math help please?

Question

I admit, I really blow at math. Please help me out and work these problems out for me. Pretty please? Thanks.

Given that jmx=1/dq, solve for d

Given that 1=q/asuz, solve for s

Given that 1/y=1/bhjz. solve for j.


solve for x

13/9=4/3x + 9/27

13/28 = 6/4x + 20/16

6/x = 84/48 - 46/36

23/42 = 41/24 + 13/5x

13/24 = 54/32 + 6/2x

Answer 1

That was fun.. haven't done that in years.. funny how algebra is never useful once your out of school..
Changing Fractions to decimal.

13/9=4/3x + 9/27
4 *13/9 = 4 *4/3x+.33
4*1.44=3x+.33
5.76=3x+.33
3/5.76=3x/3+.33
.33-1.92=x+.33-.33
Solution is 1.62=x
___________________________________________

13/28=6/4x+20/16
1.25-.46=1.5x+1.25-1.25
1.5/.79=1.5x/1.5
Solution is 1.89=x
____________________________________________

6/x=84/48-46/36
6/x=1.75-1.27
6*6x=.48*6
Solution is: x=2.88
____________________________________________

23/42=41/24+13/5x
1.70-.54=1.70+13/5x-1.70
1.16/13.5=13/5x/13.5
2.60/1.16=x
Solution is 2.24 = x
_____________________________________________

13/24=54/32+6/2x
1.68-.54=1.68+6/2x-1.68
6/2/1.14=6/2x/6/2
Solution is 3=x


Hey everyone needs a break in math.. especially when your eyes start to swim in numbers!
Good luck.. Back check the answer by plugging the solution in as your x and check the work :)

Answer 2

Alright, I'm not going to solve all these problems for you, but I will tell you how to do them yourself.

For the first set:
Invert both sides, then divide both sides but the numerator coefficients and multiply both side by the denominator coefficients.

For instance; 1 = q/asuz
Invert both sides: 1 = asuz/q (1 is the same inverted)
Divide both sides by the numerator coefficients: 1/auz = s/q
Multiply both side by the denominator coefficients: q/auz = s

For the second set;
Isolate the term involving x, evaluate the numerical parts, then invert both sides, and divide both sides by x's coefficient.
For instance: 13/9 = 4/3x + 9/27
Isolate x: 13/9 - 9/27 = 4/3x
Evaluate fractions: 48/27 = 4/3x
Invert both sides: 27/48 = 3x/4
Divide both sides by x's coefficient: 27/48 * 4/3 = x
108/144 = x
Simplify: 3/4 = x

See...easy.

Answer 3

solve for d, given that jmx = 1/dq
working: (mutiply both sides by dq): dqjmx = 1
(divide both sides by qjnx): d=1/qjmx


solve for s, given that 1 = q/asuz
working: (multiply both sides by asuz): asuz = q
(divide both sides by auz): s = q/auz

solve j, given that 1/y = 1/bhjz
(multiply both sides by bhjz): bhjz/y = 1
(muitiply both sides by y):bhjz = y
(divide both sides by bhz): j = y/bhz

solve for x:
13/9 = 4/3x + 9/27
13/9 - 9/27 = 4/3x
10/9 = 4/3x
(muitiply both sides by 3x)
3x(10/9) = 4
30x/9 = 4
30x = 36
5x = 6
x = 6/5


13/28 = 6/4x + 20/16
13/28 - 20/16 = 6/4x
-352/448 = 6/4x
(multiply both sides by 4x)
4x(-352/448) = 6
-1408x/448 = 6
-1408x = 2688
x = 2688/-1408

hopefully you should find the last three simple following the other methods I have used.

Answer 4

i do one and explain

you can multiply both sides with same amount,
you can divide both sides by same amount,
you can add to both sides same amount,
you can subtract from both sides the same amount,

if you do that the equation = doesnt change
for iinstance
3 = 3
3*5 = 3*5
3 + 2 = 3+2

With characters it isnt different ...

jmx=1/dq
jmxq = 1/d ( multiply with q )
jmxqd = 1/1 ( multiply with d )
d = 1/(jmxq) ( divide by jmxq )
DONE.

Answer 5

This is obviously a school assignment in which you should do yourself. I does people no good to skate through life by having everyone fix their problems for them. You should study and answer the questions your self.

Answer 6

use this: