I:2x – y + z = 7
II:x + 2y – 5z = -1
III:x – y = 6
you can first use III and substitute x or y to eliminate it in I and II (in this case x):
III: x=6+y
-> substitute x in I and II:
I': 2*(6+y)-y+z=7
12+2y-y+z=7
y+z=-5
II':(6+y)+2y-5z=-1
3y+6-5z=-1
3y-5z=-7
then you can subtract II' from 3*I' to eliminate y:
3*I'=3y+3z=-15
3*I'-II':
3y+3z-(3y-5z)=-15-(-7)
8z=-8
z=-1
insert z in II' to calculate y:
3y-5z=-7
3y+5=-7
3y=-12
y=-4
insert y into III to calculate x:
x-(-4)=6
x+4=6
x=2
so the solution is
x=2
y=-4
z=-1
4/(1+3) or
Addition is the first step for two reasons.
1) the Order of Operations requires you do operations in parentheses first
2) the division cannot be performed until you know what you're dividing
Step-by-step explanation:
1. 1*1/6= 1 /6=1.67
2. 9*7/10= 63/10= 6.3
3. 7*4/8= 28/8= 3.5
4. 1/2 of 2= 1/2*2= 0.5*2= 1
5. 1/12 of 2= 1/12*2= 8.33*2= 16.66
6.2/6 of 2= 2/6*2= 3.33*2= 6.66
7. 1/3 of 5= 1/3*5= 3.33*5= 16.65
8. 3/10 of 8= 3/10*8= 0.3*8= 2.4
Hope it's help you ♥️
She needs 8 more dollars.
if you multiply 3 times 5 equals 15
and 4 times 3 equals 12
then add that which is 27
35 minus 27 is 8.
