All categories

4 years ago

The product of two whole numbers is 416 and their sum is 42. What are the two numbers?

Mathematics

1 answer:

olga55 [171]4 years ago

7 0

Answer:

The two numbers are <em>16</em> and <em>26</em>.

Step-by-step explanation:

We can solve this question using 2 simultaneous equations based on the given information from the question.

Let number 1 = x

Let number 2 = y

xy = 416 -> ( 1 )

x + y = 42 -> ( 2 )

We can use either substitution or elimination to solve simultaneous equations. For this question, we will use substitution as it is the easier and shorter option.

Make y the subject in ( 2 ):

x + y = 42 -> ( 2 )

y = 42 - x -> ( 3 )

Substitute ( 3 ) into ( 1 ):

xy = 416 -> ( 1 )

x ( 42 - x ) = 416

42x - x^2 = 416

-x^2 + 42x - 416 = 0

- [ x^2 - 42x + 416 ] = 0

- [ x^2 - 16x - 26x + 416 ] = 0

- [ x ( x - 16 ) - 26 ( x - 16 ) ] = 0

- ( x - 16 ) ( x - 26 ) = 0

x = 16 -> ( 4 ) , x = 26 -> ( 5 )

Substitute ( 4 ) into ( 3 ):

y = 42 - x -> ( 3 )

y = 42 - ( 16 )

y = 26

Substitute ( 5 ) into ( 3 ):

y = 42 - x -> ( 3 )

y = 42 - ( 26 )

y = 16

Therefore:

x = 16 , y = 26

x = 26 , y = 16

The two numbers are 16 and 26.

You might be interested in

Let us have four distinct collinear points $a,$ $b,$ $c,$ and $d$ on the cartesian plane. the point $c$ is such that $\dfrac{ab}

Deffense [45]

Start with a line segment connecting two points, A and B. $\dfrac{DA}{BA}=3$ means DA is 3 times longer than BA. Clearly, D cannot fall between A and B because that would mean DA is shorter than BA. So there are two possible locations where D can be placed on the line relative to A and B.

But with $\dfrac{DB}{BA}=2$ , or the fact that DB is 2 times longer than BA, we can rule out one of these positions; referring to the attachment, if we place D to the left of A, then DB would be 4 times longer than BA.

Finally, $\dfrac{AB}{CB}=\dfrac12$ , so that CB is 2 times longer than AB. Again we have two possible locations for point C (it cannot fall between A and B), but one of them forces C to occupy the same point as D. However, A, B, C, D are distinct, so C must fall to the left of A.

Now let $d$ be the length of AB. Then the length of CD in terms of $d$ is $4d$ . We have the coordinates of C and D, and the distance between them is $\sqrt{(4-0)^2+(0-4)^2}=4\sqrt2$ . So

$4d=4\sqrt2\implies d=\sqrt2$

The slope of the line through C and D is

$\dfrac{0-4}{4-0}=-1$

and so the equation of the line through these points is

$y-4=-(x-0)\implies x+y=4$

So the coordinates of A are $(x,y)=(x,4-x)$ . The distance between C and A is $d=\sqrt2$ , so we have

$\sqrt{(x-0)^2+(4-x-4)^2}=\sqrt{2x^2}=|x|\sqrt2=\sqrt2\implies|x|=1$

Since A falls to the right of C (in the $x,y$ plane, not just in the sketch), we know to take the positive value $x=1$ . Then the $y$ coordinate is $y=4-1=3$ .