Intermediate algebra problems..dividing fractions with variables??

Question

Im having so much trouble with this stuff. I know in class he said to find the common denominator and multiply everything else to get that. But how would I do this?? Oh and those long lines are supposed to be division --> ____

1. 36a^2 + 24 -3
_____ ____ ___
b^2c bc^3 7bc


2. 1 - 1
__ __
x 5
___________
1 + 1
__ __
x+3 x
(thats supposed to be 1/x - 1/5 divided by 1/x+3 + 1/x)

3. This one i cant figure out..i thought move x to one side?
Solve ax+b=cx+ d for x

That first problem got messed up but is supposed to be 36a^2/b^2c + 24?bc^3 - 3/7bc

Answer 1

(1)
36a^2/b^2c + 24/bc^3 - 3/7bc

I guess the lowest common denominator is 7b^2c^3
cos you have a b^2c and a bc^3 and a 7bc

So... the first one you need to multiply by 7bc^2, the next one gets multiplied by 7b and the last one needs an extra c^2

((36a^2.7bc^2) + (24.7b) - (3c^2)) / 7b^2c^3

(2)
(1/x - 1/5) / (1/(x+3) + 1/x)
Okay - so you work out what the top fraction is going to be and then you work out what the bottom fraction is going to be, and then you turn the bottom fraction upside down and multiply by it (like how dividing by 1/2 is the same as multiplying by 2)

((5-x)/5x) / ((x+x+3)/(x(x+3))

= (5-x)/5x X x(x+3)/(2x + 3)

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

(3)
ax + b = cx + d

Then
ax = cx + d - b if you subtract b from both sides

Likewise
ax - cx = d - b if you now subtract cx from both sides

and to go further
x(a-c) = d-b

And so now you can get x all by itself by dividing both sides by (a-c)

Answer 2

#1:

... Okay, first of all you need to figure out what the lowest common denominator would be. For this problem, it would be 7b²c³. Now multiply each term top and bottom by an amount that would get you to this number. For the 1st term, multiply in 7c², for the 2nd 7b, and for the 3rd bc². It should end up being...

(252a²c² + 168b - 3bc²)/7b²c³

#2:

... We'll treat the fraction in the numerator separately from the one in the denominator. On top, the LCD is 5x, so that becomes...

(5-x)/5x

... On the bottom, the LCD is x(x + 3), so that becomes...

(x + (x + 3))/x(x + 3)

... Or just simply...

(2x + 3)/x(x + 3)

... Dividing by a number is the same as multiplying by its reciprocal, so we can now put it together as...

(5-x)(x)(x + 3)/5x(2x + 3)

... the x cancels out, leaving...

(5-x)(x + 3)/5(2x + 3)

#3:

... Solving for x, I'll subtract (cx - b) from both sides, leaving...

ax - cx = d - b

x(a - c) = d - b

x = (d - b)/(a - c)

Answer 3

the answer's undemanding, yet no longer so undemanding as subtracting 5 from correct and bottom. first, flow (u+5) to the RHS, which elements us 5 = -2(u+5) defactorizing the equation, 5 = -2u-10 Rearrange it, 15=-2u And for this reason in the top you've, u = -15/2 = -7.5, no longer -2.