Question

What two positive numbers whose difference is 7 and whose product is 294

Answer 1

To find two positive numbers whose difference is 7 and whose product is 294, we can set up a system of equations. The numbers are 21 and 14.

Step-by-step explanation:

To find two positive numbers whose difference is 7 and whose product is 294, we can set up a system of equations. Let's call the two numbers x and y. From the given information, we have the equations:

x - y = 7

xy = 294

From the first equation, we can express x in terms of y as x = y + 7. Substituting this into the second equation, we get (y + 7)y = 294. Simplifying, we have y^2 + 7y - 294 = 0. We can factor this equation as (y - 14)(y + 21) = 0. Therefore, either y - 14 = 0 or y + 21 = 0. Solving each equation, we find that y = 14 or y = -21. Since we are looking for positive numbers, the answer is y = 14. Substituting this into x = y + 7, we find x = 14 + 7 = 21. Therefore, the two positive numbers whose difference is 7 and whose product is 294 are 21 and 14.

Answer 2

Step-by-step explanation:

We will use x and y as our positive numbers. We need to write 2 equations from this information. The first is that the difference between the numbers is 7:

x - y = 7 is the equation for that.

The second is that the product of the 2 numbers is 294:

xy = 294

Let's begin by solving the first equation for x:

x = 7 + y

Now sub that in for x in the second equation:

(7 + y)y = 294 and

`7y+y^2=294`

This is a quadratic, so let's put it into standard form, setting it equal to 0:

`y^2+7y-294=0`

We have to factor this now to find out what values of y satisfy this equation.

Think: "what 2 number multiply to equal -294 and add to equal 7. To find these numbers we find all the factors of 294, which are:

1, 294

2, 147

3, 98

6, 49

7, 42

14, 21

It looks like 14 and 21 will work. If we make the 14 negative, then 21 - 14 = 7. And if the 14 is negative, then -14 * 21 = -294. So the signs are correct. The equation rewritten using those values in place of 7y is:

`y^2+21y-14y-294 = 0` and we factor by grouping:

`(y^2+21y)-(14y-294)=0`

Out of the first set of parenthesis we can pull out a y, and from the second set we can pull out a 14:

y(y + 21) - 14(y + 21) = 0

What's common now is the factor (y + 21). So factor THAT out, and you're left with:

(y + 21)(y - 14) = 0

Set each of these expression equal to 0 and solve for y:

y + 21 = 0 so

y = -21 and

y - 14 = 0 so

y = 14

Since we were told that both of the numbers are positive, then y has to be 14.

Sub in y = 14 to find x:

x = 7 + 14 so

x = 21

There you go!