84 lines
		
	
	
	
		
			2.2 KiB
		
	
	
	
		
			Markdown
		
	
	
	
	
	
			
		
		
	
	
			84 lines
		
	
	
	
		
			2.2 KiB
		
	
	
	
		
			Markdown
		
	
	
	
	
	
---
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tags: [algebra]
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---
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## Equivalent equations
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> Two equations are equivalent if they have the same
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> [solution](Algebra%20key%20terms.md#678811) set.
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We know from the distributive property of multiplication that the equation
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$a \cdot (b + c )$ is equivalent to $a \cdot b + a \cdot c$. If we assign values
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to the variables such that $b$ is $5$ and $c$ is $2$ we can demonstrate the
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equivalence that obtains in the case of the distributive property by showing
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that both $a \cdot (b + c )$ and $a \cdot b + a \cdot c$ have the same solution:
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$$ 2 \cdot (5 + 2) = 14 $$
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$$ 2 \cdot 5 + 2 \cdot 2 =14 $$
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When we substitute $a$ with $2$ (the solution) we arrive at a true statement
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(the assertion that arrangement of values results in $14$). Since both
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expressions have the same solution they are equivalent.
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## Creating equivalent equations
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We can create equivalent equations by adding, subtracting, multiplying and
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dividing the _same quantity_ from both sides of the equation (i.e. either side
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of the $=$ symbol). Adding or subtracting the same quantity from both sides
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(either side of the $=$ ) of the equation results in an equivalent equation.
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### Demonstration with addition
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$$ x - 4 = 3
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$$
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The [solution](Algebra%20key%20terms.md#678811) to this equation is $7$
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$$ x -
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4 (+4) = 3 (+ 4) $$
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Here we have added $4$ to each side of the equation. If $x = 7$ then:
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$$ 7 - 4 (+ 4) = 7 $$
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and:
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$$ 3 + 4 = 7 $$
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### Demonstration with subtraction
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$$ x + 4 = 9
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$$
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The [solution](Algebra%20key%20terms.md#678811) to this equation is $5$.
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$$ x +
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4 (-4) = 9(-4) $$
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Here we have subtracted $4$ from each side of the equation. If $x = 5$ then:
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$$ 5 + 4 (-4) = 5 $$
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and
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$$ 9 - 4 = 5 $$
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### Demonstration with multiplication
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$$x \cdot 2 = 10  $$
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The [solution](Algebra%20key%20terms.md#678811) to this equation is $5$.
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$$
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(x \cdot 2) \cdot 3 = 10 \cdot 3 $$ Here we have multiplied each side of the
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equation by $3$. If $x =5$ then
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$$ (5 \cdot 2) \cdot 3 = 30$$
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$$ 10 \cdot 3 = 30$$
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### Demonstration with division
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$$x \cdot 3 = 18  $$
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The [solution](Algebra%20key%20terms.md#678811) to this equation is $6$.
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$$\frac{x
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\cdot 3}{3} = \frac{18}{3}
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$$
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Here we have divided each side of the equation by $3$. If $x$ is 6, then
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$$\frac{6
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\cdot 3}{3} = 6$$
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$$\frac{18}{3} = 6 $$
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