How do you solve 2x-3y=8 and x+y=11 using substitution?

Our equations are as follows:
`2x-3y=8`
`x+y=11`

Since the second equation is much simpler, we can subtract `y` from both sides to get an value for `x` in terms of `y`.

Subtracting `y` from both sides of the second equation, we get:

`x=11-y`

We can now substitute this into the first equation:
`2(11-y)-3y=8`, which simplifies to:

`22-2y-3y=8`, which can be simplified more to:

`22-5y=8`

We can solve for `y` now, by subtracting `22` from both sides and dividing both sides by `-5.` We get:

`-5y=-14`

`y=14/5`

We can plug this into either equation to solve for `x.` Let's plug in `14/5` in the first equation, `2x-3y=8`:

`2x-3(14/5)=8`

`2x-42/5=8` (We can change 2x to `10/5x` and 8 into `40/5` to have a common denominator).

`10/5x-42/5=40/5`

Let's add `42/5` to both sides:

`10/5x=82/5`

We can multiply both sides by the reciprocal of `10/5` to solve for `x`.

`x=82/5(5/10)`

The `5s` cancel, and we're left with `82/10` or `x=8.2.`

`2x-3y=8`
`x+y=11`

We need to solve `x+y=11` for `x`

`x+y=11`

Subtract `y` from both sides

`x+y-y=11-y`

`x=11-y`

now substitute `-y+11` for `x` in `2x -3y=8`

`2x - 3y =8`

`2(-y+11)-3y=8`

Use the distributive property

`(2)(-y)+(2)(11)-3y=8`

`-2y + 22 - 3y =8`

`-5y + 22 = 8`

`-5y = 8-22`

#-5y= -14

Divide both sides by `-5`

`(-5y)/-5=(-14)/-5`

`y= 14/5`

Now substitute `14/5` for `y` in `x=-y+11`

`x=-y+11`

`x=-14/5+11`

`x=(-14)/5 + 11/1`

`x=(-14)/5+11/5`

`x=(-14+55)/5`

`x=41/5`

Answer: `x=41/5` and y`14/5`

Isolate `y` in the second equation by subtracting `x` on both sides:

`x+y=11 => y=11-x`

Substitute `y` in the first equation with the expression `11-x` then solve for `x`, then solve for `y`:

`2x - 3y = 8`

`2x - 3(11-x) = 8`

`2x - 33 + 3x = 8`

`5x - 33 = 8`

`5x = 41`

`x = 41/5`

`x = 8.2`

So

`=> y = 11 - (41/5)`

`y = 55/5 - 41/5`

`y = 14/5`

Or

`y = 11 - (8.2)`

`y = 2.8`

Check your solution: insert your values in each of the given equations and verify that they satisfy the system:

`2(8.2) - 3(2.8) = 8 " " {"true"}`

`(8.2) + (2.8) = 11 " "{"true"}`

Your solution is correct.