Solving a system by comparison and elimination

When both equations are written as $\displaystyle {y}={\ldots}$, set the two expressions equal – this is the comparison method.

• Where the lines meet, both have the same $\displaystyle y$, so $\displaystyle {3x-4}={x+2}$.
• We solve: $\displaystyle {2x}={6}$, $\displaystyle {x}={3}$.
• We substitute in one of the equations: $\displaystyle {y}={3+2}={5}$.
• The solution of the system: $\displaystyle (3,5)$ – the meeting point on the graph.

Solve by comparison.
In part 1 the comparison’s equation is written already: solve it, work out $\displaystyle y$ and write the solution of the system.
In part 2 solve the whole system.

$\displaystyle {\begin{cases}y=2x+5\\ y=-x-1\end{cases}}$

Two equations can be added: left side to left side, right side to right side. When the coefficients of one unknown are opposite, that unknown drops out in the sum.

• $\displaystyle y$ has opposite coefficients, 1 and $\displaystyle -1$. We add: $\displaystyle {(x+y)+(x-y)}={9+3}$, that is $\displaystyle {2x}={12}$, so $\displaystyle {x}={6}$.
• We substitute in the first equation: $\displaystyle {6+y}={9}$, so $\displaystyle {y}={3}$.
• The solution of the system: $\displaystyle (6,3)$.

When the coefficients of one unknown are equal, subtract the equations – and that unknown drops out.

• $\displaystyle y$ has the same coefficient, 2. We subtract the second equation from the first: $\displaystyle {(3x+2y)-(x+2y)}={16-8}$, that is $\displaystyle {2x}={8}$, so $\displaystyle {x}={4}$.
• We substitute in the second equation: $\displaystyle {4+2y}={8}$, $\displaystyle {2y}={4}$, so $\displaystyle {y}={2}$.
• The solution of the system: $\displaystyle (4,2)$.
• Careful: every term of the second equation is subtracted, so write it in brackets.

When the coefficients are neither equal nor opposite, first multiply an equation by a number so that they are – this is the elimination method. Multiplying both sides by the same number does not change the equation’s solutions.

• $\displaystyle y$ has the coefficients 3 and $\displaystyle -1$. We multiply the second equation by 3: $\displaystyle {12x-3y}={30}$. Now the coefficients of $\displaystyle y$ are opposite.
• We add it to the first equation: $\displaystyle {14x}={42}$, so $\displaystyle {x}={3}$.
• We substitute in $\displaystyle {4x-y}={10}$: $\displaystyle {12-y}={10}$, so $\displaystyle {y}={2}$.
• The solution of the system: $\displaystyle (3,2)$. Check: $\displaystyle {2\cdot3+3\cdot2}={12}$ and $\displaystyle {4\cdot3-2}={10}$.

Solve by adding or subtracting the equations.

Solve by elimination: first multiply one of the equations by a number.

Every method solves every system, but for each system one of them is shorter:

• both equations in the form $\displaystyle {y}={\ldots}$ – comparison;
• an isolated unknown, or an unknown with coefficient 1 or $\displaystyle -1$ – substitution;
• equal or opposite coefficients – adding or subtracting the equations;
• otherwise – elimination, multiplying first.

Solve by the method that looks shortest to you.