Question

Some help me out this is ALG 1 problem. Steps are appreciated

Answer 1

when two numbers are added together, the result is 45. Twice the greater number is 6 more than 5 times the lesser. What are the numbers?

Answer 2

Number 1 = x
Number 2 = y
----------
x + y = 45
(let y be greater, it doesn't matter as long as only one is greater) 2y = 5x +6

Answer 3

Solve the system

Answer 4

Do you know how?

Answer 5

wait does this work?

Answer 6

x+y=45
2x+6+5y=45

x=greater
y=lesser

Answer 7

wait nevermind

Answer 8

can you continue in your way?

Answer 9

If x is lesser, than just switch the x and y in the second equation I gave you

Answer 10

*greater

Answer 11

yeah, could you explain your way?

Answer 12

2x = 5y + 6
divide each side by 2

x = (5/2)y+3
------------
x + y = 45
(5/2)y+3 + y = 45
(5/2)y + y + 3 = 45
(7/2)y +3 = 45
-3 on both sides
(7/2)y = 42
divide both sides by (7/2)
y = 12
---------
x + 12 = 45
subtract 12 on both sides

x = 33
-------
(33, 12)

Answer 13

Thank you so much :DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD

Answer 14

Just substitute x in the terms of y for y to solve for y, then use that to solve for x