x + y = 20
30x + 25y = 550
Using the first equation, we can find a temporary value of x.
x = 20 - y
Now that we have a value of x, we can plug it into the second equation to find the exact value of y.
30x + 25y = 550
30(20 - y) + 25y = 550
Distributive property.
600 - 30y + 25y = 550
Combine like terms.
600 - 5y = 550
Add 5y to both sides.
600 = 550 + 5y
Subtract 550 from both sides.
50 = 5y
Divide both sides by 5.
y = 10.
Now we can plug the value of y into the original equation to find the value of x.
10 + x = 20
Subtract 10 from both sides.
x = 10
The value of x is equal to 10, and the value of y is also equal to 10.
solution is (x, y ) = (10, 10)
given
x + y = 20 → (1)
30x + 25y = 550 → (2)
from (1) y = 20 - x → (3)
substitute y = 20 - x in (2)
30x + 25(20 - x) = 550
30x + 500 - 25x = 550
5x + 500 = 550 ( subtract 500 from both sides )
5x = 50 ( divide both sides by 5 )
x = [tex]\frac{50}{5}[/tex] = 10
substitute x = 10 in (3) : y = 20 - 10 = 10
thus x = 10, y = 10 is the solution
