Two numbers that have a sum of 28 and a product of 160. What are the two numbers?
x+y=28 ---->y = 28-x
xy=160
x(28-x) = 160
-x^2 + 28x - 160 = 0
x^2 - 28x + 160 = 0
(x-20)(x-8) = 0
x = 20 or x = 8
if x=20 or x=8
if x = 20 , y = 8
if x = 8 , y = 20
the two numbers are 20 and 8
To find the two numbers, we can set up a system of equations based on the given information.
Let's call the two numbers x and y.
From the given information, we have two pieces of information:
1. The sum of the two numbers is 28: x + y = 28
2. The product of the two numbers is 160: xy = 160
Now we can solve this system of equations.
We can start by using the first equation to solve for one variable in terms of the other. Let's solve for x by subtracting y from both sides of the equation:
x = 28 - y
Next, we can substitute this expression for x into the second equation:
(28 - y)y = 160
Multiplying the terms out, we have:
28y - y^2 = 160
Rearranging the equation in standard quadratic form:
y^2 - 28y + 160 = 0
Now we can factor this quadratic equation:
(y - 20)(y - 8) = 0
Setting each factor equal to zero, we get:
y - 20 = 0 or y - 8 = 0
Solving for y, we have two solutions:
y = 20 or y = 8
Now we can substitute these values of y back into the expression for x to find the corresponding values of x:
For y = 20:
x = 28 - y
x = 28 - 20
x = 8
For y = 8:
x = 28 - y
x = 28 - 8
x = 20
Therefore, the two numbers are 8 and 20.
To find the two numbers, we can use algebraic equations. Let's assume the two numbers are x and y.
We are given two pieces of information:
1. The sum of the two numbers is 28, so the equation is: x + y = 28.
2. The product of the two numbers is 160, so the equation is: x * y = 160.
We have a system of two equations with two variables. To solve this system, we can use substitution or elimination.
Let's solve by substitution:
1. Rearrange the equation x + y = 28 to solve for x in terms of y: x = 28 - y.
2. Substitute this expression for x in the second equation: (28 - y) * y = 160.
3. Simplify the equation: 28y - y^2 = 160.
4. Rearrange the equation by moving all terms to one side: y^2 - 28y + 160 = 0.
5. Factorize the equation: (y - 20)(y - 8) = 0.
6. Set each factor equal to zero and solve for y:
y - 20 = 0, so y = 20.
y - 8 = 0, so y = 8.
Now that we have the solutions for y, substitute them back into the original equation x + y = 28 to find x:
For y = 20: x + 20 = 28, x = 8.
For y = 8: x + 8 = 28, x = 20.
Therefore, the two numbers are 8 and 20.