half of the sum of two number is 7,while one third of their product is 16,find the numbers
x+y)/2 = 7
xy/3 = 16
y=48/x
x+48/x = 14
x^2-14x+48 = 0
(x-6)(x-8) =0
The numbers are 6 and 8
That's all
(x+y)/2 = 7
xy/3 = 16
y=48/x
x+48/x = 14
x^2-14x+48 = 0
(x-6)(x-8) = 0
x and y are 6 and 8
Thanks
thanks for solving my question
1.A ,2.B
To find the two numbers, let's assign variables to represent them. Let's say the first number is 'x' and the second number is 'y'.
According to the given information:
1. "Half of the sum of two numbers is 7", which can be expressed as (x + y)/2 = 7.
2. "One third of their product is 16", which can be expressed as (x * y)/3 = 16.
Let's solve these two equations step by step:
1. From the first equation, we can multiply both sides by 2 to get (x + y) = 14.
2. From the second equation, we can multiply both sides by 3 to get (x * y) = 48.
Now, we have a system of equations:
x + y = 14 (Equation 1)
x * y = 48 (Equation 2)
To solve this system, we can use substitution or elimination. Let's solve it using substitution:
1. Rearrange Equation 1 to get x = 14 - y.
2. Substitute this value of x into Equation 2: (14 - y) * y = 48.
3. Simplify the equation: 14y - y^2 = 48.
4. Rewrite the equation in the standard quadratic form: y^2 - 14y + 48 = 0.
5. Factor the quadratic equation to get (y - 6)(y - 8) = 0.
6. Set each factor equal to zero: y - 6 = 0 or y - 8 = 0.
7. Solve for y: y = 6 or y = 8.
Now that we have the possible values for y, let's substitute them back into Equation 1 to find the corresponding values of x:
For y = 6:
x + 6 = 14
x = 14 - 6
x = 8
For y = 8:
x + 8 = 14
x = 14 - 8
x = 6
Therefore, the two numbers could be (6, 8) or (8, 6).