algebra II help?

Question

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

solve for each variable and plz leave a step by step answer! thanks alot, i think algebra II is really hard

thanks for all the help!!

Answer 1

Using the third equation, you can see that z = x + y + 1.

Substituting into the first two equations:

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

Simplifying:

3x + y = 5
4x - y = 2

Adding the two equations:

7x = 7
x = 1

Substituting x into the first of the two equations that I have containing only x's and y's:

3(1) + y = 5
y = 2

Substituting x and y into the original equation that I derived:

z = (1) + (2) + 1
z = 4

(x, y, z) = (1, 2, 4)

You can verify that this is a solution by substituting values into the original 3 x 3 system.

Answer 2

[Edit: I messed up down in Step 7, so skip this answer. Lots of others have done it right.]

Okay, I'll do this one. To keep track, we'll number the equations:

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

We're going to eliminate z first. Equation (4) will be Eq. 1 minus Eq. (3). Be careful to change the signs when you subtract:

(4) 3x + y = 5

Now, Equation 5 will be Eq. (2) plus two times Eq. (3). Be careful to double everything in Eq. (3) when you add:

(5) 4x - y = 2

Add Eq. (4) and (5) together to get:

(6) 7x = 7 ==> x = 1 (Answer)

Substitute this into Eq (4) to get

(7) 3x + y = 21 + y = 5 ==> y = -16 (Answer)

Substitute (6) and (7) into (3) to get

(8) x + y - z = 1 - 16 - z = -1 ==> z = -14 (Answer)

Answer 3

There are several ways to solve this system of equations. I will show the cancellation method.

Multiply first eq by 2 and add to the second to cancel z;
Multiply third eq by 2 and add to the second to cancel z;

you should now have two eqs

10x + y = 12
4x - y = 2

Now do the same thing again: Add these equs together

14x = 14; x = 1

Put this x into one of the above equs to get y. (y = 2) Then put x and y into one of the original equs to get z (z = 4).

There are other approaches; you can choose what equs to multiply and add or subtract and which variables to cancel out.

Answer 4

(1) 4x + 2y - z = 4
(2) 2x - 3y + 2z = 4
(3) x + y - z = -1 you can add 1, 2 & 3 & get
7x=7
x=1
now you have
4+2y-z=4
(4) 2y-z=0

2-3y+2z=4
(5) -3y+2z=2
1+y-z=-1
(6) y-z=-2

subtract 6 from 4
y=2
so now

2-z=-2
-z=-4
z=4

x-1
y=2
z=4

Answer 5

you can use matrixt
[4 2 -1][x] =[4]
[2 -3 2][y] [4]
[1 1 -z][z] [-1]
remember Ax=b to find x
x = inv(A)b
or you can
multiply 3rd equation by 2 and add to 2nd equation
4x - y = 2
mutliply 1st equation by 2 and add to 2nd equation
10x + y = 12
add these two equation together
14x = 14
x = 1 => y = 2 => z = 4