6 3 Solving Systems Using Elimination: A Step-by-Step Guide
Solving systems of equations is a fundamental skill in algebra that allows us to find the point where two or more equations intersect. And one of the most effective methods for solving these systems is elimination, also known as the addition method. This technique involves manipulating the equations to eliminate one variable, making it easier to solve for the remaining variables. Which means whether you're tackling homework problems or preparing for exams, mastering elimination can significantly simplify your approach to linear systems. In this article, we’ll explore how to solve systems using elimination, break down the process into clear steps, and provide practical examples to reinforce your understanding.
What Is a System of Equations?
A system of equations consists of two or more equations that share the same variables. The goal is to find the values of these variables that satisfy all equations simultaneously. Here's one way to look at it: consider the system:
2x + 3y = 7
4x - y = 5
Here, we need to find the values of x and y that make both equations true. The elimination method is particularly useful when the coefficients of one variable are opposites or can be made opposites through multiplication.
Steps to Solve Systems Using Elimination
Step 1: Align the Equations
Write the equations in standard form (Ax + By = C) and align them vertically so that corresponding variables and constants are in the same column. For example:
2x + 3y = 7
4x - y = 5
Step 2: Multiply to Create Opposite Coefficients
If the coefficients of one variable are not already opposites, multiply one or both equations by a constant to create them. To give you an idea, in the system above, the coefficients of y are +3 and -1. To eliminate y, multiply the second equation by 3:
2x + 3y = 7
12x - 3y = 15
Step 3: Add or Subtract the Equations
Add or subtract the equations to eliminate one variable. In this case, adding the two equations eliminates y:
(2x + 12x) + (3y - 3y) = 7 + 15
14x = 22
x = 22/14 = 11/7
Step 4: Solve for the Remaining Variable
Once one variable is eliminated, solve the resulting equation for the remaining variable. Here, we found x = 11/7.
Step 5: Substitute Back to Find the Other Variable
Plug the value of the solved variable into one of the original equations to find the other variable. Using the first equation:
2(11/7) + 3y = 7
22/7 + 3y = 7
3y = 7 - 22/7
3y = 49/7 - 22/7 = 27/7
y = 27/21 = 9/7
Step 6: Check the Solution
Substitute both values into the original equations to ensure they satisfy both. For the second equation:
4(11/7) - 9/7 = 44/7 - 9/7 = 35/7 = 5
Since both equations hold true, the solution is correct: x = 11/7 and y = 9/7.
Scientific Explanation: Why Does Elimination Work?
The elimination method is rooted in the principle that if two equations are true, their sum or difference is also true. On the flip side, by strategically combining equations, we reduce the system to a simpler form without changing the solution set. On top of that, this works because:
- Addition Property of Equality: Adding the same value to both sides of an equation maintains equality. - Multiplication Property of Equality: Multiplying both sides of an equation by the same non-zero constant preserves equality.
When we multiply an equation by a constant, we create an equivalent equation. Practically speaking, adding or subtracting these equivalent equations eliminates a variable, allowing us to solve for the remaining variables. This method is particularly efficient when dealing with systems where coefficients are easily aligned through multiplication.
Real-World Applications
Systems of equations using elimination often arise in real-life scenarios. For example:
- Mixing Solutions: A chemist might mix two solutions with different concentrations to achieve a desired concentration. Also, - Budgeting: Determining how many items of two different prices can be purchased with a fixed budget. - Physics Problems: Calculating forces or velocities when two equations describe the same scenario.
Consider a problem where two friends buy tickets for a concert. One buys 3 adult tickets and 2 child tickets for $42, while the other buys 2 adult tickets and 3 child tickets for $39. To find the price of each ticket, we set up a system:
3a + 2c =
### Completing theTicket‑Price Example
Continuing from the system
\[
\begin{cases}
3a + 2c = 42\\[2pt]
2a + 3c = 39
\end{cases}
\]
we can eliminate one variable by making the coefficients of *a* (or *c*) equal.
Multiplying the first equation by 2 and the second equation by 3 gives
\[
\begin{aligned}
6a + 4c &= 84 \quad\text{(1′)}\\
6a + 9c &= 117 \quad\text{(2′)}
\end{aligned}
\]
Subtracting (1′) from (2′) eradicates *a*:
\[
(6a+9c)-(6a+4c)=117-84 \;\Longrightarrow\; 5c = 33 \;\Longrightarrow\; c = \frac{33}{5}=6.6 .
\]
Now substitute \(c = 6.6\) into the original first equation:
\[
3a + 2(6.6) = 42 \;\Longrightarrow\; 3a + 13.Consider this: 2 = 42 \;\Longrightarrow\; 3a = 28. In practice, 8 \;\Longrightarrow\; a = 9. 6 .
Thus the adult ticket costs **\$9.60** and the child ticket costs **\$6.60**.
A quick verification using the second equation:
\[
2(9.Consider this: 6) = 19. 6) + 3(6.2 + 19.
which matches the given total, confirming the solution.
### Why Elimination Is Effective
The technique hinges on the fact that adding or subtracting equations does not alter the truth of the statements involved. In real terms, by scaling one or both equations, we create *equivalent* forms whose left‑hand sides contain matching terms. On top of that, once the remaining variable is known, back‑substitution yields the other unknown. When those terms are combined, the corresponding variable disappears, leaving a single‑variable equation that is straightforward to solve. This systematic reduction is especially handy when the coefficients are integers or simple fractions, because the required scaling is usually minimal.
This is where a lot of people lose the thread.
### Additional Real‑World Situations
1. **Mixture Problems** – In a laboratory, a technician may need to combine a 30 % acid solution with a 50 % acid solution to produce 200 mL of a 40 % acid mixture. Setting up the mass balance for acid and for total volume leads to a two‑equation system that can be solved by elimination.
2. **Travel Planning** – Suppose two cyclists travel different distances at distinct speeds. If the total distance covered in a given number of hours is known for each cyclist, the system of equations can be solved to determine each rider’s speed.
3. **Economics** – When analyzing supply and demand, the equilibrium price and quantity often satisfy two linear relationships. Solving the simultaneous equations provides the market‑clearing values.
### Conclusion
The elimination method offers a clear, step‑by‑step pathway to resolve systems of linear equations. By strategically adding or subtracting appropriately scaled equations, one variable is removed, simplifying the problem to a single‑variable computation. The solved variable is then substituted back to retrieve the remaining unknown, and a final check guarantees that the pair satisfies every original equation. This approach not only underpins many textbook problems but also translates directly to practical scenarios—from laboratory chemistry to budgeting and transportation planning—making it an indispensable tool for anyone working with quantitative relationships.