The product of two numbers is 21 and their sum is 10, what are the numbers
If you are looking for whole numbers, I can think of only two numbers so that
3 x 7 = 21, and it so happens that 3+7 = 10
How easy was that?
Eq1: x * y = 21.In Eq1, replace X with
y = 21/x.
Eq2: x + y = 10.
In Eq2, replace Y with 21/x and solve for X:
x + 21/x = 10.
Multiply both sides by X:
x^2 + 21 = 10x,
x^2 - 10x + 21 = 0,
(x-3)(x- 7) = 0,
X = 3, or X = 7.
In Eq1, replace X with 3 and solve for Y:
3 + Y = 10,
Y = 7.
y
= 7.
Note: The first line of my response should be Eq1: x * y = 21.
To solve this problem, we can use a system of equations. Let's represent the two numbers as x and y.
The first equation we can write is: x * y = 21. This equation represents the product of the two numbers.
The second equation is: x + y = 10. This equation represents the sum of the two numbers.
Now, we have a system of two equations with two variables. To solve for x and y, we can use the substitution method.
Rearrange the second equation to solve for one variable. For example, let's solve for x:
x = 10 - y.
Now substitute the value of x in the first equation:
(10 - y) * y = 21.
Multiplying the terms, we get:
10y - y^2 = 21.
Rearrange this equation by moving all the terms to one side:
y^2 - 10y + 21 = 0.
Now we have a quadratic equation. Factor or solve for y by using the quadratic formula:
y = (-b ± √(b^2 - 4ac)) / 2a.
In our case, a = 1, b = -10, and c = 21.
Plugging in these values, we have:
y = (-(-10) ± √((-10)^2 - 4 * 1 * 21)) / (2 * 1).
Simplifying this expression, we get:
y = (10 ± √(100 - 84)) / 2.
y = (10 ± √16) / 2.
Now, evaluate the two possible solutions for y:
y1 = (10 + 4) / 2 = 14 / 2 = 7,
y2 = (10 - 4) / 2 = 6 / 2 = 3.
Now that we have the values for y, substitute them back into the equation x = 10 - y to find x:
For y = 7: x = 10 - 7 = 3.
For y = 3: x = 10 - 3 = 7.
Therefore, the two numbers are 3 and 7.