Alexandra L.

Currently unavailable: for regular students

Aeronautical and Aerospace Engineering (Bachelors) - Leeds University

5.0

30 completed lessons

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#### Ratings & Reviews

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#### Qualifications

Further MathsA-level (A2)A
PhysicsA-level (A2)A
ChemistryA-level (A2)A
MathematicsA-level (A2)A*

#### General Availability

Pre 12pm12-5pmAfter 5pm
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#### Subjects offered

SubjectQualificationPrices
ChemistryA Level£22 /hr
Further MathematicsA Level£22 /hr
MathsA Level£22 /hr
PhysicsA Level£22 /hr
ChemistryGCSE£20 /hr
MathsGCSE£20 /hr

### Simplify 5(x-1)^2/25(x^2-1)

The first thing to notice in this fraction is that the constants 5 and 25 can be simplified by cancelling down straight away. 5/25 = 1/5 The expression thus becomes (x-1)2/5(x2-1).

Next we focus on the bracket contents. Examining the denominator shows us that it can be factorised - it's the difference of two squares! You can tell this because both terms x2 and 1 are perfect squares. Therefore the denominator can be broken up into its factors (x+1)(x-1) and the overall expression now looks like (x-1)2/5(x-1)(x+1).

Because the numerator and denominator both have a factor of (x-1), we can cancel out one of the (x-1) brackets for both the top and bottom. This gives us (x-1)/5(x+1). We know this cannot be simplified further because (x-1) and (x+1) are not multiples of each other and there are no other constants to cancel.

Finally, we can multiply out the brackets in the denominator and we get (x-1)/(5x+5). Be careful that you multiply each term in (x+1) by 5. 5(x+1) is not the same as (5x+1); the 1 needs to be multiplied by 5 as well.

The first thing to notice in this fraction is that the constants 5 and 25 can be simplified by cancelling down straight away. 5/25 = 1/5 The expression thus becomes (x-1)2/5(x2-1).

Next we focus on the bracket contents. Examining the denominator shows us that it can be factorised - it's the difference of two squares! You can tell this because both terms x2 and 1 are perfect squares. Therefore the denominator can be broken up into its factors (x+1)(x-1) and the overall expression now looks like (x-1)2/5(x-1)(x+1).

Because the numerator and denominator both have a factor of (x-1), we can cancel out one of the (x-1) brackets for both the top and bottom. This gives us (x-1)/5(x+1). We know this cannot be simplified further because (x-1) and (x+1) are not multiples of each other and there are no other constants to cancel.

Finally, we can multiply out the brackets in the denominator and we get (x-1)/(5x+5). Be careful that you multiply each term in (x+1) by 5. 5(x+1) is not the same as (5x+1); the 1 needs to be multiplied by 5 as well.

1 year ago

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