5(x+3)/6=4+3(x-1)/5 ANSWER: x = 27/7. Could someone please show me the steps as to how this final answer was derived? Thank you.

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- Nov 2nd 2012, 05:55 AMMathClownBasic Linear Algebra Help
5(x+3)/6=4+3(x-1)/5 ANSWER: x = 27/7. Could someone please show me the steps as to how this final answer was derived? Thank you.

- Nov 2nd 2012, 06:23 AMPaulAdrienMauriceDiracRe: Basic Linear Algebra Help
$\displaystyle \frac{5(x+3)}{6}=4+\frac{3(x-1)}{5}$

$\displaystyle \frac{5(x+3)}{6}-\frac{3(x-1)}{5}=4$

$\displaystyle \frac{25(x+3)-18(x-1)}{30}=4$

$\displaystyle 25x+75-18x+18=120$

$\displaystyle 7x+93=120$

$\displaystyle 7x=27$

$\displaystyle x=\frac{27}{7}$ - Nov 2nd 2012, 06:41 AMMathClownRe: Basic Linear Algebra Help
- Nov 2nd 2012, 07:07 AMPaulAdrienMauriceDiracRe: Basic Linear Algebra Help
Sorry, I kind of skipped some parts.

$\displaystyle \frac{5(x+3)}{6}=4+\frac{3(x-1)}{5}$

Now, we have to movie the terms that have variable x in them, to try to isolate it, so we subtract both sides by $\displaystyle \frac{3(x-1)}{5}$:

$\displaystyle \frac{5(x+3)}{6}-\frac{3(x-1)}{5}=4$

Next we find common denominator of these two fractions, which is 30:

$\displaystyle \frac{25(x+3)-18(x-1)}{30}=4$

Next we distribute $\displaystyle 25(x+3)=25x+25*3=25x+75$ and $\displaystyle -18(x-1)=-18x-18(-1)=-18x+18$, and we multiply both sides by 30 to get rid of the denominator on the left side and we get 120 on the right side by multiplying 30 to 4.

$\displaystyle 25x+75-18x+18=120$

Next we subtract 93 to both sides.

$\displaystyle 7x+93=120$

$\displaystyle 7x=27$

...and finally, divide by 7 to both sides.

$\displaystyle x=\frac{27}{7}$

I hope this is more clear. - Nov 2nd 2012, 07:09 AMHallsofIvyRe: Basic Linear Algebra Help
$\displaystyle \frac{5(x+ 3)}{6}= 4+ \frac{3(x- 1)}{5}$

Athough it is really the same thing, instead of getting a common denominator I would**multiply through**by 30:

(30)(5)(x+ 3)/6= 4(30)+ (30)(3)(x- 1)/5 and, since 30/6= 5 and 30/5= 6,

(5)(5)(x+ 3)= 4(30)+ 6(3)(x- 1)

25(x+ 3)= 120+ 18(x- 1).

Now we apply the "distributive law": a(b+ c)= ab+ ac. That is, 25(x+ 3)= 25x+ 25(3) and 18(x- 1)= 18x- 18(1). Of course, 25(3)= 75 so this becoes 25x+ 75= 120+ 18x- 18= (120- 18)+ 18x= 102+ 18x. Now, subtract 18 from both sides and subtract 75 from both sides to get

25x- 18x= 7x= 102- 75= 27. The last step is to divide both sides by 7: (7x)/7= x= 27/7. - Nov 2nd 2012, 08:28 AMDevenoRe: Basic Linear Algebra Help
here is how i would do it:

$\displaystyle \frac{5(x+3)}{6} = 4 + \frac{3(x-1)}{5}$

first i would get the 2nd side to "be one fraction", which means changing 4 to (1)(4) = (5/5)(4) = 20/5:

$\displaystyle \frac{5(x+3)}{6} = \frac{20}{5} + \frac{3(x-1)}{5}$

now we can add the numerators together, and we have:

$\displaystyle \frac{5(x+3)}{6} = \frac{3(x-1) + 20}{5}$

since 3(x-1) = 3x - 3, 3(x-1) + 20 = 3x - 3 + 20 = 3x + 17 (i just took the 3 from the 20: the fewer terms, the better)

$\displaystyle \frac{5(x+3)}{6} = \frac{3x - 17}{5}$

now, "cross-multiply": if a/b = c/d, then ad = bc. why does this work?

b(a/b) = b(c/d) (because a/b = c/d, so multplying them both by b still preserves the equality)

ab/b = bc/d...but ab/b = (a/1)(b/b) = (a/1)(1) = a/1 = a. so:

a = bc/d, and multplying both sides by d:

ad = (bc/d)d = (bc/d)(d/1) = (bc)(d/d) = (bc)(1) = bc, which is what i said was true up there a few lines ago.

so when we cross-multiply, we get:

$\displaystyle 5(5(x+3)) = 6(3x - 17)$ look, no fractions! (i HATE fractions, every sane person should).

we can simplify the left side a bit, before we continue:

$\displaystyle 5(5(x+3)) = 5(5x+15) = 25x + 75$

and we can also simplify the right side, as well:

$\displaystyle 6(3x + 17) = 18x + 102$ (i used a calculator to calculate 6*17...so sue me)

so now we have:

$\displaystyle 25x + 75 = 18x + 102$ this is looking better already.

subtracting 18x from both sides gives us:

$\displaystyle 7x + 75 = 102$

subtracting 75 from both sides gives us:

$\displaystyle 7x = 27$

and finally dividing both sides by 7 gives us:

$\displaystyle x = \frac{27}{7}$

it should comfort you that 3 different people using 3 different methods all give the same answer. we're all probably right. how do you make sure? - Nov 2nd 2012, 08:17 PMMathClownRe: Basic Linear Algebra Help
Thanks for all your help, guys! I will no doubt stumble across a similar question and once again ask for help. :S I've repped everyone in this thread for his useful help.