Question

How do you solve `3x+y=5` and `x-2y=11` using substitution?

Answer 1

See the entire solution process below:

Explanation:

Step 1) Solve the first equation for `y`:

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

`0 + y = 5 - 3x`

`y = 5 - 3x`

Step 2) Substitute `color(red)(5 - 3x)` for `y` in the second equation and solve for `x`:

`x - 2(color(red)(5 - 3x))= 11`

`x - (2 xx color(red)(5)) + (2 xx color(red)(3x))= 11`

`x - 10 + 6x = 11`

`7x - 10 = 11`

`7x - 10 + color(red)(10) = 11 + color(red)(10)`

`7x - 0 = 21`

`7x = 21`

`(7x)/color(red)(7) = 21/color(red)(7)`

`(color(red)(cancel(color(black)(7)))x)/cancel(color(red)(7)) = 3`

`x = 3`

Step 3) Substitute `color(red)(3)` for `x` in the solution to the first equation at the end of Step 1

`y = 5 - (3 xx color(red)(3))`

`y = 5 - 9`

`y = -4`

The solution is `x = 3` and `y = -4`

Answer 2

`x = -1/7`, `y=-39/7`. Explanation below.

Explanation:

Using the second equation, we can express `y` in terms of `x` as such:

`-2y = 11 - x => y = -11/2+x/2`.

Substituting this `y` value in the first equation:

`3x + (-11/2 + x/2) = 5 => (7x)/2 = -1/2 =>`

`=>7x = -1 => x = -1/7`.

Now, we can substitute this `x` value in the second equation to find `y` (any equation will do, but `x` is easier to evaluate than `3x`)

`-1/7 - 2y = 11 => 2y = -78/7 => y = -78/14 = -39/7`