Two positive numbers have a difference of 21. The larger number is five more than twice the smaller. Find the two numbers.

1 answer:

6 0

Let n = the smaller of the two numbers, and since the other number is 5 more than twice the smaller number n, then ...

Let 2n + 5 = the second and larger number.

Since the sum of the two unknown numbers is 26, then we can write the following equation to be solved for n as follows:
n + (2n + 5) = 26

n + 2n + 5 = 26

Collecting like-terms on the left, we get:
3n + 5 = 26

3n + 5 - 5 = 26 - 5

3n + 0 = 21

3n = 21

(3n)/3 = 21/3

(3/3)n = 21/3

(1)n = 7

n = 7

Therefore, ...
2n + 5 = 2(7) + 5
= 14 + 5
= 19

CHECK:
n + (2n + 5) = 26
7 + (19) = 26
7 + 19 = 26
26 = 26

Therefore, the two desired numbers whose sum is 26 are indeed 7 and 19.

Hope this helps :)

You might be interested in

find the zeros of 2x^2 - 16x + 27 using the quadratic formula. be sure to simplify the expression. Can someone please show me ho

____ [38]

For a quadratic of the form $f(x)=ax^2+bx+c$ , we have the quadratic formula
$x=\dfrac{-b \pm \sqrt{b^2 -4ac} }{2a}$ ,
where a is the coefficient (number before the variable) of the squared term, b is the coefficient of the linear term, and c is the constant term.

So, given $2x^2-16x+27=0$ , we can get that $a=2, \ b=-16$ , and $c=27$ . We substitute these numbers into the quadratic formula above.

$x=\dfrac{-(-16) \pm \sqrt{(-16)^2 -4(2)(27)} }{2(2)}$

$x=\dfrac{16 \pm \sqrt{(256 -216)} }{4}$

$x=\dfrac{16 \pm \sqrt{40} }{4}$

$x=\dfrac{16 \pm 2\sqrt{10} }{4}$

$x=4+ \frac{\sqrt{10}}{2}, \ x=4- \frac{\sqrt{10}}{2}$

This is our final answer.

If you've never seen the quadratic formula, you can derive it by completing the square for the general form of a quadratic. Note that the $\pm$ symbol (read: plus or minus) represents the two possible distinct solutions, except for zero under the radical, which gives only one solution.