Math Help Algebra 1?

A) x+y=62
B) x-y=36

from B) you can obtain x=36+y
substituting this result in A) gets you (36+y)+y=62
doing the operations
36+2y=62
therefore
2y=62-36
2y=26
solving for y
y=((26)/2)=13

Substituting in x=36+y
you get x=36+13=49

finally x=49 and y=13

x+y=62
x-y=36


2x=98 x=49 y=62-49=13

x + y = 62
x - y = 36-----ADD

2x = 98
x = 49

49 + y = 62
y = 13

x = 49 , y = 13

Two numbers are x and y.
x + y = 62 ... (1)
x - y = 36 ... (2)

Use elimination method.

(1) + (2)
(x + y + x - y = 62 + 36)
2x = 98
x = 49

Substitue x = 49 into (1)
x + y = 62
49 + y = 62
y = 13

The two numbers are 49 and 13

x+y=62
x-y=36

see how the 2 equatons line up underneath each other?

we can add them together

2x+y-y=62+36

2x=98

x=49

49-y=36

49=36+y

13=y

the numbers are 13 and 49

here is the solution.

let x is the 1st num and y is the 2nd num

x+y = 62 -eq1
x-y = 36 -eq2


y = 62 -x
y = 36 +x

now with y as common:

36 + x = 62 - x

2x = 26
x = 13
y = 36 +13
y = 49

answer:

x = 13
y = 49

to check:

sum of two num is 62.

13 + 49 = 62

the diff of two num is 36

49 - 13 = 36

This is a system of equations. You have two unknowns which means that you will have two equations.

X is going to represent one of the numbers and Y represents the other number.

So you have two numbers that add to give you 62
That means

x + y = 62 (equation 1)

and when you subtract these numbers you get 36
that means

x - y = 36 (equation 2)

line these up and add them

x+y=62
x-y= 36 notice that the y's cancel and you are left with

2x=98 solve for x and you get x=49 plug this back into one of the ORIGINAL equations and you get y=13. And you would represent this solution as an ordered pair (49,13).

x > y
x + y = 62
x - y = 36

x + y = 62
x = 62 - y

x - y = 36
(62 - y) - y = 36
62 - y - y = 36
-y - y = 36 - 62
-2y = -26
y = -26/-2
y = 13

x + y = 62
x + 13 = 62
x = 62 - 13
x = 49

∴ x = 49 , y = 13