Question 79185This question is from textbook College Algebra
: How do I solve this? I've tried by susbtitution and elimination without success.
(1/3)x - (3/2)y = -5
(3/4)x + (1/3)y = 11
Thank you! This question is from textbook College Algebra
You can put this solution on YOUR website! sometimes, eliminating the fractions will make things easier ... multiplying the first equation by 6 gives 2x-9y=-30
multiplying the second equation by 12 gives 9x+4y=132 ... multiplying again by 9 gives 81x+36y=1188
multiplying the first equation again by 4 gives 8x-36y=-120 ... adding equations gives 89x=1068 , so x=12 and y=6
You should clear of fractions first.
Clear the first of fractions by multiplying
through by :
2x - 9y = -30
Clear the second of fractions by multiplying
through by :
9x + 4y = 132
So we have the system
2x - 9y = -30
9x + 4y = 132
The coefficients of y are -9 and 4. The LCD of
their absolute values is 36, so
1. We need to make the -9 in the first into a -36
by multiplyng the whole first equation through by 4.
2. We need to make the 4 in the second into a 36
by multiplyng the whole second equation through by 9.
(4)2x - (4)9y = (4)(-30)
(9)9x + (9)4y = (9)(132)
which simplified becomes:
8x - 36y = -120
81x + 36y = 1188
Add term by term vertically
8x - 36y = -120
81x + 36y = 1188
------------------
89x = 1068
Divide through by 89
x = 1068/89
x = 12
Sunstitute 12 for x in
2x - 9y = -30
2(12) - 9y = -30
24 - 9y = -30
-9y = -30 - 24
-9y = -54
y = (-54)/(-9)
y = 6
So the solution is (x,y) = (12,6).
Edwin