5.3 Solve Systems of Equations by Elimination

Solving Systems of Equations by Elimination

The Elimination Method

The elimination method solves a system of equations by adding or subtracting the equations so that one variable cancels out. Once a variable is gone, you're left with a simple one-variable equation you can solve directly.

Here are the steps:

1. Set up the equations so both are in standard form ($Ax + By = C$), with like terms lined up vertically.
2. Multiply one or both equations by a constant so that the coefficients of one variable are exact opposites (like $2y$ and $-2y$). Pick whichever variable is easiest to eliminate.
3. Add the equations together. The opposite terms cancel, leaving one equation with one variable.
4. Solve that equation for the remaining variable.
5. Substitute that value back into either original equation to find the other variable.
6. Check your solution by plugging both values into both original equations. If both equations are true, you've got it.

Example:

Solve the system: $3x + 2y = 11$ and $2x - 2y = 2$

Notice the $y$-coefficients are already opposites ($+2y$ and $-2y$), so no multiplication is needed. Add the equations directly:

$3x + 2y + 2x - 2y = 11 + 2$

$5x = 13$

$x = \frac{13}{5}$

Hmm, that gives a fraction. Let's re-examine. Actually, adding these equations straight gives $5x = 13$, so $x = \frac{13}{5}$. But the original guide claimed $x = 1, y = 2$. Let's check: $3(1) + 2(2) = 7 \neq 11$. So the original example had an error.

Here's a corrected example that works cleanly:

Solve the system: $3x + 2y = 7$ and $2x - 2y = 8$

The $y$-coefficients ($+2$ and $-2$) are already opposites. Add the equations:

$3x + 2y + 2x - 2y = 7 + 8$

$5x = 15$

$x = 3$

Substitute $x = 3$ into the first equation:

$3(3) + 2y = 7$

$9 + 2y = 7$

$2y = -2$

$y = -1$

Solution: $x = 3$, $y = -1$

Check: Plug into both equations. First: $3(3) + 2(-1) = 9 - 2 = 7$ ✓ Second: $2(3) - 2(-1) = 6 + 2 = 8$ ✓

When You Need to Multiply First

Sometimes neither variable has opposite coefficients right away. In that case, multiply one or both equations by a constant before adding.

Example:

Solve: $2x + 3y = 12$ and $5x + 2y = 13$

To eliminate $y$, make the $y$-coefficients opposites. Multiply the first equation by $2$ and the second by $-3$:

$4x + 6y = 24$

$-15x - 6y = -39$

Add them:

$-11x = -15$

$x = \frac{15}{11}$

That's messy, so try eliminating $x$ instead. Multiply the first equation by $5$ and the second by $-2$:

$10x + 15y = 60$

$-10x - 4y = -26$

Add them:

$11y = 34$

$y = \frac{34}{11}$

Still not clean. The point is: the method works the same way regardless. Pick whichever variable leads to simpler arithmetic, and don't worry if the answer is a fraction.

Real-World Applications

Word problems using elimination follow a consistent pattern:

1. Identify the unknowns and assign each a variable.
2. Write two equations from the information given. You need exactly two equations for two unknowns.
3. Solve using elimination.
4. Interpret your answer in context. Make sure it's reasonable (you can't sell negative baskets).

Example:

A small business sells regular gift baskets for $\$30$ each and deluxe baskets for $\$50$ each. They sold 20 baskets total and earned $\$700$ in revenue. How many of each type did they sell?

Let $x$ = regular baskets and $y$ = deluxe baskets.

• Total baskets: $x + y = 20$
• Total revenue: $30x + 50y = 700$

Multiply the first equation by $-30$:

$-30x - 30y = -600$

Add to the second equation:

$30x + 50y - 30x - 30y = 700 - 600$

$20y = 100$

$y = 5$

Substitute back: $x + 5 = 20$, so $x = 15$.

Interpretation: The business sold 15 regular baskets and 5 deluxe baskets.

Check: $15 + 5 = 20$ ✓ and $30(15) + 50(5) = 450 + 250 = 700$ ✓

Choosing the Right Method

Not every system is best solved by elimination. Here's a quick guide for deciding:

Solve: $y = 2x + 1$ and $4x + 2y = 14$

Since the first equation already gives you $y$ in terms of $x$, substitute directly:

$4x + 2(2x + 1) = 14$

$4x + 4x + 2 = 14$

$8x = 12$

$x = \frac{3}{2}$

Then $y = 2\left(\frac{3}{2}\right) + 1 = 4$.

Solution: $x = \frac{3}{2}$, $y = 4$

Key Vocabulary

• Simultaneous equations: Another name for a system of equations solved together.
• Linear combination: The formal name for what you're doing in elimination: multiplying equations by constants and adding them to produce a new equation.
• Cancellation: What happens when opposite terms (like $3y$ and $-3y$) are added and become zero, removing that variable from the equation.