Solve the system of linear equations and check any solutions algebraically. (If there is no solution, enter NO SOLUTION. If the system is dependent, express x, y, and z in terms of the parameter a.)

x-3y+2z=21
5x-13y+12z=81

I have an answer but i'm not sure if it's right.

I have (x,y,z)=(-5a-15,-a-5,a)

Thanks, Ms.Sue! You are amazing!

To solve the system of linear equations:

Step 1: We'll start by multiplying the first equation by 5 and the second equation by 1 to make the coefficients of x the same:

5(x-3y+2z)=5(21)
5x-15y+10z=105

1(5x-13y+12z)=1(81)
5x-13y+12z=81

This results in two equations with the same coefficients for x:

5x-15y+10z=105
5x-13y+12z=81

Step 2: We'll subtract the second equation from the first equation to eliminate x:

(5x-15y+10z) - (5x-13y+12z) = 105 - 81

This simplifies to:

-2y-2z = 24

Step 3: We'll simplify this equation further by dividing both sides by -2:

(-2y-2z) / -2 = 24 / -2

This gives us:

y + z = -12

Step 4: Now, we can express one variable in terms of the other. Let's solve for y:

y = -12 - z

Step 5: We'll substitute this expression for y in the first equation of the original system:

x - 3(-12 - z) + 2z = 21

Simplifying:

x + 36 + 3z + 2z = 21
x + 5z + 36 = 21
x + 5z = -15

Step 6: Finally, we'll express x in terms of z:

x = -15 - 5z

So, the solutions to the system of linear equations can be expressed as:

x = -15 - 5z
y = -12 - z

To check the solutions algebraically, substitute the values of x, y, and z into both original equations and verify that they are satisfied.

To solve the system of linear equations, you can use the method of elimination or substitution.

Let's solve it using the method of elimination:

1) Multiply the first equation by 5, and the second equation by 1, to make the coefficients of x in both equations equal:
5(x - 3y + 2z) = 5(21)
1(5x - 13y + 12z) = 1(81)

Simplifying:
5x - 15y + 10z = 105
5x - 13y + 12z = 81

2) Subtract the second equation from the first equation:
(5x - 15y + 10z) - (5x - 13y + 12z) = 105 - 81

Simplifying:
-2y - 2z = 24

3) Now we have a new equation with y and z. Let's continue solving for them:

-2y - 2z = 24
Divide all terms by -2 to isolate y and z:
y + z = -12

4) At this point, we have one equation (y + z = -12) involving two variables. This means we have infinitely many solutions, and the system of equations is dependent.

If you want to express x, y, and z in terms of a parameter, let's choose z = t (t is a parameter).

5) Substitute z = t in the equation y + z = -12:
y + t = -12
y = -12 - t

6) Substitute the values of y and z in terms of t back into the original equations:
x - 3(-12 - t) + 2t = 21
5x - 13(-12 - t) + 12t = 81

Simplify and solve these equations to get the expressions for x, y, and z in terms of the parameter t.

Therefore, the system of equations has infinitely many solutions, and the expressions for x, y, and z in terms of the parameter a are dependent on your calculations.