Question 154668
Find all values of x satisfying the given conditions.
:
y1 = 10(2x-1), y2 = 2x+1, and y1 is 14 more than 8 times y2.
:
Use some logic here, it says:
y1 = 8(y2) + 14
:
From the 1st two given equations: 
we can substitute 10(2x-1) for y1 and 2x+1 for y2, in the above; therefore:
:
10(2x-1) = 8(2x+1) + 14
Multiply the terms inside the brackets, and solve for x
20x - 10 = 16x + 8 + 14
20x - 10 = 16x + 22
Some basic algebra:
20x - 16x = 22 + 10
4x = 32
x = {{{32/4}}}
x = 8
:
:
y1= 1/x, y2= 1/2x, y3= 1/x-1, and the sum of 3 times y1 and 4 times y2 is the product of 4 and y3(corrected this typo).
:
Write an equation for the statement:
"the sum of 3 times y1 and 4 times y2 is the product of 4 and y3"
3(y1) + 4(y2) = 4(y3)
:
Substitute: (1/x) for y1; (1/2x) for y2; and (1/(x-1)) for y3
{{{3(1/x) + 4(1/(2x)) = 4(1/((x-1)))}}}
which is:
{{{(3/x) + (4/(2x)) = 4/((x-1))}}}
2 will cancel into 4 so we have:
{{{(3/x) + (2/x) = 4/((x-1))}}}
we can add like fractions
{{{5/x = 4/((x-1))}}}
Cross multiply and find x:
4x = 5(x-1)
4x = 5x - 5
4x -5x = -5
-x = -5
x = 5
:
:
did this make sense to you now?