplication that the equation
$a \cdot (b + c )$ is equivalent to $a \cdot b + a \cdot c$. If we assign values
to the variables such that $b$ is $5$ and $c$ is $2$ we can demonstrate the
equivalence that obtains in the case of the distributive property by showing
that both $a \cdot (b + c )$ and $a \cdot b + a \cdot c$ have the same solution:

$$ 2 \cdot (5 + 2) = 14 $$

$$ 2 \cdot 5 + 2 \cdot 2 =14 $$

When we substitute $a$ with $2$ (the solution) we arrive at a true statement
(the assertion that arrangement of values results in $14$). Since both
expressions have the same solution they are equivalent.

## Creating equivalent equations

We can create equivalent equations by adding, subtracting, multiplying and
dividing the _same quantity_ from both sides of the equation (i.e. either side
of the $=$ symbol). Adding or subtracting the same quantity from both sides
(either side of the $=$ ) of the equation results in an equivalent equation.

### Demonstration with addition

$$
x - 4 = 3
$$

The [solution](Algebra%20key%20terms.md#678811) to this equation is $7$

$$
x -
4 (+4) = 3 (+ 4)
$$

Here we have added $4$ to each side of the equation. If $x = 7$ then:

$$ 7 - 4 (+ 4) = 7 $$

and:

$$ 3 + 4 = 7 $$

### Demonstration with subtraction

$$
x + 4 = 9
$$

The [solution](Algebra%20key%20terms.md#678811) to this equation is $5$.

$$
x +
4 (-4) = 9(-4)
$$

Here we have subtracted $4$ from each side of the equation. If $x = 5$ then:

$$ 5 + 4 (-4) = 5 $$

and

$$ 9 - 4 = 5 $$

### Demonstration with multiplication

$$x \cdot 2 = 10  $$ The [solution](Algebra%20key%20terms.md#678811) to this
equation is $5$.

$$
(x \cdot 2) \cdot 3 = 10 \cdot 3 $$ Here we have multiplied each side of the
equation by $3$. If $x =5$ then

$$ (5 \cdot 2) \cdot 3 = 30$$
$$ 10 \cdot 3 = 30$$

### Demonstration with division

$$x \cdot 3 = 18  $$
The [solution](Algebra%20key%20terms.md#678811) to this equation is $6$.
$$\frac{x
\cdot 3}{3} = \frac{18}{3}
$$

Here we have divided each side of the equation by $3$. If $x$ is 6, then

$$
\frac{6
\cdot 3}{3} = 6
$$

$$\frac{18}{3} = 6 $$