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QuestionSimplify these expressions:
1. 3a2−3b24a2+4b2−8ab÷3(b+a)28(a−b)2\frac{3 a^{2}-3 b^{2}}{4 a^{2}+4 b^{2}-8 a b} \div \frac{3(b+a)^{2}}{8(a-b)^{2}}
2. a−3a2−7a+12−a+3a2−9\frac{a-3}{a^{2}-7 a+12}-\frac{a+3}{a^{2}-9}

Studdy Solution

STEP 1

Assumptions1. The variables a and b are real numbers. . For the first expression, the denominator is not equal to zero, i.e., 4a+4b−8ab≠04 a^{}+4 b^{}-8 a b \neq0 and (a−b)≠0(a-b)^{} \neq0.
3. For the second expression, the denominators are not equal to zero, i.e., a−7a+12≠0a^{}-7 a+12 \neq0 and a−9≠0a^{}-9 \neq0.

STEP 2

Let's start with the first expression. We can simplify division of fractions by multiplying the first fraction by the reciprocal of the second fraction.
a2−b24a2+4b2−8ab÷(b+a)28(a−b)2=a2−b24a2+4b2−8ab×8(a−b)2(b+a)2\frac{ a^{2}- b^{2}}{4 a^{2}+4 b^{2}-8 a b} \div \frac{(b+a)^{2}}{8(a-b)^{2}} = \frac{ a^{2}- b^{2}}{4 a^{2}+4 b^{2}-8 a b} \times \frac{8(a-b)^{2}}{(b+a)^{2}}

STEP 3

Now, we can simplify the expressions in the numerator and denominator. Notice that 3a2−3b23 a^{2}-3 b^{2} is the difference of squares and can be written as 3(a−b)(a+b)3(a-b)(a+b). Similarly, a2+b2−8ab a^{2}+ b^{2}-8 a b is the square of a binomial and can be written as (a−b)2(a-b)^{2}.
3a2−3b2a2+b2−8ab×8(a−b)23(b+a)2=3(a−b)(a+b)(a−b)2×8(a−b)23(b+a)2\frac{3 a^{2}-3 b^{2}}{ a^{2}+ b^{2}-8 a b} \times \frac{8(a-b)^{2}}{3(b+a)^{2}} = \frac{3(a-b)(a+b)}{(a-b)^{2}} \times \frac{8(a-b)^{2}}{3(b+a)^{2}}

STEP 4

implify the expression by cancelling out the common terms in the numerator and denominator.
3(a−b)(a+b)4(a−b)2×8(a−b)23(b+a)2=2(a+b)(b+a)\frac{3(a-b)(a+b)}{4(a-b)^{2}} \times \frac{8(a-b)^{2}}{3(b+a)^{2}} = \frac{2(a+b)}{(b+a)}

STEP 5

Further simplify the expression by cancelling out the common terms in the numerator and denominator.
2(a+b)(b+a)=2\frac{2(a+b)}{(b+a)} =2

STEP 6

Now, let's move on to the second expression. Notice that a2−a+12a^{2}- a+12 can be factored into (a−3)(a−4)(a-3)(a-4) and a2−9a^{2}-9 can be factored into (a−3)(a+3)(a-3)(a+3).
a−3a2−a+12−a+3a2−9=a−3(a−3)(a−4)−a+3(a−3)(a+3)\frac{a-3}{a^{2}- a+12}-\frac{a+3}{a^{2}-9} = \frac{a-3}{(a-3)(a-4)}-\frac{a+3}{(a-3)(a+3)}

STEP 7

implify the expression by cancelling out the common terms in the numerator and denominator.
a−3(a−3)(a−4)−a+3(a−3)(a+3)=1a−4−1a+3\frac{a-3}{(a-3)(a-4)}-\frac{a+3}{(a-3)(a+3)} = \frac{1}{a-4}-\frac{1}{a+3}

STEP 8

Now, we can combine these two fractions into one by finding a common denominator, which is (a−4)(a+3)(a-4)(a+3).
1a−4−1a+3=a+3−(a−4)(a−4)(a+3)\frac{1}{a-4}-\frac{1}{a+3} = \frac{a+3-(a-4)}{(a-4)(a+3)}

STEP 9

implify the numerator.
a+3−(a−4)(a−4)(a+3)=7(a−4)(a+3)\frac{a+3-(a-4)}{(a-4)(a+3)} = \frac{7}{(a-4)(a+3)}So, the simplified forms of the given expressions are2 and 7(a−4)(a+3)\frac{7}{(a-4)(a+3)} respectively.

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