Question

2x +y+z=7 and 3x+2y+4z+19 and 5x +3y+2z+17 solve for x

Answer 1

do you prefer elimination or substitution ?

Answer 2

oh...and please check your equations again...is there supposed to be equal signs in the last 2 ?

Answer 3

im sorry yea =19 and =17 and we have been using elimination

Answer 4

ok...lets get started

2x + y + z = 7 --->(-4)
3x + 2y + 4z = 19
---------------
-8x - 4y - 4z = -28 --- result of multiplying by -4
3x + 2y + 4z = 19
---------------add
-5x - 2y = - 9

3x + 2y + 4z = 19
5x + 3y + 2z = 17 --->(-2)
----------------
3x + 2y + 4z = 19
-10x - 6y - 4z = - 34 --- result of multiplying by -2
---------------add
-7x - 3y = - 15

-5x - 2y = -9 --->(3)
-7x - 3y = -15 --->(-2)
----------------
-15x - 6y = -27 --- result of multiplying by 3
14x + 6y = 30 --- result of multiplying by -2
----------------add
-x = 3
x = -3

now sub -3 in for x in either equation that you eliminated z in
-5x - 2y = -9
-5(-3) - 2y = -9
15 - 2y = -9
-2y = -9 - 15
-2y = - 24
y = 12

now sub our info into any of the original equations
2x + y + z = 7
2(-3) + 12 + z = 7
-6 + 12 + z = 7
6 + z = 7
z = 7 - 6
z = 1

now check our answers
3x + 2y + 4z = 19
3(-3) + 2(12) + 4(1) = 19
-9 + 24 + 4 = 19
-9 + 28 = 19
19 = 19 (correct)

solution is : x = -3, y = 12, z = 1

any questions ?

Answer 5

No...Thank you so much...you are awesome!