How to solve system of elimination is a fundamental skill in algebra that allows you to find the values of unknown variables by systematically removing one variable at a time. The elimination method, also known as the addition‑subtraction method, works by adding or subtracting equations so that one variable cancels out, leaving a simpler equation that can be solved directly. Once that variable is known, you substitute it back into one of the original equations to find the remaining variable(s). This technique is especially useful for linear systems with two or three equations, and it forms the basis for more advanced methods such as Gaussian elimination used in higher‑dimensional problems Less friction, more output..
Understanding the Elimination Method
At its core, the elimination method relies on the principle that if two expressions are equal to the same value, their sum or difference is also equal to that value. So by manipulating the equations—multiplying them by suitable constants—you can create opposite coefficients for one of the variables. When the equations are added, that variable disappears, reducing the system’s complexity.
Key points to remember:
- Goal: Produce a new equation where one variable has a coefficient of zero.
- Tools: Multiplication of an entire equation by a non‑zero constant, and addition or subtraction of equations.
- Outcome: After elimination, solve the resulting single‑variable equation, then back‑substitute to find the other variable(s).
Step‑by‑Step Procedure
Follow these steps to solve a linear system using elimination:
- Align the equations in standard form (ax + by = c) (or (ax + by + cz = d) for three variables). Write them one under the other so that like terms are in columns.
- Choose a variable to eliminate. Look at the coefficients of each variable; pick the one that will be easiest to make opposite.
- Multiply one or both equations by appropriate constants so that the chosen variable’s coefficients are opposites (e.g., (+3) and (-3)).
- Add or subtract the equations to eliminate the chosen variable. The result is a single equation with one fewer variable.
- Solve the resulting equation for the remaining variable.
- Back‑substitute the found value into one of the original equations to solve for the other variable.
- Check your solution by substituting both values into all original equations; they should satisfy each equation.
If the system has three variables, repeat the elimination process: first eliminate one variable to obtain a two‑equation system, then eliminate a second variable to solve for the third, and finally back‑substitute upward.
Example 1: Two‑Equation System
Solve the system
[ \begin{cases} 2x + 3y = 8 \ 4x - 5y = -2 \end{cases} ]
Step 1: The equations are already aligned.
Step 2: Eliminate (x). The coefficients are 2 and 4; multiplying the first equation by (-2) gives (-4x).
Step 3: Multiply the first equation by (-2):
[ -4x - 6y = -16 ]
Step 4: Add this to the second equation:
[ (-4x - 6y) + (4x - 5y) = -16 + (-2) \ -11y = -18 ]
Step 5: Solve for (y):
[ y = \frac{-18}{-11} = \frac{18}{11} ]
Step 6: Back‑substitute into the first original equation:
[ 2x + 3\left(\frac{18}{11}\right) = 8 \ 2x + \frac{54}{11} = 8 \ 2x = 8 - \frac{54}{11} = \frac{88}{11} - \frac{54}{11} = \frac{34}{11} \ x = \frac{34}{22} = \frac{17}{11} ]
Step 7: Check:
- First equation: (2\left(\frac{17}{11}\right) + 3\left(\frac{18}{11}\right) = \frac{34}{11} + \frac{54}{11} = \frac{88}{11} = 8) ✓
- Second equation: (4\left(\frac{17}{11}\right) - 5\left(\frac{18}{11}\right) = \frac{68}{11} - \frac{90}{11} = -\frac{22}{11} = -2) ✓
Solution: (\displaystyle \left(\frac{17}{11},; \frac{18}{11}\right)).
Example 2: Three‑Equation System
Solve
[ \begin{cases} x + y + z = 6 \ 2x - y + 3z = 14 \
- x + 4y - z = -2 \end{cases} ]
Step 1: Align the equations (already done).
Step 2: Eliminate (x) from the second and third equations using the first.
- Multiply the first equation by (-2) and add to the second:
[ -2x - 2y - 2z = -12 \ ;;2x - y + 3z = 14 \ \hline -3y + z = 2 \quad\text{(Equation A)} ]
- Multiply the first equation by (+1) and add to the third:
[ x + y + z = 6 \
- x + 4y - z = -2 \ \hline 5y = 4 \quad\text{(Equation B)} ]
Step 3: Solve Equation B for (y):
[ y = \frac{4}{5} ]
Step 4: Substitute (y) into Equation A to find (z):
[ -3\left(\frac{4}{5}\right) + z = 2 \ -\frac{12}{5} + z = 2 \ z = 2 + \frac{12}{5} = \frac{10}{5} + \frac{12}{5} = \frac{22}{5} ]
Step 5: Back‑substitute (y) and (z) into the first original equation to get (x):
[ x + \frac{4}{5} + \frac{22}{5} = 6 \ x + \frac{26}{5} = 6 \ x = 6 - \frac{26}{5} = \frac{30}{5} - \frac{26}{5} = \frac{4
