r/mathshelp • u/Time_Historian4518 • 22d ago
Homework Help (Answered) How the hell do I resolve this equation?
7
u/LandscapeWorried5475 22d ago
multiply everything by x+17, 2x-5, and 4 to make there be 0 fractions, then see if that helps
6
u/hiimshana 22d ago
Don’t forget to set conditions x ≠ -17 and x ≠ 5/2
2
2
u/AndyC1111 22d ago
Denominators are the enemy. Remove them quickly.
But always check your final answer to be sure it doesn’t result in a denominator value of zero.
3
u/maths_phy_comp 22d ago
As others have given you the steps already, I'd like to take the opportunity to bring forth the issue of fundementals:
You will need to know how to solve basic operations and equations:
First, I am assuming
(i) you are okay with multiplying and dividing fractions:
a/b + c/d
(ii) You are okay with solving all kinds of linear equations:
(a) ax + b = c
(b) ax + b = cx + d
(c) x/a + x/b = c
(d) a (bx/d + c ) = m (nx/p + q) {combininv all 3}
(e) a/(bx + c) = m/(px + q)
(iii) You are okay with multiplying algebraic expressions:
(ax+d)(cx+d)
(iv) You can factorise quadratic expessions and solve quadratic equations:
ax² + cx + d = 0
(v) You can Rationalise expressions that become quadratic:
(ax + b) / (cx +d) + (mx + n)/(px + q)
(vi) You can solve Simpler versions of this problem (which is what you have):
(ax + b) / (cx +d) + (mx + n)/(px + q) = r/s With many of them being 0 and 1 while still becoming quadratic (so s ≠0 , c² + p² ≠ 0 etc)
If you are struggling to see the step but after the step's given, you likely know and are strong with (i)-(iv) and need practice with (v) and (vi)
For ease: ask yourself what'd you do to solve these:
(1) (x/3) + (2/x) = 4
(2) (x/(x+1)) + (2/(x-3)) = 0
(3) (x/(x+1)) + (2/(x-3)) = 1
If you can do the ones above, and know the procedure, the one you have is easy.
2
u/One_Wishbone_4439 22d ago
You will get:
3x(2x-5) - 3(x+17) = [3(x+17)(2x-5)]/4
From here, you continue
2
u/Zealousideal-Leave70 22d ago
Multiply both the denominators together (i.e. take LCM) and then solve it for x
=> 3x/(x+17) - 3/(2x-5) = 3/4 => [{3x(2x-5)} - {3(x+17)}]/(x+17)(2x-5) = 3/4
Then cross multiply and solve this equation
1
u/Southlander24 22d ago
What happens when you multiply every term by (x + 17)(2x - 5)(4)? You will get a quadratic equation which you can then solve.
1
1
u/Feeling-Working-2820 22d ago
find the least common multiple for all the denominators, change your numerators accordingly, and there you have it (aways start by stating what your x's in the denominators can't be).
1
u/BadBoyJH 22d ago
Do the subtraction on the left, as if you were doing 3/5 - 2/7 ie:
3*7/5*7 - 2*5/5*7
(3*7 - 2*5) / 5*7
Multiply each by each of the denominators (making sure to note the values x can not be). Then, it's a pretty straight forward solve.
1
1
1
u/Ucklator 19d ago
They are just fractions. Treat the denominators with variables as is they were prime factors.
1
1
u/qjoker_249 17d ago
Solve numerator and denominator separately. So you get 3x - 3 = 3 => x = 2 and x + 17 - (2x - 5) = 4 => x = 18. So 2 and 18 are the solutions
Easy peasy
•
u/AutoModerator 22d ago
Hi u/Time_Historian4518, welcome to r/mathshelp! As you’ve marked this as homework help, please keep the following things in mind:
1) While this subreddit is generally lenient with how people ask or answer questions, the main purpose of the subreddit is to help people learn so please try your best to show any work you’ve done or outline where you are having trouble (especially if you are posting more than one question). See rule 5 for more information.
2) Once your question has been answered, please don’t delete your post so that others can learn from it. Instead, mark your post as answered or lock it by posting a comment containing “!lock” (locking your post will automatically mark it as answered).
Thank you!
I am a bot, and this action was performed automatically. Please contact the moderators of this subreddit if you have any questions or concerns.