100 lines
2.2 KiB
Markdown
100 lines
2.2 KiB
Markdown
---
|
|
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
|
|
|
|
$$
|
|
\frac{6
|
|
\cdot 3}{3} = 6
|
|
$$
|
|
|
|
$$\frac{18}{3} = 6 $$
|