Eighth Grade Mathematics: Solving Systems Using the Elimination Method
When both equations in a system are arranged in standard form, Ax + By = C, isolating a variable can introduce awkward fractions that complicate calculation. The elimination method, also known as linear combination, provides an elegant alternative based on the Addition Property of Equality: if equal quantities are added to equal quantities, the resulting totals remain equal. By creating opposite coefficients for one of the variables, you can add the equations vertically, causing that variable to vanish completely. In this chapter, you will master the elimination method across all levels of complexity.
The Principle of Opposites in Elimination
The foundation of the elimination method rests on additive inverses: any number plus its opposite equals zero (a + (-a) = 0). If one equation contains +4y and the second equation contains -4y, adding the two equations together cancels the y-variable entirely.
The Elimination Roadmap:
Step 1: Write both equations in standard form: Ax + By = C.
Step 2: Inspect coefficients. If necessary, multiply one or both equations by chosen
constants so that one variable has OPPOSITE coefficients (e.g., +6 and -6).
Step 3: Add the two equations vertically to eliminate that variable.
Step 4: Solve the resulting single-variable equation.
Step 5: Substitute the found value into either original equation to solve for the second variable.
Step 6: Write the solution as (x, y) and verify in both original equations.
Case 1: Direct Elimination (Opposite coefficients already present)
Equation 1: 3x + 4y = 19
Equation 2: 5x - 4y = 5
Notice the coefficients of y are +4 and -4 (perfect opposites!).
Add the equations vertically:
3x + 4y = 19
+ 5x - 4y = 5
---------------
8x + 0 = 24
8x = 24 ---> x = 3
Back-substitute x = 3 into Equation 1:
3*(3) + 4y = 19
9 + 4y = 19
4y = 10 ---> y = 2.5
Solution: (3, 2.5)
Multiplying One Equation to Create Opposites
When neither variable has opposite coefficients, you can multiply one entire equation by an integer to manufacture opposites.
Solve: 2x + 3y = 13
4x - y = 5
Option: Eliminate y by multiplying the second equation by 3!
Multiply Equation 2 by 3:
3 * (4x - y = 5) ---> 12x - 3y = 15
Now add to Equation 1:
2x + 3y = 13
+ 12x - 3y = 15
----------------
14x + 0 = 28
14x = 28 ---> x = 2
Back-substitute x = 2 into 4x - y = 5:
4*(2) - y = 5 ---> 8 - y = 5 ---> y = 3
Solution: (2, 3)
Multiplying every term of an equation preserves its truth, allowing us to align coefficients effortlessly.
Multiplying Both Equations to Create Opposites
In more advanced systems, neither coefficient is a clean multiple of the other. In this case, you find the Least Common Multiple (LCM) of the target variable's coefficients and multiply BOTH equations.
Solve: 3x + 4y = 1
2x + 3y = 2
Target x: Coefficients are 3 and 2. LCM is 6.
Multiply Equation 1 by +2: 2 * (3x + 4y = 1) ---> 6x + 8y = 2
Multiply Equation 2 by -3: -3 * (2x + 3y = 2) ---> -6x - 9y = -6
Add the modified equations:
6x + 8y = 2
+ -6x - 9y = -6
----------------
0 - y = -4 ---> y = 4
Back-substitute y = 4 into 2x + 3y = 2:
2x + 3*(4) = 2 ---> 2x + 12 = 2 ---> 2x = -10 ---> x = -5
Solution: (-5, 4)
Notice that we multiplied the second equation by a negative (-3) so that the coefficients of x became +6 and -6, which add to 0.
Chapter Exercises
Exercise 1: Solve the system of equations using elimination: 4x + 3y = 24 and 2x - 3y = -6. Show the vertical addition and verification steps.
Exercise 2: Solve the system 3x + 5y = 21 and x - 2y = -4 using elimination by multiplying one equation. Explain which variable you chose to eliminate and why.
Exercise 3: Solve the system 4x + 3y = 9 and 5x + 4y = 11 by multiplying both equations. Show the multiplication factors chosen, the addition of equations, and back-substitution.
Solutions and Explanations
Solution 1: In the system 4x + 3y = 24 and 2x - 3y = -6, the y-terms have coefficients +3 and -3, which are exact additive inverses. Add the two equations vertically: (4x + 2x) + (3y - 3y) = 24 + (-6). This yields 6x + 0 = 18. Solving for x: 6x = 18 gives x = 3. Now back-substitute x = 3 into the first equation: 4(3) + 3y = 24, which simplifies to 12 + 3y = 24. Subtract 12 from both sides: 3y = 12. Divide by 3: y = 4. The solution is (3, 4). Verifying in the second equation: 2(3) - 3*(4) = 6 - 12 = -6. The solution is confirmed.
Solution 2: In the system 3x + 5y = 21 and x - 2y = -4, the variable x in the second equation has a coefficient of 1. It is easiest to eliminate x by multiplying the entire second equation by -3, creating a coefficient of -3x to cancel with the +3x in the first equation. Multiplying the second equation by -3 gives -3(x) - 3(-2y) = -3(-4), which simplifies to -3x + 6y = 12. Now add this to the first equation: (3x - 3x) + (5y + 6y) = 21 + 12. This eliminates x, giving 11y = 33. Dividing by 11 yields y = 3. To find x, substitute y = 3 into the original second equation: x - 2(3) = -4, giving x - 6 = -4. Adding 6 to both sides yields x = 2. The solution is (2, 3). Checking in the first equation: 3(2) + 5(3) = 6 + 15 = 21, verifying the result.
Solution 3: In the system 4x + 3y = 9 and 5x + 4y = 11, choose to eliminate x. The coefficients of x are 4 and 5, with a least common multiple of 20. To create opposite coefficients of +20 and -20, multiply the first equation by 5 and the second equation by -4. Multiplying the first equation by 5 yields 5(4x + 3y = 9) ---> 20x + 15y = 45. Multiplying the second equation by -4 yields -4(5x + 4y = 11) ---> -20x - 16y = -44. Now add the two expanded equations vertically: (20x - 20x) + (15y - 16y) = 45 + (-44). This produces -y = 1, so y = -1. Now back-substitute y = -1 into the first original equation to solve for x: 4x + 3(-1) = 9, which gives 4x - 3 = 9. Adding 3 to both sides gives 4x = 12, so x = 3. The solution is (3, -1). Checking in the second equation: 5(3) + 4*(-1) = 15 - 4 = 11. The solution is verified.