Student B:
1/2 / 4
= 1/2 x 1/4
= 1/8
student A:
2/3 / 6
= 2/3 x 1/6
= 1/9
student C:
4/5/ 8
= 4/5 x 1/8
= 1/10
answer
student B: 1/8 of a mile per minute
student A: 1/9 of a mile per minute
student C: 1/10 of a mile per minute
Answer:
x = 3.25
y =4.75
Step-by-step explanation:
In order to Solve the following system of equations below algebraically using substitution method we say that;
let;
8x - 4y = 7
..................... equation 1
x + y = 8.......................... equation 2
from equation2
x + y = 8.......................... equation 2
x = 8 - y.............................. equation 3
substitute for x in equation 1
8x - 4y = 7
..................... equation 1
8(8-y) - 4y = 7
64-8y-4y=7
64-12y=7
collect the like terms
64-7 = 12y
57= 12y
divide both sides by the coefficient of y which is 12
57/12 = 12y/12
4.75 = y
y =4.75
put y = 4.75 in equation 3
x = 8 - y.............................. equation 3
x = 8 -4.75
x = 3.25
to check if your answer is correct, put the value of x and y in either equation 1 or 2
from equation 2
x + y = 8.......................... equation 2
3.25 + 4.75 =8
8=8.................... proved
Hello from MrBillDoesMath!
Answer:
x = 1/2 (1 +\- i sqrt(23))
Discussion:
x \3x - 2 = (x/3)*x - 2 = (x^2)/3 - 2 (*)
1 \3x - 4 = (1/3)x - 4 (**)
(*) = (**) =>
(x^2)/3 -2 = (1/3)x - 4 => multiply both sides by 3
x^2 - 6 = x - 12 => subtract x from both sides
x^2 -x -6 = -12 => add 12 to both sides
x^2-x +6 = 0
Using the quadratic formula gives:
x = 1/2 (1 +\- i sqrt(23))
Thank you,
MrB
How do you need to write it? Do you mean in word form?
Call the number : x
(1/3)x + 5 = x-6
=> (1/3)x + 5 - (x-6) = 0
=> (1/3)x + 5 - x + 6 = 0
Group (1/3)x with -x, 5 with 6
=> [(1/3)x - x] + (5+6) = 0
=> (-2/3)x + 11 = 0
=> (-2/3)x = -11
=> x = -11 : (-2/3) = 33/2.
Recheck : 1/3 x 33/2 + 5 = 33/6 + 5 = 63/6 = 21/2
33/2 - 21/2 = 12/2 = 6 (21/2 is 6 less than 33/2, satisfied.)
