Unit 5 · Topic 01 · Linear Equations
Arjun stared at (2x+1)/3 − (x−2)/4 = 1 and said equations with fractions in them were a 'different kind of problem.'
Meera said they weren't — they just needed one extra first step.
That one step turned the whole thing back into the equations he already knew how to solve.
Arjun had solved plenty of equations like 2x+3=11, but froze when he saw (2x+1)/3 − (x−2)/4 = 1. "This has fractions in it," he said. "It's different." Meera disagreed: "It's the exact same kind of equation, just dressed up. Clear the fractions first, and it turns back into what you already know."
She found the LCM of the denominators (3 and 4), which is 12, and multiplied EVERY term on both sides by 12: 12×(2x+1)/3 − 12×(x−2)/4 = 12×1. That simplifies to 4(2x+1) − 3(x−2) = 12, since 12/3=4 and 12/4=3. "No more fractions," she said. "Now just expand and solve like normal."
Arjun expanded: 8x+4−3x+6=12, giving 5x+10=12, so 5x=2, x=2/5. He checked by substituting back into the ORIGINAL fractional equation, confirming both sides matched.
Meera then showed him a bracket-heavy equation without fractions: 3(x−2) − 2(2x+1) = 5. "Expand the brackets first, being careful with the signs," she said. 3x−6−4x−2=5, giving −x−8=5, so −x=13, meaning x=−13. "Notice the negative sign on x at the end — you still need one more step, dividing both sides by −1, to get x alone."
Arjun tried a mixed one: 2(x+3)/5 = 4. He cleared the fraction first by multiplying both sides by 5: 2(x+3)=20. Then expanded: 2x+6=20, so 2x=14, x=7.
By the end, Arjun had a reliable extra first step for any equation with fractions: multiply every term on both sides by the LCM of all the denominators to clear them completely, THEN proceed exactly as with any ordinary linear equation — expand brackets, collect like terms, and isolate the variable.
To solve an equation containing fractions, first multiply EVERY term on both sides by the LCM of all the denominators — this clears the fractions completely, turning it into an ordinary linear equation.
Always expand brackets carefully before collecting terms, distributing any negative sign to every term inside the bracket it multiplies.
After clearing fractions and expanding brackets, solve exactly as any linear equation: collect variable terms on one side, constants on the other, then isolate the variable.
Verify the solution by substituting it into the ORIGINAL equation (with fractions/brackets intact), not the simplified version — this catches any error made while clearing fractions.
Solve (3x−1)/2 + (x+2)/3 = 5 by first clearing the fractions using the LCM of 2 and 3.
Solve 4(x−1) − 2(3x+2) = 10, being careful to distribute the negative sign correctly through the second bracket.
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