linear combo method help? ><?

Question

-2x + y + z = - 2
5x + 3y + 3z = 71
4x - 2y = 1

i know theres many type of way to solve it
but im still reallie confused if im doing it rite ><
any suggestions on making it easy?
or wut do u recomend that is easy way to solve it?

Answer 1

Thanks for a very simple question. Follow my step below.

Firstly, given the equation:
-2x + y + z = -2
5x + 3y + 3z = 71
4x - 2y = 1

Then, mark as (1), (2), and (3):
-2x + y + z = -2.....(1)
5x + 3y + 3z = 71.....(2)
4x - 2y = 1.....(3)

Next, we use elimination method from equation (1) and equation (2):
Times with 5: 5(-2x + y + z = -2)
Times with 2: 2(5x + 3y + 3z = 71)

-10x + 5y + 5z = -10.....(3)
10x + 6y + 6z = 142.....(4)

Later, equation (3) + equation (4):
11y + 11z = 132.....(5)

Next, we use elimination method from equation (1) and equation (3):
Times with 2: 2(-2x + y + z = -2)
Times with 1: 1(4x - 2y = 1)

-4x + 2y + 2z = -4.....(5)
4x - 2y = 1.....(6)

Later, equation (5) + equation (6):
2z = -3
z = -3 ÷ 2
z = -1.5

Then, we substitute z = -3/2 into equation (5):
11y + 11z = 132
11y + 11(-1.5) = 132
11y + (-16.5) = 132
11y - 16.5 = 132
11y = 132 + 16.5
11y = 148.5
y = 148.5 ÷ 11
y = 13.5

Lastly, we substitute y = 13.5 into equation (3) or equation (6):
4x - 2y = 1
4x - 2(13.5) = 1
4x - 27 = 1
4x = 1 + 27
4x = 28
x = 28 ÷ 4
x = 7

Therefore the answer should be:
x = 7, y = 13.5, and z = -1.5

I hope you got the best answer for your question. Have a nice day!

Answer 2

This particular system happens to be easy to solve. Notice that if you multiply the first equation by 3 and subtract it from the second equation, you get an equation in x only, so you can solve for x. Next, note that the third equation involves only x and y. You now know x, so you can use this equation to determine y. Finally, you can use either the first or the second equation, with the now-known values of x and y, to find z.

But this is an unusual situation. Things are seldom this easy. For a 3 by 3 system, for ease I'd probably use Cramer's Rule. It's not the most efficient way to solve a system, but it is relatively easy and you do it the same way no matter what the system is.