Linear Equations in Two Variables — Telangana (SSC) Class 9 Mathematics Solutions (Free)

A linear equation in two variables has infinitely many solutions — any pair (x, y) that satisfies the equation counts. We find solutions by choosing a value for one variable and solving for the other. Choose x = 0: 3(0) + 2y = 18 → 2y = 18 → y = 9. Solution: (0, 9) Choose x = 2: 3(2) + 2y = 18 → 6 + 2y = 18 → 2y = 12 → y = 6. Solution: (2, 6) Choose x = 4: 3(4) + 2y = 18 → 12 + 2y = 18 → 2y = 6 → y = 3. Solution: (4, 3) Choose x = 6: 3(6) + 2y = 18 → 18 + 2y = 18 → 2y = 0 → y = 0. Solution: (6, 0) Four solutions: (0, 9), (2, 6), (4, 3), (6, 0) — plotting these on a graph, they would all lie on the same straight line, since every solution to a linear equation in two variables lies on that equation's graph line.

Using the elimination method — since one equation has +y and the other has −y, adding the two equations directly eliminates y: 3x + y = 10 ... (1) x − y = 2 ... (2) Adding (1) and (2): (3x + x) + (y − y) = 10 + 2 4x = 12 x = 3 Substitute x = 3 into equation (2): 3 − y = 2 y = 1 Verification: 3(3) + 1 = 10 ✓, and 3 − 1 = 2 ✓ The solution is x = 3, y = 1.

Starting equation: 3x + 4y − 12 = 0 Step 1: Isolate the y term. 4y = −3x + 12 Step 2: Divide every term by 4. y = (−3/4)x + 3 This is now in the form y = mx + c, where: Slope (m) = −3/4 y-intercept (c) = 3 The negative slope means the line falls (decreases) as x increases, and the y-intercept of 3 tells us the line crosses the y-axis at the point (0, 3).

To check if a point is a solution of a system, substitute its coordinates into BOTH equations — it must satisfy every equation in the system, not just one. Checking x + 2y = 5 with (3, 1): 3 + 2(1) = 3 + 2 = 5 ✓ (matches) Checking 2x − y = 5 with (3, 1): 2(3) − 1 = 6 − 1 = 5 ✓ (matches) Since (3, 1) satisfies BOTH equations, it IS a solution of the system.

To solve graphically, find at least two points on each line, plot them, and identify where the two lines cross. For x + y = 7: if x = 0, y = 7 → point (0, 7). If x = 7, y = 0 → point (7, 0). For x − y = 3: if x = 3, y = 0 → point (3, 0). If x = 0, y = −3 → point (0, −3). Plotting both lines and drawing them out, they intersect at a single point. We can verify this algebraically: adding the two equations, (x+y) + (x−y) = 7+3, gives 2x = 10, so x = 5; substituting back, 5 + y = 7 gives y = 2. The point of intersection — and the solution to the system — is (5, 2), which a correctly drawn graph would show as the single point where both lines cross.

Since (2, 1) is a solution of both equations, substitute x = 2, y = 1 into each. From the first equation: a(2) + b(1) = 8, so 2a + b = 8 ... (1) From the second equation: 2a(2) − b(1) = 7, so 4a − b = 7 ... (2) Adding (1) and (2) to eliminate b: (2a + 4a) + (b − b) = 8 + 7 6a = 15 a = 2.5 Substitute a = 2.5 into equation (1): 2(2.5) + b = 8 5 + b = 8 b = 3 Verification: check equation (2): 4(2.5) − 3 = 10 − 3 = 7 ✓ a = 2.5, b = 3.

Using the point-slope form of a line: y − y₁ = m(x − x₁), where m is the slope and (x₁, y₁) is the known point. Given: slope m = 3, point (2, 3). y − 3 = 3(x − 2) y − 3 = 3x − 6 y = 3x − 3 Rearranging into the form ax + by + c = 0: 3x − y − 3 = 0 Verification: substituting the original point (2, 3): 3(2) − 3 − 3 = 6 − 3 − 3 = 0 ✓ The equation of the line is 3x − y − 3 = 0.

Step 1: Solve the simpler equation for one variable. From 3x − y = 9: y = 3x − 9 Step 2: Substitute this expression for y into the OTHER equation, 2x + 3y = 17: 2x + 3(3x − 9) = 17 2x + 9x − 27 = 17 11x = 44 x = 4 Step 3: Substitute x = 4 back into y = 3x − 9: y = 3(4) − 9 = 12 − 9 = 3 Verification: 2(4) + 3(3) = 8 + 9 = 17 ✓, and 3(4) − 3 = 12 − 3 = 9 ✓ The solution is x = 4, y = 3.