Two integers with a product of +30 and a sum of -13?

Answer 1

Integer 1 is x
Interger 2 is y

xy = 30
x + y = -13

Solve the sum for one variable...
x = -13 - y

Put that in the product one.

xy = 30
(-13 - y)(y) = 30
-13y - y^2 = 30
0 = y^2 + 13y + 30
0 = (y + 3)(y + 10)
y = -3 or -10
So, x is also -3 or -10.

So... Your two integers are (-3) and (-10)

Answer 2

this appears to be a simple substitution problem; in other words two equations and two "unknowns".

The two unknowns are the two integers. Call one "X" and one "Y". The problem states the product is "30" and sum is -13. Therefore:

X*Y = 30

and

X + Y = -13

Next: substitute

if X + Y = -13 then Y = -13 - X

X(-13 - X) = 30
-X^2 -13X -30 = 0

the equation is now in the form "AX^2 + BX + C = 0" and the answer is:

X = (-B +/- SQRT(B^2 - 4AC))/2A

(13 +/- SQRT(49)) / -2

X = (13 + 7)/-2 and (13 - 7)/ -2
X = -10 and X = -3

Y = -13 - X so

Y = -13 - (-10) = -3
Y = -13 - (-3) = -10

I suppose the answer could be stated:

if X = -13 then Y = -3
or
if X = -3 then Y = -10

What makes this problem interesting is that, technically it has 4 answers but practically speaking, one only can use two of them at a time.

Answer 3

Let the 2 intergers be x and y.

So x + y = -13
This means x = -13 - y

Also xy = 30
Substituting, (-13 - y) (y) = 30
Expanding, -13y - y^2 = 30
Bring to right side, y^2 + 13y + 30 = 0
Factorizing, (y + 10) (y + 3) = 0
So y = -3 or -10

If y = -3, x = -13 - (-3) = -10
If y = -10, x = -13 - (-10) = -3

So the two integers are -3 and -10

Answer 4

xy = 30- - - - - - - - -Equation 1
x + y = - 13- - - - - -Equation 2
- - - - - - - - - -

Substitute method equation 2

x + y = - 13

x + y - x = - x - 13

y = - x - 13

Substitute the y value into equaion 1

- - - - - - - - - - - - - -

xy = 30

x(- x - 13) = 30

- x² - 13x = 30

- x² - 13x - 30 = 30 - 30

- x² - 13x - 30 = 0

- ( - 1)(x²) - (- 1)(13x) - (- 1)(30) = 0

- ( - x²) - (- 13x) - (- 30) = 0

x² + 13x + 30 = 0

x² + 10x + 3x + 30 = 0

x(x + 10) + 3(x + 10) = 0

(x + 3)(x + 10)

- - - - - - - - - -

Roots

x + 3 = 0

x + 3 - 3 =0 - 3

x = - 3

Insert the x value into equaton 1

- - - - - - - -

Roots

x + 10 = 0

x + 10 - 10 0 - 10

x = - 10. . . .<=. .not used

- - - - - - -

Equation 1 solving for y

xy = 30

- 3y = 30

- 3y / - 3 = 30 / - 3

y = 30 / - 3

y = - 10

Insert the y value into equation 1

- - - - - - - - - - - - - - - -

Check for equation 1

xy = 30

- 3(- 10) = 30

- (- 30) = 30

30 = 30

- - - - - - - - -

check for equation 2

x + y = - 13

- 3 + (- 10) = - 13

- 3 - 10 = - 13

- 13 = - 13

- - - - - - - - -

The equations balance

The solution set { - 3, - 10 }


- - - - - - - - - -s-

Answer 5

Hello

so we have XY = 30
And X+Y = -13

Lets use substituion ---- x=-13-y

Now plug into first equation: (-13-y)y = 30 === -13y-y^2 = 30.

Simplify: 0 = y^2 +13y + 30

Factor to get: (y+10)(y+3).

Thus the solutions are -10 and -3.

Hope this helps

Answer 6

xy = 30
x + y = - 13
y = - 13 - x = - (13 + x)
- x.(13 + x) = 30
-13x - x² = 30
x² + 13x + 30 = 0
(x + 10).(x + 3) = 0
x = - 10,x = - 3
When x = - 10, y = -3
When x = - 3 , y = - 10
The two integers are - 10 and - 3

Answer 7

-10 and -3

Answer 8

The way to do this problem is:

x + y = -13
x*y=30

Solve the 1st equation for x:
x = -13-y
and substitute into the 2nd equation

(-13-y)*y=30

Solve for y and then substitue the value for y into the 1st equation and find x.

Answer 9

-10, -3

Answer 10

-10 and -3 work nicely.