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

1 Answer
Mar 22, 2018

Answer:

#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