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Linear Equations in Two Variables

Both roads, one answer

Two badges and a flag cost ₹90. Five badges and three flags cost ₹235. Let x be a badge's price and y a flag's price. That gives 2x + y = 90 and 5x + 3y = 235. Walk both roads. A road is one method.

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Start: 2x + y = 90 and 5x + 3y = 235.
Step 1 of 4
  1. Start: 2x + y = 90 and 5x + 3y = 235.
  2. Substitution road: y = 90 − 2x, then x = 35.
  3. Elimination road: scale by 3, subtract, x = 35 too.
  4. Both roads land on (35, 20).

The rule

Rule: Substitution and elimination both describe the same two lines. The two roads cannot land on two different answers.

The first equation gives y = 90 − 2x. Put that in the second equation: 5x + 3(90 − 2x) = 235. That is 270 − x = 235, so x = 35. Then y = 90 − 70 = 20. For elimination, scale the first equation by 3: 6x + 3y = 270. Subtract the second equation. The y-terms cancel, and 6x − 5x = x while 270 − 235 = 35. So x = 35 again.

Checking (35, 20) in both original equations
EquationAt (35, 20)Claim says
2x + y = 9070 + 20 = 9090
5x + 3y = 235175 + 60 = 235235

Badges cost ₹35 each. Flags cost ₹20 each.

A letter with no number in front makes substitution easy. Coefficients that are easy to match make elimination easy. Here both roads are easy.

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This track

  1. 0 Start here
  2. 1 One claim with two letters
  3. 2 Checking a pair twice
  4. 3 All the pairs make a line
  5. 4 Two lines and one shared spot
  6. 5 The three things two lines can do
  7. 6 Checkpoint: claims, pairs and lines
  8. 7 The substitution walk
  9. 8 Scale the equations, then take one away
  10. 9 Both roads, one answer
  11. 10 What falls out when nothing is left
  12. 11 Checkpoint: the two algebra roads
  13. 12 The big picture
  14. 13 Track quiz

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