2 x the first number increased by 4 x the 2nd number = 32. The difference of 2 #'s is 10 find the two #'s?

Answer 1

This is a systems of equations.

Let x be the first number
Let y be the second number

The first equation is:

2x + 4y = 32

The second equation is:

x-y =10

So, the easiest way to solve this is to mulitply the second equation by 4 on both sides:

4x - 4y = 40

Now, add both equations, and you get:

6x = 72 or
x = 12

Now, substitute into the second equation:

12 - y = 10 or
y=2

Answer 2

the first number is 12. the second number is 2.

2a + 4b = 32
- 4b - 4b

2a = 32 - 4b
/ 2 / 2

a = 16 - 2b

a - b = 10
+ b + b

a = 10 + b

16 - 2b = 10 + b
+ 2b + 2b

16 = 10 + 3b
- 10 - 10

6 = 3b
/ 3 /3

b = 2

2a + 4(2) =32
- 8 - 8

2a = 24
/ 2 /2

a = 12

Answer 3

Let the 2 numbers be represented by "X" and "Y"
Thus 2x + 4y = 32.....equation 1
x - y =10 .....equation 2
That is a Simultaneous equation
Then multiply eqn 2 by 4 to eliminate y; 4x - 4y = 40...equation 3
2x + 4y = 32
4x - 4y = 40
---------------
6x = 72
x = 72/6
x = 12
Then substitute 12 for x in equation 2
x - y = 10
12 - y = 10
y = 2

thus the numbers are 2 and 12

Answer 4

x=first number
y=second number

Given:
2x + 4y = 32
x - y = 10

From the second equation:
x = 10 + y

Substitute into the first equation and solve for y:
2(10 + y) + 4y = 32
20 + 2y + 4y = 32
6y = 12
y = 2

Substitute into either of the Given equations and solve for x:
x - 2 = 10
x = 12

x = 12, y = 2

Answer 5

2a + 4b = 32
(a + 2b = 16)

|a - b| = 10


a = 12
b = 2