IB Mathematics Analysis and Approaches Question Type 1: Finding a Unique Solution to a System of Equations Algebraically Using Elimination or Substitution Exercises

Solve the system algebraically for $x$, $y$, and $z$:

$$\{\begin{array}{l}2x+y-z=3\\ x+3y+2z=14\\ 3x-y+z=4\end{array}$$

Solve the system algebraically for $x$ and $y$:

$$\{\begin{array}{l}3x+4y=-2\\ 5x-2y=14\end{array}$$

Solve the system algebraically for $x$, $y$, and $z$:

$$\{\begin{array}{l}x+y+z=6\\ x-y+z=2\\ 2x+z=5\end{array}$$

Solve the system algebraically for $x$ and $y$:

$$\{\begin{array}{l}2x+3y=13\\ 5x-3y=7\end{array}$$

Solve the system algebraically for $x$ and $y$:

$$\{\begin{array}{l}2x+y=7\\ 3x+y=10\end{array}$$

Solve the system algebraically for $x$, $y$, and $z$:

$$\{\begin{array}{l}4x-y+2z=1\\ -2x+3y+z=9\\ x+y-3z=-8\end{array}$$

Solve the system algebraically for $x$ and $y$:

$$\{\begin{array}{l}y=3x-8\\ 2x+y=7\end{array}$$

Solve the system algebraically for $x$, $y$, and $z$:

$$\{\begin{array}{l}0.5x+y-0.5z=1\\ 1.5x-y+0.5z=2\\ x+2y+z=7\end{array}$$

Adult tickets cost $12 and child tickets cost $8. If $120$ tickets were sold for a total of $1200, solve for the numbers of adult tickets $x$ and child tickets $y$ using the system:

$$\{\begin{array}{l}x+y=120\\ 12x+8y=1200\end{array}$$

Solve the system algebraically for $x$ and $y$:

$$\{\begin{array}{l}0.4x+0.2y=3\\ 1.2x-0.1y=5\end{array}$$

Solve the system algebraically for $x$ and $y$:

$$\{\begin{array}{l}y-2x=7\\ 3y+x=-1\end{array}$$

Solve the system algebraically for $x$ and $y$:

$$\{\begin{array}{l}\frac{1}{2}x+\frac{1}{3}y=5\\ x-y=1\end{array}$$

Solve the system algebraically for $x$ and $y$:

$$\{\begin{array}{l}x+y=7\\ x-y=1\end{array}$$