Two methods, one point
Some pairs are quick by one method and slow by the other.
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2x + 3y = 65 and x + y = 25.- Two facts:
2x + 3y = 65andx + y = 25. - Eliminate x:
y = 15. - Substitute:
x = 10. - Answer the question: 4 samosas + 2 chai = ₹70.
The rule
Substitution: write one letter using the other. Then replace that letter in the other equation.
Elimination: make the coefficients of one letter match. A coefficient is the number beside a letter. Subtract, or add if the signs are opposite. That letter cancels.
Example
Let x be a samosa's price, in rupees. Let y be a chai's price. The facts are x + y = 25 and 2x + 3y = 65.
From x + y = 25 write x = 25 − y. Replace x in 2x + 3y = 65: 2(25 − y) + 3y = 65. That is 50 + y = 65, so y = 15. Then x = 25 − 15 = 10.
Eliminate instead. Double x + y = 25: 2x + 2y = 50. Subtract it from 2x + 3y = 65: y = 15. Then x = 10. Both methods give the same pair.
Now 4 samosas and 2 chai cost 4×10 + 2×15 = 40 + 30 = ₹70.
Take a new pair: 4x + 3y = 30 and 4x + y = 18. The x-terms match. Subtract: 2y = 12. So y = 6. Then 4x + 6 = 18, so 4x = 12 and x = 3.
If nothing matches, multiply one equation first. Pick a number that makes the coefficients of one letter match. Multiply every term of one equation by that number. Then add or subtract.
| The pair offers | The quick method |
|---|---|
| A letter standing alone | Substitution |
| Matching coefficients | Elimination |
| Neither | Scale, then eliminate |