Question

3x-5y+z=8, 4y-z=10 ,7x+y=4

Answer 1

I would solve by elimination....are you familiar with that method ?

Answer 2

x = 11/5, y = -57/5, z = -278/5

Answer 3

can you guys solve step by step

Answer 4

wait...you can't use elimination because the 2nd equation has no x value...you have to use substitution

Answer 5

come on help me this out please

Answer 6

you could use cramers rule - matrices.

Answer 7

whats that rule omg.. never heard about that

Answer 8

Solve the following system:
{3 x-5 y+z = 8 | (equation 1)
4 y-z = 10 | (equation 2)
7 x+y = 4 | (equation 3)
Swap equation 1 with equation 3:
{7 x+y+0 z = 4 | (equation 1)
0 x+4 y-z = 10 | (equation 2)
3 x-5 y+z = 8 | (equation 3)
Subtract 3/7 × (equation 1) from equation 3:
{7 x+y+0 z = 4 | (equation 1)
0 x+4 y-z = 10 | (equation 2)
0 x-(38 y)/7+z = 44/7 | (equation 3)
Multiply equation 3 by 7:
{7 x+y+0 z = 4 | (equation 1)
0 x+4 y-z = 10 | (equation 2)
0 x-38 y+7 z = 44 | (equation 3)
Swap equation 2 with equation 3:
{7 x+y+0 z = 4 | (equation 1)
0 x-38 y+7 z = 44 | (equation 2)
0 x+4 y-z = 10 | (equation 3)
Add 2/19 × (equation 2) to equation 3:
{7 x+y+0 z = 4 | (equation 1)
0 x-38 y+7 z = 44 | (equation 2)
0 x+0 y-(5 z)/19 = 278/19 | (equation 3)
Multiply equation 3 by 19:
{7 x+y+0 z = 4 | (equation 1)
0 x-38 y+7 z = 44 | (equation 2)
0 x+0 y-5 z = 278 | (equation 3)
Divide equation 3 by -5:
{7 x+y+0 z = 4 | (equation 1)
0 x-38 y+7 z = 44 | (equation 2)
0 x+0 y+z = -278/5 | (equation 3)
Subtract 7 × (equation 3) from equation 2:
{7 x+y+0 z = 4 | (equation 1)
0 x-38 y+0 z = 2166/5 | (equation 2)
0 x+0 y+z = -278/5 | (equation 3)
Divide equation 2 by -38:
{7 x+y+0 z = 4 | (equation 1)
0 x+y+0 z = -57/5 | (equation 2)
0 x+0 y+z = -278/5 | (equation 3)
Subtract equation 2 from equation 1:
{7 x+0 y+0 z = 77/5 | (equation 1)
0 x+y+0 z = -57/5 | (equation 2)
0 x+0 y+z = -278/5 | (equation 3)
Divide equation 1 by 7:
{x+0 y+0 z = 11/5 | (equation 1)
0 x+y+0 z = -57/5 | (equation 2)
0 x+0 y+z = -278/5 | (equation 3)
Collect results:
Answer: |
| {x = 11/5
y = -57/5
z = -278/5

That is elimination.

Answer 9

wow so nice and useless

Answer 10

spam.

Answer 11

y = -7x + 4

now sub -7x + 4 in for y in the 2nd equation
4y - z = 10
4(-7x + 4) - z = 10
-28x + 16 - z = 10
-28x - z = 10 - 16
-28x - z = -6
-z = 28x - 6
z = -28x + 6

now sub -7x + 4 in for y and -28x + 6 in for zin the 2st equation
3x - 5y + z = 8
3x - 5(-7x + 4) -28x + 6 = 8
3x + 35x - 20 - 28x + 6 = 8
38x - 28x - 14 = 8
10x = 8 + 14
10x = 22
x = 22/10 = 11/5

now sub 11/5 in for x in the 3rd equation
7x + y = 4
7(11/5) + y = 4
77/5 + y = 4
y = 4 - 77/5
y = 20/5 - 77/5
y = - 57/5

now sub -57/5 in for y in the 2nd equation
4y - z = 10
4(-57/5) - z = 10
-228/5 - z = 10
-z = 10 + 228/5
-z = 50/5 + 228/5
-z = 278/5
z = -278/5

check the answers
3x - 5y + z = 8
3(11/5) - 5(-57/5) - 278/5 = 8
33/5 + 285/5 - 278/5 = 8
318/5 - 278/5 = 8
40/5 = 8
8 = 8 (correct)

jprulz is correct......x = 11/5, y = -57/5, z = -278/5

Answer 12

wow nice work bros love you guys all

Answer 13

glad to help :)

Answer 14

thank you so much texaschic XP