1. Find the product of the solutions of this equation:
1 / x+1 - 6-x/6x = 0
2. If y is jointly proportional to x and z and y = 4 when x = 8 and z = 1, what is the value of y when x = 24 and z = 3?
For this one I think I'm solving it wrong, because I keep getting the answer as y=12, and that's not even one of my choices.
If you mean
1/(x+1) - (6-x)/(6x) = 0 then
1/(x+1) = (6-x)/(6x)
6x = (6-x)(x+1)
6x = -x^2+5x+6
x^2 + x - 6 = 0
product of roots = -6
y = kxz
4 = k(8)(1)
k = 1/2, so
y = xz/2
y = 24*3/2 = 36
Ohhh I get it now! Thanks alot for clearing it up
1. To find the product of the solutions to the equation
1 / (x + 1) - (6 - x) / (6x) = 0
We need to solve the equation first to find the values of x.
To begin, let's simplify the equation by finding a common denominator for the fractions on both sides:
(6 - x) has the factor of x in the denominator, so we multiply the first fraction by x / x:
[(x / x) / (x + 1)] - [(6 - x) / (6x)] = 0
Next, multiply each term by the common denominator, (x + 1) * (6x), to eliminate the fractions:
x(6 - x) - (x + 1)(6 - x) = 0
Simplify:
6x - x^2 - 6 + x - 6x - x^2 = 0
Combine like terms:
2x^2 - 12 = 0
Divide both sides by 2 to simplify:
x^2 - 6 = 0
Now we have a quadratic equation. To solve for x, we can factor it:
(x - √6)(x + √6) = 0
This gives us two solutions:
x = √6 and x = -√6
Now, to find the product of the solutions, we multiply them:
Product of solutions = (√6) * (-√6) = -6
Therefore, the product of the solutions to the equation is -6.
2. If y is jointly proportional to x and z, we can express this relationship as:
y = k * x * z
where k is the constant of proportionality.
To find the value of k, we can use the given information.
When x = 8, z = 1, and y = 4, we can substitute these values into the equation:
4 = k * 8 * 1
Simplify:
4 = 8k
Divide both sides by 8 to solve for k:
k = 4/8 = 1/2
Now that we have the value of k, we can use it to find y when x = 24 and z = 3:
y = (1/2) * 24 * 3
Simplify:
y = 36
Therefore, when x = 24 and z = 3, the value of y is 36.