##### Solving systems of linear equations

label Algebra
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5

y-x+7         7   8   9   10

y=9-x         9   8   7  6

Solutions to both equations is (ordered pair)

Sep 12th, 2015

Thank you for the opportunity to help you with your question!

Chrissy,

I assume that the first line is actually y = x + 7. If not, then you don't have a SYSTEM of linear equations.

First off, if y = x+7 and y = 9-x, then (since y = y)

x + 7 = 9 - x. Adding x to both sides yields 2*x + 7 = 9. Subtracting 7 from both sides yields 2x = 2, so x = 1. Since x = 1 is not one of the available values, this means that one or the other (or both) of the two equations is misstated.

Let's try assuming that y - x = 7. Adding x to both sides yields y = 7 + x. So if y = y, then 7 + x = 9 - x and so

adding x to both sides yields 7 + 2x = 9, and 2x = 2, as before.

You need to check the expressions in column 1. Let me know what you discover and we can solve the problem.

Please let me know if you need any clarification. I'm always happy to answer your questions.
Sep 12th, 2015

it is y=x+7.. I type fast.

Sep 12th, 2015

Ok. If the first equation is y = x+7 and the second equation is y = 9-x, then (as I showed earlier), x must equal 1. But 1 is not an available value for x. Or is it?

Can you check to see if the number 10 is actually a 1 followed by a zero.

If it isn't then none of the possible (x,y) pairs will satisfy both equations.

Sep 12th, 2015

It is a "10"

Sep 13th, 2015

The none of  available ordered pairs will be a solution to the system. You could try all of them, but I wouldn't recommend it. The mathematics is airtight. Best wishes.

Sep 13th, 2015

...
Sep 12th, 2015
...
Sep 12th, 2015
Oct 19th, 2017
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle