Question

X/2+y=4/5. x+y/2=7/10

Answer 1

The answer is x = 2/5, y = 3/5 or (2/5, 3/5)

First, in order to avoid fractional values during calculation, multiply both equations by 10 to simplify.

1. 10 (x/2 + y) = 10 (4/5) ⇒ 5x + 10y = 8
2. 10 (x + y/2) = 10 (7/10) ⇒ 10x + 5y = 7

Multiply the 1st equation by 10 and 2nd equation by 5.

3. 10 (5x + 10y) = 10 (8) ⇒ 50x + 100y = 80

4. 5 (10x + 5y) = 5 (7) ⇒ 50x + 25y = 35

Subtract : 3rd equation - 4th equation

• 50x + 100y - 50x - 25y = 80 - 35
• 75y = 45
• y = 3/5

Now, substitute for x in the 1st equation to find y.

• x/2 + 3/5 = 4/5
• x/2 = 1/5
• x = 2/5

Answer 2

x = 2/5

y = 3/5

Explanation:

**Disclaimer** Hi there! I assumed the question to be a system of equations. The following answer is a method for solving a system of equations. Thus, if it is not, please let me know and I will modify my answer.

Given information:

`(x)/(2)+y=(4)/(5)`

`x+(y)/(2) =(7)/(10)`

Eliminate fractions by multiplying 10 on both sides of the first equation:

`10*(x)/(2)+10*y=10*(4)/(5)`

`5x+10y=8`

Eliminate fractions by multiplying 10 on both sides of the second equation:

`10*x+10*(y)/(2) =10*(7)/(10)`

`10x+5y=7`

Current system:

`5x+10y=8`

`10x+5y=7`

Multiply the second equation by 2:

`5x+10y=8`

`20x+10y=14`

Subtract the first equation from the second equation:

`(20x+10y)-(5x+10y)=14-8`

`20x+10y-5x-10y=6`

`(10y-10y)+20x-5x=6`

`0+15x=6`

`\boxed{x=(2)/(5) }`

Substitute the x value back to one of the equations to get the y value:

`x+(y)/(2) =(7)/(10)`

`((2)/(5)) +(y)/(2) =(7)/(10)`

`((2)/(5)) +(y)/(2)-(2)/(5) =(7)/(10)-(2)/(5)`

`(y)/(2) =(3)/(10)`

`(y)/(2)*2 =(3)/(10)*2`

`\boxed{y=(3)/(5) }`

Hope this helps!! :)

Please let me know if you have any questions