Chapter 13 · II · Algebra

Two Variables, One Line Worksheet

8 questions Linear equations and domains Graphs and slope Substitution and elimination Consistent and inconsistent pairs

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Foundation

Graphical solution and nature of solutions · 1 practice mark

Question 1

The pair x + y = 3 and 2x + 2y = 8 has:

  1. One solution
  2. Infinitely many solutions
  3. No solution
  4. Exactly two solutions
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(c) No solution

  1. The second equation reduces to x + y = 4.
  2. These are distinct parallel lines. No ordered pair can have both sums 3 and 4.

Consistency · 1 practice mark

Question 2

The pair x + 2y = 5 and 2x + 4y = 10 has:

  1. No solution
  2. Exactly one solution
  3. Infinitely many solutions
  4. Exactly two solutions
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(c) Infinitely many solutions

  1. The second equation is twice the first.
  2. The graphs coincide; every point on their common line satisfies both equations.

Practice

Two algebraic methods · 3 practice marks

Question 3

Solve x + y = 9 and x − y = 3 by elimination, then verify by substitution.

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x = 6, y = 3.

  1. Add the equations to eliminate y: 2x = 12, giving x = 6.
  2. From x + y = 9, y = 3.
  3. Check 6 + 3 = 9 and 6 − 3 = 3. Substitution of x = y + 3 into the first also gives 2y + 3 = 9.

Challenge

Modelling pairs of equations · Substitution method · Graphical solution and nature of solutions · 4 practice marks

Question 4

Solve x + y = 7 and y = 2x + 1 by substitution. State what the solution means on a graph.

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x = 2, y = 5; the two lines meet at (2,5).

  1. Substitute y = 2x + 1 into x + y = 7.
  2. Then 3x + 1 = 7, giving x = 2.
  3. Use y = 2x + 1 to get y = 5; verify both equations.
  4. A common ordered-pair solution is the intersection of the two graphs.

Modelling pairs of equations · Elimination method · 4 practice marks

Question 5

A shop sells 2 notebooks and 3 pens for ₹90, and 3 notebooks and 2 pens for ₹110. Use elimination to find each price.

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Notebook ₹30; pen ₹10.

  1. Let notebook price be n and pen price p. Then 2n + 3p = 90 and 3n + 2p = 110.
  2. Multiply the first equation by 3 and the second by 2: 6n + 9p = 270 and 6n + 4p = 220.
  3. Subtract to obtain 5p = 50, so p = 10.
  4. Substitute: 2n + 30 = 90, so n = 30. Check both purchase totals.

Slope, intercept and graphs including restricted domains · Graphical solution and nature of solutions · 4 practice marks

Question 6

Graph y = x + 1 and y = −x + 5 using three points per line. Find their intersection. Explain what changes if x and y must be non-negative integers.

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The lines intersect at (2,3); the restricted graphs consist only of allowed lattice points.

  1. For y = x + 1 use (0,1), (2,3), (4,5). For y = −x + 5 use (0,5), (2,3), (4,1).
  2. Plot and join each set for the real-variable graphs; they cross at (2,3).
  3. Verify 3 = 2 + 1 and 3 = −2 + 5.
  4. For integer counts retain only points with both coordinates non-negative integers. The common point is still allowed, but intermediate non-integer points are excluded.
Graph of y equals x plus one and y equals minus x plus five, meeting at two comma three(2,3)y = x + 1y = −x + 5xy0

Modelling pairs of equations · Graphical solution and nature of solutions · Substitution method · Elimination method · 4 practice marks

Question 7

Classify these pairs as uniquely solvable, consistent dependent, or inconsistent: (i) x + y = 4, x − y = 2; (ii) x + y = 4, 2x + 2y = 8; (iii) x + y = 4, x + y = 6. Describe the graphs and solve (i) by either algebraic method.

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(i) Unique: (3,1). (ii) Infinitely many solutions. (iii) No solution.

  1. For (i), add the equations to get 2x = 6; x = 3 and y = 1. Two distinct non-parallel lines intersect once.
  2. For (ii), the second equation is twice the first. Their graphs coincide and every point on x + y = 4 is a solution.
  3. For (iii), the same left side cannot equal both 4 and 6. The graphs are distinct parallel lines.
  4. Pairs (i) and (ii) are consistent; pair (iii) is inconsistent.

Restricted domains · 4 practice marks

Question 8

One adult ticket costs ₹50 and one child ticket ₹30. List every non-negative integer solution of 50a + 30c = 150. Explain why the full real-variable line is not the ticket-count graph.

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(a,c) = (0,5) or (3,0).

  1. Divide by 10: 5a + 3c = 15.
  2. Non-negative counts require a = 0,1,2 or 3; compute c = (15 − 5a)/3.
  3. Only a = 0 and a = 3 give non-negative integer c, namely 5 and 0.
  4. The real-variable line also contains fractional and negative coordinates that cannot represent ticket counts. Plot only the two admissible lattice points for this total.