# {x-3y+4z=26 2x+y+z=3 -2x+3y-3z=-23} Solve the given system of equations. If the system has no solution, say that it is inconsistent. ?

### 5 Answers

- Φ² = Φ+1Lv 72 weeks ago
{x-3y+4z=26 2x+y+z=3 -2x+3y-3z=-23}

Rewrite as: ❶: y+z=3-2x, ❷: 3y-3z=-23+2x, and ❸: x=26+3y-4z

From ❶ and ❷ we get:

❹: y = ((3-2x)×-3-1×(-23+2x))/(1×-3-1×3) = (-7-2x)/3

❺: z = (1×(-23+2x)-(3-2x)×3)/(1×-3-1×3) = (16-4x)/3

Substituting ❹ and ❺ into ❸ we get:

❻: x = 26+3(-7-2x)/3-4(16-4x)/3 = (10x-7)/3 = 1

Substituting ❻ into ❹ and ❺ yields:

❼: y = (-7-2×1)/3 = -3 and ❽: z = (16-4×1)/3 = 4

Solution point: (1, -3, 4)

- az_lenderLv 72 weeks ago
Another approach, elimination:

Add the 2nd & 3rd equations, obtaining:

4y - 2z = -20.

Double the 1st eqn and add it to the 3rd, obtaining:

-3y + 5z = 29.

Now multiply these two new equations so that the "y" coefficients will cancel one another out:

12y - 6z = -60;

-12y + 20z = 116.

Adding these together gives 14z = 56,

so z = 4; back-substitution gives y = -3 and x = 1.

Check by using any of the original equations,

for example:

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

yes, that works.

The solution by llaffer is wrong, does not check out.

- PuzzlingLv 72 weeks ago
x - 3y + 4z = 26 [eq 1]

2x + y + z = 3 [eq 2]

-2x + 3y - 3z = -23 [eq 3]

Add eq 1 and 3:

-x + z = 3 [eq 4]

Triple eq 2 and add it to 1:

7x + 7z = 35

Divide by 7:

x + z = 5 [eq 5]

Add to eq 4:

2z = 8

z = 4

Substitute in eq 5:

x + 4 = 5

x = 1

Substitute in eq 2:

2(1) + y + 4 = 3

y = -3

I'll leave the step of verifying to you.

Answer:

x = 1

y = -3

z = 4

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- llafferLv 72 weeks ago
x - 3y + 4z = 26 and 2x + y + z = 3 and -2x + 3y - 3z = -23

Start by solving one equation for one variable then substitute it into the other two. I'll solve the first equation for x in terms of y and z:

x - 3y + 4z = 26

x = 26 + 3y - 4z

Substitute:

2x + y + z = 3 and -2x + 3y - 3z = -23

2(26 + 3y - 4z) + y + z = 3 and -2(26 + 3y - 4z) + 3y - 3z = -23

52 + 6y - 8z + y + z = 3 and -52 - 6y + 8z + 3y - 3z = -23

7y - 7z = -49 and -3y + 5z = 29

Now solve the first equation for y in terms of z and substitute into the last:

7y - 7z = -49

y - z = -7

y = z - 7

-3y + 5z = 29

-3(z - 7) + 5z = 29

-3z + 21 + 5z = 29

2z = 28

z = 14

Now that we have z, solve for y, then go back to x:

y = z - 7

y = 14 - 7

y = 7

x = 26 + 3y - 4z

x = 26 + 3(7) - 4(14)

x = 26 + 21 - 56

x = -9

The solution to this system of equations is:

x = -9, y = 7, z = 14