To solve this system of equations, we can use the method of substitution or elimination. Let's use the elimination method here:
Given the system of equations:
1. \(x + y + 3z = 4\)
2. \(-2x - y - z = 6\)
3. \(2x + 3y - 7z = 13\)
Let's start by adding the first and second equations to eliminate \(y\):
Adding equations 1 and 2:
\(x + y + 3z - 2x - y - z = 4 + 6\)
\(-x + 2z = 10\)
\(x - 2z = -10\)
Now, let's multiply the second equation by 2 and add it to the third equation to eliminate \(x\):
Multiplying equation 2 by 2:
\(-4x - 2y - 2z = 12\)
Adding the modified equation 2 and equation 3:
\(-4x - 2y - 2z + 2x + 3y - 7z = 12 + 13\)
\(-2x + y - 9z = 25\)
Substituting the value of \(x\) from the first equation into this last equation:
\(-10 + y - 9z = 25\)
\(y - 9z = 35\)
Now we have a system of two equations with two variables:
1. \(x - 2z = -10\)
2. \(y - 9z = 35\)
Solve the second equation for \(y\):
\(y = 35 + 9z\)
Substitute this value of \(y\) into the first equation:
\(x - 2z = -10\)
\(x = -10 + 2z\)
Now we have expressions for \(x\) and \(y\) in terms of \(z\). We can substitute these into any of the original equations to solve for \(z\). Let's use the first equation:
\(x + y + 3z = 4\)
\((-10 + 2z) + (35 + 9z) + 3z = 4\)
\(26z + 25 = 4\)
\(26z = -21\)
\(z = -\frac{21}{26}\)
Now that we have the value of \(z\), we can substitute it back into the expressions for \(x\) and \(y\) to find their values:
\(x = -10 + 2z = -\frac{52}{26} = -2\)
\(y = 35 + 9z = \frac{234}{26} = 9\)
So, the solution to the system of equations is \(x = -2\), \(y = 9\), and \(z = -\frac{21}{26}\).