systems of equations with 3 variables - addion method

Question

I am stuck on solving the following systems of equations with 3 variables. The textbook asks to use the addition method so can we please stick to that.

${5x -y = 3}$

${3x + z = 11}$

${y - 2z = -3}$

I am used to systems of equations where each equation has at least one instance of the variable e.g. ${x + y + z = 1}$ but in each of the above, one of the variables is omitted in each equation.

Could somebody explain what to do in this situation? Should I multiply both sides by one of the variables to balance it up?

Answer 1

$5x-y=3$
$3x+z=11$
$y-2z=-3$
Multiply first equation by $3$ and second one by $(-5)$.
Add both.
$15x-3y=9$
$-15x-5z=-55$
Adding, $3y+5z=46$
$3(2z-3)+5z=46$
$11z-9=46$
$z=5$, $y=7$, $x=2$
Also you can use Cramer's rule-
$\frac{\begin{vmatrix}3 \ -1 \ \ \ \ \ 0 \\ 11 \ \ \ \ \ 0 \ \ \ \ \ 1 \\ -3 \ \ \ \ \ \ 1 \ -2\end{vmatrix}}{\begin{vmatrix}5 \ -1 \ \ \ \ \ 0 \\ 3 \ \ \ \ \ 0 \ \ \ \ \ 1 \\ 0 \ \ \ \ \ \ 1 \ -2\end{vmatrix}}=x$ and so on.
$x=\frac {D_x}{D}; y=\frac{D_y}{D}; z=\frac{D_z}{D}$

Answer 2

$${5x -y = 3} \tag{a}$$

$${3x + z = 11}\tag{b}$$

$${y - 2z = -3}\tag{c}$$

Adding (a) and (c) we get $$ 0=3-3=5x -y+y - 2z=5x-2z\tag{d}$$

Adding (d) and 2 times (b) shows

$$ 22=6x + 2z+5x-2z=11x \qquad\implies\qquad \boxed{\phantom{\int} x = 2\phantom{\int}}$$

Substituting $x=2$ in (d) gives

$$11 = 3x+z = 6+z \qquad \implies \qquad \boxed{\phantom{\int}z = 5\phantom{\int}}$$

And substituting $x=2$ in (a) leads to

$$3 = 5x-y = 10-y \qquad \implies \qquad \boxed{\phantom{\int}y=7\phantom{\int}}$$