Question

Use substitution to solve the following system of linear equations and fill in the following blanks:

Answer 1

Answer:

$x=-4$

$y=3$

$z=-5$

Step-by-step explanation:

Given:

$x+y+z=-6$

$x-6y-7z=-29$

$-7y-5z=4$

Solve for $x$ in the 1st equation:

$x+y+z=-6$

$x+y=-z-6$

$x=-y-z-6$

Substitute the value of $x$ into the 2nd equation and solve for $z$:

$x-6y-7z=-29$

$(-y-z-6)-6y-7=-29$

$-7y-z-13=-29$

$-7y-z=-16$

$-z=-16+7y$

$z=16-7y$

Substitute the value of $z$ into the 3rd equation and solve for $y$:

$-7y-5z=4$

$-7y-5(16-7y)=4$

$-7y-80+35y=4$

$28y-80=4$

$28y=84$

$y=3$

Plug $y=3$ into the solved expression for $z$ and evaluate to solve for $z$:

$z=16-7(3)$

$z=16-21$

$z=-5$

Plug $z=-5$ into the solved expression for $x$ and evaluate to solve for $x$:

$x=-(3)-(-5)-6$

$x=-3+5-6$

$x=2-6$

$x=-4$

Therefore:

$x=-4$

$y=3$

$z=-5$