Class 9 Maths Ganita Manjari Chapter 13 MCQ – Two Variables, One Line

The coefficients a and b may be positive, negative or zero, but they cannot both be zero. If both were zero, the variables would disappear from the equation. One zero coefficient is allowed, as in x = 3.

The equation x = 3 can be written as x + 0y − 3 = 0. It has the required form ax + by + c = 0, with a = 1 and b = 0. Squares and products of variables make the other equations non-linear.

Multiplying every term by 2 removes the denominators and gives x − 8 = 3y. Bringing all terms to the left gives x − 3y − 8 = 0. The coefficients are a = 1 and b = −3, while c = −8.

The mangoes cost ₹60x because their price per kilogram is multiplied by x kilograms. Similarly, the bananas cost ₹50y. Adding the two costs gives the total payment, so the correct equation is 60x + 50y = 280.

Substitute x = 2 and y = 1 into the equation. The left-hand side becomes 2 × 2 + 3 × 1 = 7, equal to the right-hand side. None of the other listed pairs makes the left-hand side equal to 7.

Rearranging gives y = (12 − 3x)/2. Every real value chosen for x gives a corresponding value of y. Since infinitely many real values of x are possible, the equation has infinitely many solutions, all lying on one straight line.

Multiplying both sides of an equation by the same non-zero number preserves its solutions. Multiplying 3x + 4y = 7 by 2 gives 6x + 8y = 14. Dividing this new equation by 2 recovers the original equation.

Every point on the x-axis has y-coordinate zero. Substituting y = 0 gives 4x = 12, so x = 3. Therefore, the intersection point is (3, 0). Setting x = 0 instead would locate the intersection with the y-axis.

The origin has coordinates (0, 0). Substituting these values into ax + by + c = 0 gives c = 0. Thus, the line passes through the origin exactly when its constant term is zero, provided it is a valid linear equation.

Use slope = (y₂ − y₁)/(x₂ − x₁). Here, the change in y is −7 − 5 = −12, while the change in x is 4 − (−2) = 6. Therefore, the slope is −12/6 = −2.

To read the slope and y-intercept, first write the equation as y = mx + d. From 3x + 2y = 6, we get y = −(3/2)x + 3. Hence, the slope is −3/2 and the y-intercept is 3.

The slope −3 means that y decreases by 3 for every increase of 1 in x. Therefore, an increase of 2 in x changes y by −3 × 2 = −6. The constant 4 does not affect this change.

Every point on this line has y-coordinate 5, while x can vary. Thus, the line is horizontal and parallel to the x-axis. There is no vertical change between two distinct points, so its slope is zero.

All points on the line have x-coordinate −2, so the line is vertical and parallel to the y-axis. The horizontal change between any two distinct points is zero. Calculating slope would require division by zero, so the slope is undefined.

The slope is rise/run = 1/12. If the required horizontal distance is r cm, then 18/r = 1/12. Multiplying across gives r = 18 × 12 = 216 cm. This is the horizontal distance, not the sloping length.

A line with slope 2 has the form y = 2x + d. Since it passes through (3, 7), we get 7 = 2 × 3 + d, so d = 1. Therefore, its equation is y = 2x + 1.

A common solution must satisfy both equations. Adding the equations eliminates y and gives 3x = 9, so x = 3. Substituting into x + y = 7 gives y = 4. Also, 2 × 3 − 4 = 2 confirms the second equation.

Replace x in the second equation with y + 2. This gives 3(y + 2) + 2y = 16, or 5y + 6 = 16. Hence, y = 2 and x = y + 2 = 4. Both original equations are satisfied.

Multiplying the first equation by 2 gives 4x + 6y = 26. Multiplying the second by 3 gives 9x − 6y = 0. Adding cancels +6y and −6y, leaving 13x = 26. This is the elimination method.

The second equation is obtained by multiplying the first by 3. Therefore, both describe the same line and have exactly the same solutions. Every point on this line satisfies both equations, giving infinitely many common solutions.

Doubling the first equation gives 2x + 4y = 8, while the second requires the same expression to equal 12. Both conditions cannot hold together. Therefore, there is no common solution, and the graphs are distinct parallel lines.

The point of intersection lies on both graphs, so its coordinates satisfy both equations. In an ordered pair, the first coordinate gives x and the second gives y. Therefore, the unique solution is x = −1 and y = 3.

Both lines have slope 2, but their y-intercepts are different: 1 and −3. For every x-value, their y-values differ by 4. They therefore never meet and are distinct parallel lines, so their equations have no common solution.

A unique solution exists when the coefficient ratios are unequal. Here, 4/2 ≠ p/2 gives p ≠ 4. At p = 4, subtracting twice the second equation from the first gives the impossible statement 4 = 0, so that value is excluded.

Let x and y be the numbers of ₹50 and ₹100 notes. Then x + y = 25 and 50x + 100y = 2000. Dividing the second equation by 50 gives x + 2y = 40. Subtracting the first gives y = 15.

The extra 5 km costs ₹220 − ₹155 = ₹65, so the charge per kilometre is ₹13. The distance charge for 10 km is ₹130. Subtracting it from ₹155 leaves ₹25 as the fixed charge.

Let the tens and units digits be x and y. Then x + y = 9 and x − y = 3. Adding gives 2x = 12, so x = 6 and y = 3. The number is 10x + y = 63.

The angles of a triangle add to 180°, so x + y + 50 = 180. Substitute y = 4x to get 5x = 130. Therefore, x = 26 and y = 104. The angles 26°, 104° and 50° add to 180°.

Let the common reading be t, so F = C = t. Then t = (9/5)t + 32. Multiplying by 5 gives 5t = 9t + 160, hence −4t = 160 and t = −40. Thus, −40 °C equals −40 °F.

Let the call time be x minutes. Equal costs give 50 + 0.20x = 30 + 0.30x. Rearranging gives 20 = 0.10x, so x = 200. At this call time, both plans cost ₹90 for the month.