Recall to always convert the mixed fractions to "improper" fractions first,
I got y=2/3x + 10 , i hope that’s right!
Answer:
x=8
y=16
Step-by-step explanation:
2x+2y=48
3x+y=40
Ther is many ways to solve this, one of them is clearance:
WE CLEAR AN UNLOCKED (X) AND REPLACE IT IN THE OTHER
x= (40-y)/3
2[(40-y)/3] +2y=48
(80-2y)/3 +2y=48
(80-2y+6y)/3=48
80+4y=48(3)
4y=144
y= 144/4
y=16
x=(40-y)/3
x=(40-16)/3
x=24/3
x=8
Manipulate the first equation to
y = 22-7z
Then substitute in the second to find z
8(22-7z) + 7z = 127
176-56z+7z = 127
-49z = -49
z = 1
So y = 22-7(1) = 22-7 = 15
y = 15; z = 1
Answer:
Try c am sure it is correct
