Question

How do you solve `1 / 5x + 1/3y = - 1 / 15` and `1 / 5x + 2y = 8 / 5` using substitution?

Answer 1

We can solve each equation for `1/5x` and set them equal to each other:

`1/5x+1/3y=-1/15=>1/5x=-1/15-1/3y`

`1/5x+2y=8/5=>1/5x=8/5-2y`

`:.-1/15-1/3y=8/5-2y`

We can now solve for `y`. Let's first multiply through by 15:

`15(-1/15-1/3y)=15(8/5-2y)`

`-1-5y=24-30y`

`25y=25`

`y=1`

And now we can substitute back into one of the original equations to solve for `x`:

`1/5x+2(1)=8/5`

`1/5x=8/5-2`

`1/5x=8/5-10/5=-2/5`

`x=-2`

Which gives us the point `(-2,1)` which we can see on the graph:

graph{(1/5x+1/3y+1/15)(1/5x+2y-8/5)=0}

Answer 2

`x=-2 and y=1`

Explanation:

I would usually simplify the equations to get rid of the fractions first, but in this case we have identical terms in each equation, so I will solve for those first.

`1/5x = -1/15-1/3y" and "1/5x =8/5-2y`

`1/5x = 1/5x`

We can substitute the other sides of each equation for `1/5x`

`-1/15-1/3y = 8/5-2y" "xx15`

`-1-5y= 24-30y`

`30y-5y = 24+1`

`25y = 25`

`y =1`

Substitute into `1/5x =8/5-2y`

`1/5x = 8/5 -2(1)" "xx 5`

`x = 8-10`

`x =-2`

Check:
`1/5x +1/3y = -1/15" " xx5`

`x+5/3y = -1/3`

`-2 +1 2/3 = -1/3`

`-1/3 = -1/3`