Question

How do you solve the following linear system: `3x + 5y = -1, 2x − 5y = 16`?

Answer 1

`x = 3, y = -2`

Explanation:

Starting with the first equation, we need to make either x or y the subject (I chose x)

`3x + 5y = -1`

`x = (-5y-1)/3`

`2x - 5y = 16`

`2*(-5y-1)/3 - 5y = 16`

`(-10y-2)/3 - 5y = 16`

`-10y-2 - 15y = 48`

`-25y = 50`

`y = -2`

Now we simply substitute `(y = -2)` into the second equation...

`2x-5y = 16`

`2x-5(-2)=16`

`2x + 10 = 16`

`2x = 6`

`x = 3`

Finally, we should also verify both answers by substituting both of them into the other equation again.

`3x + 5y = -1`

#3(3) + 5(-2) = -1

`9 - 10 = -1`

`-1=-1`, so therefore our answers are correct!

Answer 2

See a solution process below:

Explanation:

Step 1) First, solve each equation for `5y`:

• Equation 1:

`3x + 5y = -1`

`3x - color(red)(3x) + 5y = -1 - color(red)(3x)`

`0 + 5y = -1 - 3x`

`5y = -1 - 3x`

• Equation 2:

`2x - 5y = 16`

`2x - 5y + color(red)(5y) - color(blue)(16) = 16 - color(blue)(16) + color(red)(5y)`

`2x - 0 - color(blue)(16) = 0 + 5y`

`2x - 16 = 5y`

`5y = 2x - 16`

Step 2) Because both equations are equal on the left side we can now equate the right sides of the equation and solve for `x`:

`-1 - 3x = 2x - 16`

`-1 + color(blue)(16) - 3x + color(red)(3x) = 2x + color(red)(3x) - 16 + color(blue)(16)`

`15 - 0 = (2 + color(red)(3))x - 0`

`15 = 5x`

`15/color(red)(5) = (5x)/color(red)(5)`

`3 = (color(red)(cancel(color(black)(5)))x)/cancel(color(red)(5))`

`3 = x`

`x = 3`

Step 3) Substitute `3` for `x` in either of the solved equations in Step 1 and solve for `y`:

`5y = 2x - 16` becomes:

`5y = (2 xx 3) - 16`

`5y = 6 - 16`

`5y = -10`

`(5y)/color(red)(5) = -10/color(red)(5)`

`(color(red)(cancel(color(black)(5)))y)/cancel(color(red)(5)) = -2`

`y = -2`

The Solution Is:

`x = 3` and `y = -2`

Or

`(3, -2)`

Answer 3

`x = 3 and y =-2`

Explanation:

Notice that the terms in `y` in the two equations are are additive inverses. The sum of additive inverses is `0`
Add the equations together to eliminate the `y` terms.

`color(white)(xxxxxxxx)3xcolor(blue)(+5y) = -1" " A`
`color(white)(xxxxxxxx)2xcolor(blue)(-5y) = 16" "B`

`A+B:color(white)(xxx)5xcolor(blue)(+0y) = 15`
`color(white)(xxxxxxxxx)xcolor(white)(xxx) = 3`

Substitute in A:
`color(white)(xxxxxxxx)3(3)+5y = -1" " A`
`color(white)(xxxxxxxxxx)9+5y = -1`
`color(white)(xxxxxxxxxxxxx)5y = -10`
`color(white)(xxxxxxxxxxxxxx)y = -2`

Check in B:
`color(white)(xxxxx)2(3)-5(-2) = 16" "B`
`color(white)(xxxxxxxxxx)6+10 = 16`
`color(white)(xxxxxxxxxxxxx)16 = 16`

The values are correct.