the sum of two numbers is 38. when 8 is added
to twice one of the number, the result is 5 times
the other number. find the number
x+y=38
8+2x=5y
8*x=2xy/8
8x/8=2y/8 8x/8-5y/8=3xy/8
x+y=38
2x/8-5y/8
40x-16y/64=24/1
40x-16
x+y = 38
8+2x = 5y
Now just solve for x and y
40x-16y=24/x+y=38
40x-16y=0
fom equl
x=31-y
substitute eqn3in 2
40x-16y=0
40(38-y)-16y
78y-38=40=16y
78y-38=40y
-38=40-78y
-38y=-78y
divide both sides by 38
38y/38=78y/38
y=2.05
To solve this problem, we can set up a system of two equations with two variables. Let's call the two numbers x and y.
From the information given, we know that the sum of two numbers is 38. So, the first equation is:
x + y = 38
The second piece of information states that when 8 is added to twice one of the numbers, the result is 5 times the other number. Mathematically, this can be written as:
8 + 2x = 5y
Now we have a system of two equations:
x + y = 38
8 + 2x = 5y
To find the values of x and y that satisfy both equations, we can use the method of substitution or elimination.
Let's solve the system using the method of substitution:
From the first equation, we get:
y = 38 - x
Substituting this expression into the second equation, we get:
8 + 2x = 5(38 - x)
Simplifying the equation:
8 + 2x = 190 - 5x
Rearranging terms:
7x = 182
Dividing both sides by 7:
x = 26
Now that we have the value of x, we can substitute it back into the first equation to find y:
y = 38 - 26
y = 12
Therefore, the two numbers are x = 26 and y = 12.