Solve the following system of linear equations in three variables 1x-2y+4=0;1y-1z+1=0;2z+3x=14 - Mathematics

System of Linear Equations Solver

Question

Solve the following system of linear equations in three variables:

$$ \frac{1}{x} - \frac{2}{y} + 4 = 0 $$
$$ \frac{1}{y} - \frac{1}{z} + 1 = 0 $$
$$ \frac{2}{z} + \frac{3}{x} = 14 $$

Solution

This system is not linear in \(x, y, z\). We can make it linear by substituting variables. Let \( p = \frac{1}{x} \), \( q = \frac{1}{y} \), and \( r = \frac{1}{z} \). Rearranging the given equations, we get:

$$ p - 2q = -4 $$ (1)
$$ q - r = -1 $$ (2)
$$ 3p + 2r = 14 $$ (3)

Multiply equation (2) by 2 and add it to equation (1) to eliminate \(q\):

Multiply (2) by 2: $$ 2(q-r) = 2(-1) \implies 2q - 2r = -2 $$

$$ p - 2q = -4 $$
$$ 2q - 2r = -2 $$
(1)
$$ p - 2r = -6 $$
(4)

Now add equation (3) and (4) to eliminate \(r\) and solve for \(p\):

$$ 3p + 2r = 14 $$
$$ p - 2r = -6 $$
(3)
(4)
$$ 4p = 8 $$
$$ p = 2 $$

Substitute \(p=2\) into equation (1) to find \(q\):

$$ (2) - 2q = -4 $$ $$ -2q = -6 $$
$$ q = 3 $$

Substitute \(q=3\) into equation (2) to find \(r\):

$$ (3) - r = -1 $$ $$ -r = -4 $$
$$ r = 4 $$

Finally, substitute the values of \(p, q, r\) back to find \(x, y, z\):

Since \( p = \frac{1}{x} \), then \( 2 = \frac{1}{x} \implies x = \frac{1}{2} \)

Since \( q = \frac{1}{y} \), then \( 3 = \frac{1}{y} \implies y = \frac{1}{3} \)

Since \( r = \frac{1}{z} \), then \( 4 = \frac{1}{z} \implies z = \frac{1}{4} \)

The final solution is:

$$ x = \frac{1}{2}, \quad y = \frac{1}{3}, \quad z = \frac{1}{4} $$