Linear Equations in Two Variables — Practice

Question 1

Express 5y = 3x − 7 in standard form ax + by + c = 0 and write a, b, c.

Accepted Answer

Step 1 — Rearrange to bring all terms to one side.
  5y = 3x − 7
  ⇒ 3x − 5y − 7 = 0 (move 5y to RHS, then everything to LHS).

Step 2 — Read off coefficients.
  Compare with ax + by + c = 0.
  a = 3, b = −5, c = −7.

✦ Answer: 3x − 5y − 7 = 0; a = 3, b = −5, c = −7.

Question 2

Is (2, −1) a solution of 3x + 2y = 4?

Accepted Answer

Step 1 — Substitute x = 2, y = −1 into the LHS.
  LHS = 3(2) + 2(−1) = 6 − 2 = 4.

Step 2 — Compare with RHS.
  RHS = 4. LHS = RHS.

✦ Answer: YES, (2, −1) is a solution.

Question 3

Is (1, 3) a solution of 2x − y + 1 = 0?

Accepted Answer

Step 1 — Substitute x = 1, y = 3.
  2(1) − 3 + 1 = 2 − 3 + 1 = 0.

Step 2 — Compare to RHS = 0.
  Both equal 0.

✦ Answer: YES, (1, 3) is a solution.

Question 4

Write the equation of the x-axis.

Accepted Answer

Step 1 — Property of the x-axis.
  Every point on the x-axis has y-coordinate 0 (regardless of x).

Step 2 — Express as an equation.
  y = 0 (or equivalently 0·x + 1·y + 0 = 0).

✦ Answer: y = 0.

Question 5

Equation of the horizontal line through (4, −2)?

Accepted Answer

Step 1 — Horizontal lines have constant y-value.
  Through (4, −2), the y-coordinate is −2.

Step 2 — Write the equation.
  y = −2.

✦ Answer: y = −2.

Question 6

Find four solutions of 2x + 3y = 12.

Accepted Answer

Step 1 — Pick convenient x-values and solve for y.
  x = 0: 3y = 12 → y = 4. Solution (0, 4).
  x = 3: 6 + 3y = 12 → y = 2. Solution (3, 2).
  x = 6: 12 + 3y = 12 → y = 0. Solution (6, 0).
  x = −3: −6 + 3y = 12 → y = 6. Solution (−3, 6).

Step 2 — Verify by substitution (sample check).
  (3, 2): 2(3) + 3(2) = 6 + 6 = 12 ✓.

✦ Answer: Four solutions are (0, 4), (3, 2), (6, 0), (−3, 6). Infinitely many more exist.

Question 7

If (3, k) is a solution of 4x + 3y = 18, find k.

Accepted Answer

Step 1 — Substitute x = 3, y = k.
  4(3) + 3k = 18.
  12 + 3k = 18.

Step 2 — Solve.
  3k = 6 → k = 2.

✦ Answer: k = 2.

Verification: (3, 2) → 4(3) + 3(2) = 12 + 6 = 18. ✓

Question 8

Find the x- and y-intercepts of 3x − 4y = 12 and sketch the line.

Accepted Answer

Step 1 — y-intercept: set x = 0.
  3(0) − 4y = 12 → −4y = 12 → y = −3.
  Point: (0, −3).

Step 2 — x-intercept: set y = 0.
  3x − 4(0) = 12 → 3x = 12 → x = 4.
  Point: (4, 0).

Step 3 — Sketch.
  Plot (0, −3) and (4, 0) on the Cartesian plane.
  Draw the straight line through them, extended in both directions.
  Verify with a third point e.g. x = 8: 3(8) − 4y = 12 → y = 3. Point (8, 3). Should lie on the line.

✦ Answer: x-intercept = 4, y-intercept = −3. Line passes through (4, 0) and (0, −3).

Question 9

A line passes through (0, 5) and (5, 0). Find its equation.

Accepted Answer

Step 1 — Recognise the two points as intercepts.
  (5, 0) is the x-intercept → x-intercept p = 5.
  (0, 5) is the y-intercept → y-intercept q = 5.

Step 2 — Use the intercept form x/p + y/q = 1.
  x/5 + y/5 = 1.

Step 3 — Multiply through by 5 to clear fractions.
  x + y = 5.

Step 4 — Write in standard form.
  x + y − 5 = 0.

✦ Answer: x + y = 5 (or equivalently x + y − 5 = 0).

Question 10

Find three solutions of 2x + y = 0 and identify what's special.

Accepted Answer

Step 1 — Compute solutions.
  x = 0: y = 0. Point (0, 0).
  x = 1: y = −2. Point (1, −2).
  x = −2: y = 4. Point (−2, 4).

Step 2 — Notice (0, 0) is the origin.
  The constant term in 2x + y = 0 is c = 0, so the line passes through the origin.

✦ Answer: (0, 0), (1, −2), (−2, 4). The line passes through the origin (because c = 0).

Question 11

If (4, 2) is a solution of px + 3y = 14, find p.

Accepted Answer

Step 1 — Substitute x = 4, y = 2.
  p(4) + 3(2) = 14.
  4p + 6 = 14.

Step 2 — Solve.
  4p = 8 → p = 2.

✦ Answer: p = 2.

Verify: 2(4) + 3(2) = 8 + 6 = 14 ✓.

Question 12

A taxi charges ₹50 booking fee plus ₹15 per km. Let x = distance in km and y = total fare. Write the linear equation, find three solutions, and identify the y-intercept and what it represents in this context.

Accepted Answer

Step 1 — Translate the verbal description.
  Total fare = booking + per-km × distance.
  y = 50 + 15x.

Step 2 — Rewrite in standard form.
  15x − y + 50 = 0.

Step 3 — Find three solutions.
  x = 0: y = 50 → (0, 50). [Fare for 0 km = just the booking fee.]
  x = 4: y = 50 + 60 = 110 → (4, 110).
  x = 10: y = 50 + 150 = 200 → (10, 200).

Step 4 — Read the y-intercept.
  y-intercept = 50 (when x = 0).
  In context: ₹50 is the BOOKING FEE — the cost when you haven't moved an inch.

Step 5 — Read the slope (gradient).
  For every 1 km, fare increases by ₹15. This is the per-km rate.

✦ Answer: Equation y = 50 + 15x; solutions (0, 50), (4, 110), (10, 200); y-intercept = 50 = booking fee.

Real-world note: x must be ≥ 0 (no negative distance) and is typically discrete (km is continuous but the meter often rounds to 100m chunks).

Question 13

Show that the line passing through (1, 1), (2, 2) and (3, 3) is y = x, and find the point on this line whose x-coordinate is 5.

Accepted Answer

Step 1 — Spot the pattern.
  For each point, y = x. Hypothesise the line y = x.

Step 2 — Verify all three points satisfy y = x.
  (1, 1): 1 = 1 ✓
  (2, 2): 2 = 2 ✓
  (3, 3): 3 = 3 ✓
  All three points lie on y = x.

Step 3 — Argue uniqueness.
  Three collinear points uniquely determine a straight line. The three given are collinear (all satisfy y = x), so the line is exactly y = x.

Step 4 — Find the point with x = 5.
  y = x = 5. Point: (5, 5).

✦ Answer: The line is y = x; the point with x = 5 is (5, 5).

Question 14

Plan A: ₹100 fixed + ₹2/min talktime. Plan B: ₹0 fixed + ₹3/min. For what talktime are the two plans equally expensive?

Accepted Answer

Step 1 — Translate each plan to a linear equation.
  Let x = talktime in minutes, y = total cost in rupees.
  Plan A: y = 100 + 2x.
  Plan B: y = 3x.

Step 2 — Set the costs equal.
  100 + 2x = 3x.

Step 3 — Solve for x.
  100 = 3x − 2x → x = 100.

Step 4 — Find the common cost.
  Substitute x = 100 into either plan.
  Plan A: y = 100 + 200 = 300.
  Plan B: y = 300. ✓ (Both agree as expected.)

Step 5 — Interpret.
  At 100 minutes both plans cost ₹300.
  Below 100 min: Plan B is cheaper (no booking fee).
  Above 100 min: Plan A is cheaper (lower per-min rate amortises the fixed fee).

✦ Answer: At 100 minutes both plans cost ₹300. This is the BREAK-EVEN POINT — exactly the intersection of the two lines.

Geometric interpretation: y = 100 + 2x and y = 3x are two lines on the Cartesian plane; they intersect at (100, 300).

Practice — Linear Equations in Two Variables

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