The sum of two numbers is 42. One number is 2 times as large as the other. What are the numbers

Question

Find the larger and smaller number.

Answer 1

Hey Mason. Let one of these numbers be x, and the other be y. Without loss of generality, we see that

x + y = 42 and 2x = y, since the numbers sum to 42, and one of them is twice the other. Substituting 2x for y in the first equation, we see that x + 2x = 42 ⇒ 3x = 42 ⇒ x = 16. But y = 2x = 2(16) = 32. So the two numbers are 16 and 32.

In terms of linear algebra, this can be represented as a system of two linear equations:

x + y = 42

2x - y = 0.

Hope this helps!

Answer 2

Let the two numbers be x and y.

x + y = 42 is one equation.

y = 2x is the other equation, and putting both variables on one side of the equation gives us 2x - y = 0.

This is a system of linear equations:

x + y = 42

2x - y = 0

Let M be this matrix:

1 1

2 -1

Let A be this column vector:

x

y

Let B be the column vector

42

0

Then we get this matrix equation:

MA = B.

If the determinant of M is not zero, then M has an inverse matrix M^-1, and we can our matrix equation on the left by M^-1, giving us

M^-1MA = M^-1B

But M^-1M = I, so

A = M^-1B

So, what is the determinant of M? M =

1 1

2 -1

so det(M) = 1*(-1) - 2*1 = -1 - 2 = -3 ≠ 0

Thus, a unique inverse of M exists. Let's find it using Cramer's rule.

1 1 | 1 0

2 -1 | 0 1

Adding row 1 to row 2,

1 1 | 1 0

3 0 | 1 1

Subtracting 1/3 of row 2 from row 1,

0 1 | 2/3 -1/3

3 0 | 1 1

Multiplying row 2 by 1/3

0 1 | 2/3 -1/3

1 0 | 1/3 1/3

Interchanging the two rows,

1 0 | 1/3 1/3

0 1 | 2/3 -1/3

Thus, M^-1 is

1/3 1/3

2/3 -1/3

Let's check our work. M^-1M =

1/3 1/3 _x 1 1 ₌ 1 0

2/3 -1/3 2 -1 0 1

Our work checks out. Then A = M^-1B =

1/3 1/3 _x 42 ₌ 42/3 ₌ 14 ₌ x

2/3 -1/3 0 2*42/3 28 y

So x = 14 and y = 28. Checking our solution, substituting into our two equations:

14 + 28 = 42

2*14 - 28 = 0

Thus, x = 14 and y = 28.

There are easier ways of doing this problem, but this approach will work in any number of dimensions, not just in 2 dimensions as in this problem.