Question

How do you solve `6m + n = 21` and `m - 8n = 28` using substitution?

Answer 1

`n = -3, " "m = 4`

Explanation:

Take the second equation:

`m - 8n = 28`

If we add `8n` to both sides, we get:

`m = 8n + 28`

Now we have two equivalent expressions: `color(red)m` and `color(blue)(8n+28)`.

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Now, look back at the first equation:

`6m + n = 21`

To solve by substitution, we need to replace (or substitute) one equivalent expression for the other so that we can solve this equation. Let's use the expressions from earlier:

`6color(red)m + n = 21`

Remember that `color(red)m = color(blue)(8n + 28)`, so we can do this:

`6(color(blue)(8n + 28)) + n = 21`

Now, we can use simple algebra to solve for `n`.

`6(8n+28) + n = 21`

`6*8n + 6*28 + n = 21`

`48n + 168 + n = 21`

`49n + 168 = 21`

`49n = -147`

`n = -3`

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Finally, let's plug `n` back into one of the equations and solve for `m` to finish the problem. I'm using equation 1 to solve for `m`, but either one will work just fine.

`6m + n = 21`

Remember from above that `color(orange)n = color(limegreen)(-3)`

`6m + color(orange)n = 21`

`6m + color(limegreen)((-3)) = 21`

`6m - 3 = 21`

`6m = 24`

`m = 4`

So now we have both parts of our solution! Here's the final solution:

`n = -3, " "m = 4`

Final Answer