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Describe the role of axiom systems in algebra? - Answers
An axiom in algebra is the stepping stone to solving equations. In order to solve and equation you know how to use the commutative, associative, distributive, transitive and equalilty axiom to solve the basic steps. For example: if you want an equation in the form y = mx + b, given 6x - 3y = 9 you must subtract 6x from both sides giving: -3y = 9-6x. Then you divide by -3 to get y = -3 + 2x. But the equation is not in the from y = mx + b. So we use the commutative property to switch the -3 + 2x and make it 2x - 3. Now it become y = 2x -3. and it is in the form y = mx + b. This manipulation could not be perfromed unless tahe student knew the commutative property. Once the axiom is know the algebraic manipulations fall into place.
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Describe the role of axiom systems in algebra? - Answers
An axiom in algebra is the stepping stone to solving equations. In order to solve and equation you know how to use the commutative, associative, distributive, transitive and equalilty axiom to solve the basic steps. For example: if you want an equation in the form y = mx + b, given 6x - 3y = 9 you must subtract 6x from both sides giving: -3y = 9-6x. Then you divide by -3 to get y = -3 + 2x. But the equation is not in the from y = mx + b. So we use the commutative property to switch the -3 + 2x and make it 2x - 3. Now it become y = 2x -3. and it is in the form y = mx + b. This manipulation could not be perfromed unless tahe student knew the commutative property. Once the axiom is know the algebraic manipulations fall into place.
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Describe the role of axiom systems in algebra? - Answers
An axiom in algebra is the stepping stone to solving equations. In order to solve and equation you know how to use the commutative, associative, distributive, transitive and equalilty axiom to solve the basic steps. For example: if you want an equation in the form y = mx + b, given 6x - 3y = 9 you must subtract 6x from both sides giving: -3y = 9-6x. Then you divide by -3 to get y = -3 + 2x. But the equation is not in the from y = mx + b. So we use the commutative property to switch the -3 + 2x and make it 2x - 3. Now it become y = 2x -3. and it is in the form y = mx + b. This manipulation could not be perfromed unless tahe student knew the commutative property. Once the axiom is know the algebraic manipulations fall into place.
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