# How do you solve this system of equations: 6x + y = 7, y + z = 2 , and 4x - z = 4?

Jun 29, 2018

color(maroon)(x = 1/2, y = 4, z = -2

#### Explanation:

$6 x + y = 7 \text{ Eqn (1)}$

$y + z = 2 \text{ Eqn (2)}$

$4 x - z = 4 \text{ Eqn (3)}$

$z = 4 x - 4 \text{ from Eqn (3)}$

Substituting the value of z in terms of x in Eqn (2),

$y + 4 x - 4 = 2 \text{ or " 4x + y = 6 " Eqn (4)}$

Subtracting Eqn (4) from (1),

$6 x + y - 4 x - y = 7 - 6$

$2 x = 1 ' \mathmr{and} x = \left(\frac{1}{2}\right)$

Substituting value of x in Eqn (3),

$4 \cdot \frac{1}{2} - 4 = z \text{ or } z = - 2$

Substituting value of z in Eqn (2),

$y + \left(- 2\right) = 2 \text{ or } y = 4$

Jun 29, 2018

$x = \frac{1}{2}$, $y = 4$ and $z = - 2$

#### Explanation:

$6 x + y - \left(4 x - z\right) - \left(y + z\right) = 7 - 4 - 2$

$2 x = 1$, so $x = \frac{1}{2}$

Hence,

$6 - \frac{1}{2} + y = 7$ or $3 + y = 7$, so $y = 4$

Thus,

$4 + z = 2$, so $z = - 2$