Precal final review solve

Question

2 equations That are giving me difficulty.

 

Solve

a)  (2x+3)/(x+5)  +  (1)/(2)   = (7)/(2x+10)

 

 

 

 

b)(x²+4x+3)/(x²+5x+6)     +   (x²-5x+6)/(x²-3x+2)   =  (x²+1x-6)/(x²+1x-2)

Answer 1

For these problems, we'll want to find common denominators. First, we'll factor everything. Then, we'll change all denominators to the common denominator and solve.

 

 

 

(a)

 

The denominators are x+5, 2, and 2x+10 = 2(x+5). The common denominator is therefore 2(x+5). Multiply the first fraction by 2/2 and the second fraction by (x+5)/(x+5):

 

(2(2x+3)) / (2(2x+5)) + (x+5) / (2(2x+5)) = 7 / (2(2x+5))

(4x+6) / (2(2x+5)) + (x+5) / (2(2x+5)) = 7 / (2(2x+5))

 

Combine the left-hand fractions into one:

 

(4x+6+x+5) / (2(2x+5)) = 7 / (2(2x+5))

(5x+11) / (2(2x+5)) = 7 / (2(2x+5))

 

Multiply both sides by 2(2x+5)) to get rid of the denominators:

 

5x+11 = 7

5x = -4

x = -4/5

 

 

 

(b)

 

Factoring everything gives us:

 

((x+3)(x+1)) / ((x+3)(x+2)) + ((x-3)(x-2)) / ((x-2)(x-1)) = ((x+3)(x-2)) / ((x+2)(x-1))

(x+1) / (x+2) + (x-3) / (x-1) = ((x+3)(x-2)) / ((x+2)(x-1))

 

The denominators are (x+2), (x-1), and (x+2)(x-1). The common denominator is therefore (x+2)(x-1). Multiply the first fraction by (x-1)/(x-1) and the second fraction by (x+2)/(x+2):

 

((x+1)(x-1)) / ((x+2)(x-1)) + ((x-3)(x+2)) / ((x+2)(x-1)) = ((x+3)(x-2)) / ((x+2)(x-1))

 

Combine the left-hand fractions into one:

 

((x+1)(x-1) + (x-3)(x+2)) / ((x+2)(x-1)) = ((x+3)(x-2)) / ((x+2)(x-1))

(x² + x - x - 1 + x² + 2x - 3x - 6) / ((x+2)(x-1)) = ((x+3)(x-2)) / ((x+2)(x-1))

(2x² - x - 7) / ((x+2)(x-1)) = ((x+3)(x-2)) / ((x+2)(x-1))

 

Multiply both sides by (x+2)(x-1) to get rid of the denominators:

 

2x² - x - 7 = x² + x - 6

x² - 2x - 1 = 0

 

Use the quadratic formula to find that x = 1±√(2).

Answer 2

Here's a) for you.

(2x+3)/(x+5)+1/2=7/(2x+10)

(2x+3)/(x+5)+(x+5)/2(x+5)=7/(2x+10)

2(2x+3)/2(x+5)+(x+5)/2(x+5)=7/2(x+5)

[2(2x+3)+(x+5)]/2(x+5)=7/2(x+5)

if the denominators are equal the numerators must be equal

2(2x+3)+(x+5)=7

4x+6+x+5=7

5x+11=7

5x=7-11

5x=-4

x=-4/5