Question

{3x+y-4z=-30
{3x+2y+2z=-8
{5x+5y+z=-26

(Give the answer to each equation)

Answer 1

To solve this system of equations, we can use the method of elimination.

First, let's eliminate the variable x by multiplying the first equation by -1 and adding it to the second equation:
-1(3x + y - 4z) = -1(-30)
-3x - y + 4z = 30

3x + 2y + 2z = -8

Adding the two equations together eliminates x:
0 + y + 6z = 22
y + 6z = 22

Next, let's eliminate x again by multiplying the first equation by -5 and adding it to the third equation:
-5(3x + y - 4z) = -5(-30)
-15x - 5y + 20z = 150

5x + 5y + z = -26

Adding the two equations together eliminates x:
0 - 10y + 21z = 124
-10y + 21z = 124

Now we have a system of two equations with two variables:
y + 6z = 22
-10y + 21z = 124

We can solve this system using substitution or elimination. For simplicity, let's use substitution:

Solve the first equation for y:
y = 22 - 6z

Substitute this value of y into the second equation:
-10(22 - 6z) + 21z = 124
-220 + 60z + 21z = 124
81z = 344
z = 344/81

Substitute this value of z into the first equation to solve for y:
y + 6(344/81) = 22
y + 2064/81 = 22
y = 22 - 2064/81

Finally, substitute the values of y and z into any of the original equations to solve for x. Let's use the first equation:
3x + (22 - 2064/81) - 4(344/81) = -30
3x + 22 - 2064/81 - 1376/81 = -30
3x - 3440/81 = -30
3x = -30 + 3440/81
x = (-30 + 3440/81)/3

So the solutions to the system of equations are:
x = (-30 + 3440/81)/3
y = 22 - 2064/81
z = 344/81