Welcome to the fifth and final lesson of Solving Systems by Elimination! Across the previous four lessons, you learned to add equations to cancel opposites, subtract to cancel matches, scale one equation to create a match, and scale both equations using the least common multiple. With those skills, you can handle any system that has integer coefficients.
But what about equations whose coefficients are fractions like 21​ or decimals like 0.75? Working directly with them makes every arithmetic step slower and more error-prone. In this lesson, you will learn a quick preparatory step that converts those messy coefficients into integers before elimination begins. Once that conversion is done, the rest of the process is exactly what you have been doing all along.
Why Clear Before Eliminating?
Imagine trying to find the LCM of 43​ and 65​ so you can align two equations for elimination. It can be done, but every calculation along the way — scaling, subtracting, simplifying — becomes harder to manage and easier to mess up. A much better strategy is to remove the fractions first by multiplying each equation by a single, carefully chosen number.
This produces an equivalent equation — one with the same solutions — whose coefficients are all integers. The same idea works for decimals: one well-chosen multiplier shifts every decimal point out of sight. Once both equations in the system have been cleared, you are back on familiar ground and can apply any of the elimination techniques from the earlier lessons.
Choosing the Right Multiplier
The multiplier you need depends on whether the equation contains fractions or decimals.
For fractions, find the least common denominator (LCD) of every fraction in the equation. The LCD is the smallest number that all denominators divide into evenly. For example, in the equation
3x​+2y​=4
the denominators are 3 and 2, so the LCD is 6. Multiplying every term by 6:
6⋅3x​+6⋅2y​=6⋅4⟹2x+3y=24
For decimals, multiply by the power of ten that removes all decimal points. In the equation
0.3x+0.25y=4.5
the term 0.25 has the most decimal places (two), so we multiply every term by 102=100:
100(0.3x)+100(0.25y)=100(4.5)⟹30x+25y=450
In both cases the critical rule is the same: every term on both sides must be multiplied. Skipping even one term throws off the balance of the equation.
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Solving a System with Fractional Coefficients
Applying the Same Idea to Decimals
The Most Common Clearing Mistake
Conclusion and Next Steps
Clearing fractions and decimals is not a new solving method — it is a preparation step that transforms a messy equation into one with integer coefficients. For fractions, multiply every term by the LCD; for decimals, multiply every term by the appropriate power of ten. After clearing, apply the elimination techniques you already know: align, cancel, solve, back-substitute, and verify.
With this lesson, you have completed all five topics in Solving Systems by Elimination. The practice exercises ahead will put your clearing skills to the test — from identifying the right multiplier and catching missed terms, all the way to solving full systems with fractional and decimal coefficients. Jump in and see how quickly those messy numbers disappear!
Equation 1: denominators 3 and 2, so the LCD is 6. Multiply every term by 6:
6⋅3x​+6⋅2y​=6⋅4⟹2x+3y=24
Equation 2: the only denominator is 2, so the LCD is 2. Multiply every term by 2:
2⋅23x​+2⋅2y​=2⋅11⟹3x+y=22
The system is now entirely integers:
2x+3y=243x+y=22
Step 2 — Eliminate a variable. We target y. Multiply the second equation by 3 so both y-coefficients become 3:
2x+3y=249x+3y=66
Step 3 — Subtract to cancel 3y:
(9x+3y)−(2x+3y)=66−247x=42⟹x=6
Step 4 — Back-substitute. Plug x=6 into the cleared equation 3x+y=22:
18+y=22⟹y=4
Step 5 — Verify in the original fractional equations:
Equation
Substitution
Result
3x​+2y​=4
36​+24​=2+2
4=4 ✓
23x​+2y​=11
218​+24​=9+2
11=11 ✓
The solution is (x,y)=(6,4). After the clearing step, every remaining step was standard elimination — the same process you have practiced throughout this course.
Decimals appear naturally whenever measurements or money are involved. Suppose two customers at a deli order cheese and salami by the pound:
Customer A: 1.5 lb of cheese and 2 lb of salami for $27.00.
Customer B: 2 lb of cheese and 1.5 lb of salami for $25.50.
Letting c = price per pound of cheese and s = price per pound of salami:
1.5c+2s=27(Customer A)2c+1.5s=25.50(Customer B)
Step 1 — Clear decimals. The most decimal places in either equation is one (the 25.50 constant has a trailing zero in the hundredths place, but 1.5 already dictates we need at least 10). Multiplying every term of each equation by 10 removes all decimal points:
15c+20s=27020c+15s=255
All six terms happen to be divisible by 5, so we can simplify to keep the numbers small:
3c+4s=544c+3s=51
Step 2 — Eliminate c. Neither coefficient on c divides evenly into the other, so we scale both equations. The LCM of 3 and 4 is 12:
Each pound of cheese costs $6.00 and each pound of salami costs $9.00. The clearing step turned decimal coefficients into integers, and from there the problem was no different from the systems we solved in earlier lessons.
The single most frequent error when clearing is forgetting to multiply one of the terms. For example, given 2x​+3y=7, a learner might multiply only the fraction by 2 and write x+3y=7 instead of the correct x+6y=14. The integer term 3y and the constant 7 must also be multiplied for the equation to remain equivalent.
A good habit is to count the terms before and after you multiply. If the original equation has three terms (left and right sides combined), you should see three multiplications. Any term that looks unchanged from the original is a red flag worth double-checking.
