Time remaining:

# System of equations with three equations and three variables

label Algebra
account_circle Unassigned
schedule 0 Hours
account_balance_wallet \$5

I have to use elimination to figure out this problem, I can do it with 2 equations and 2 variables but I can't understand the 3

-3x + y - z = -10

-4x + 2y +3y = -1

2x + 3y -2z = = -5

Nov 19th, 2017

You should try to eliminate one of the variables. For example, rearranging the first equation, express z as a function of the other two variables:

-3x + y - z = -10

z = 10 -3x + y

Armed with z = (10 - 3x + y) you can now substitute that into the other two equations where you see z, and simplify. Assuming the second equation is -4x + 2y +3z = -1 (i.e. a typo in the question):

-4x + 2y +3(10 - 3x + y) = -1

-13x + 5y = -31

And in the last equation:

2x + 3y -2(10 - 3x + y) = = -5

8x + y = 15

Repeat the process now by using that second equation to find y in terms of x:

y = 15 - 8x

And sub that in to the first bolded equation:

-13x + 5(15 - 8x) = -31

-53x = -106

x = 2

Now plug that into 8x + y = 15 to give:

y = -1

And plug both of these back into your z = (10 - 3x + y):

z = 10 - 6 + (-1)

z = 3

You can use this approach for any number n of linear equations - use one equation to get a single term as a function of the others, then plug that in to the remaining equations to get n-1 equations, each now with that term removed. Keep going until you have one equation with one term, then you can start substituting it back in step by step to find the remaining terms!

Mar 20th, 2015

...
Nov 19th, 2017
...
Nov 19th, 2017
Nov 20th, 2017
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle