# REASONING WITH PROPERTIES FROM ALGEBRA

## About "Reasoning with properties from algebra"

Reasoning with properties from algebra :

Many properties from algebra concern the equality of real numbers.

Several of these are summarized in the following list.

## Algebraic properties of equality

Let a, b and c be real numbers.

If a = b, then a + c  =  b + c

Subtraction Property :

If a = b, then a - c  =  b - c

Multiplication Property :

If a = b, then ac  =  bc

Division Property :

If a = b and c ≠ 0, then a ÷ c  =  b ÷ c

Reflexive Property :

For any real number a, a = a

Symmetric Property :

If a = b, then b = a

Transitive Property :

If a = b and b = c, then a = c

Substitution Property :

If a = b, then a can be substituted for b in any equation or expression.

Property of equality along with other properties from algebra, such as the distributive property,

a(b+c)  =  ab + ac

can be used to solve equations.

For instance, let us solve the equation given below.

3(x + 2)  =  2x + 8

Apply Distributive property on the left side of the equation.

3x + 6  =  2x + 8

Subtraction property :

Subtract 2x from each side of the equation.

x + 6  =  8

Subtract 6 from each side of the equation.

x  =  2

## Reasoning with properties from algebra - Examples

Example 1 :

Solve 7x - 2  =  4x + 9 and write a reason for each step.

Solution :

Given :

7x - 2  =  4x + 13

Subtract property of equality :

Subtract 4x from each side.

3x - 2  =  13

Addition property of equality :

Add 2 to each side.

3x  =  15

Division property of equality :

Divide both sides by 2.

x  =  5

Example 2 :

Solve 52y - 3(12 + 9y)  =  64 and write a reason for each step.

Solution :

Given :

52y - 3(12 + 9y)  =  64

Distributive Property :

Distribute 3 to 12 and 9y.

52y - 36 - 27y  =  64

Simplify :

25y - 36  =  64

Addition property of equality :

Add 36 to each side.

25y  =  100

Division property of equality :

Divide both sides by 25.

y  =  4

Example 3 :

When we do exercise every day, we should find our target heart rate. This is the rate at which we achieve an effective workout while not placing too much strain on our heart. Our target heart rate r (in beats per minute) can be determined from our age a (in years) using the equation a = 220 - 10r/7.

(i) Solve the formula for r and write a reason for each step.

(ii) Use the result to find the target heart rate for a 16 year old.

(iii) Find the target heart rate for the following ages :

20, 30, 40, 50 and 60

What happens to the target heart rate as a person gets older ?

Solution (i) :

Given :

a = 220 - 10r/7

Addition property of equality :

Add 10r/7 to each side.

a + 10r/7  =  220

Subtraction property of equality :

Subtract a from each side.

10r/7  =  220 - a

Multiplication property of equality :

Multiply both sides by 7/10.

r  =  7/10 ⋅ (220 - a)

Solution (ii) :

To find the target heart rate for a 16 year old, substitute a = 16.

r  =  7/10 ⋅ (220 - 16)

Simplify :

r  =  7/10 ⋅ 204

r  =  1428 / 10

r  =  142.8

The target hear rate for a 16 year old is about 142.8 beats per minute.

Solution (ii) :

To find the target heart rate for a 16 year old, substitute a = 16.

r  =  7/10 ⋅ (220 - 16)

Simplify :

r  =  7/10 ⋅ 204

r  =  1428 / 10

r  =  142.8

The target hear rate for a 16 year old is about 142.8 beats per minute.

Solution (iii) :

The table given below shows the heart rate for ages 20, 30, 40, 50 and 60.

From the above table, it is clear that the target heart rate appears to decrease as a person gets older

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