RS Aggarwal Class 9 Mathematics Solutions Chapter 8 Linear Equations in Two Variables

Question 1. (i) The given equation is x = 5. Take two solutions of the given equation as x = 5, y = 1 and x = 5, y = -1. Thus we get the following table:

Answer: Plot the points P(5,1) and Q(5,-1) on graph paper. Join PQ. The line PQ forms the needed graph.
In simple words: When x always stays at 5, no matter what value y takes, the graph is a vertical line passing through the point where x = 5.

Exam Tip: Remember that equations of the form x = constant always produce vertical lines parallel to the y-axis.

Question 1. (ii) The given equation is y = -2. Take two solutions of the given equation as x = 1, y = -2 and x = 2, y = -2.

Answer: Plot the points P(1,-2) and Q(2,-2) on graph paper. Join PQ. The line PQ gives you the required graph.
In simple words: When y always stays at -2, the graph is a horizontal line running parallel to the x-axis.

Exam Tip: Equations of the form y = constant always produce horizontal lines parallel to the x-axis.

Question 1. (iii) The given equation is x + 6 = 0, which gives x = -6. Let x = -6 & y = 1 and x = -6 & y = -1.

Answer: Plot the points P(-6,1) and Q(-6,-1) on graph paper. Join PQ. The line PQ shows the needed graph.
In simple words: Since x must equal -6, the line is vertical and passes through all points where the x-coordinate is -6.

Exam Tip: When an equation simplifies to x = constant, always look for a vertical line on the graph.

Question 1. (iv) The given equation is x + 7 = 0, which gives x = -7. Let x = -7, y = 2 and x = -7, y = 1.

Answer: Plot the points P(-7,2) and Q(-7,1) on graph paper. Join PQ. The line PQ is the graph you need.
In simple words: The x-value stays fixed at -7, so all points lie on a vertical line 7 units to the left of the y-axis.

Exam Tip: Vertical lines always have undefined slope and represent equations where x is held constant.

Question 1. (v) When y = 0, this represents the x-axis.
Answer: The equation y = 0 describes all points where the y-coordinate is zero, forming the horizontal axis of the coordinate plane.
In simple words: The x-axis is the line where every y value is 0.

Exam Tip: Know that y = 0 and x = 0 correspond to the two main axes of the coordinate system.

Question 1. (vi) When x = 0, this represents the y-axis.
Answer: The equation x = 0 describes all points where the x-coordinate is zero, forming the vertical axis of the coordinate plane.
In simple words: The y-axis is the line where every x value is 0.

Exam Tip: The axes are special reference lines; x = 0 is always the y-axis and y = 0 is always the x-axis.

Question 2. The given equation is y = 3x. Putting x = 1, y = 3(1) = 3. Putting x = 2, y = 3(2) = 6.

Answer: Plot the points (1,3) and (2,6) on graph paper and join them to get the needed graph.
In simple words: For this equation, whenever you pick a value for x, multiply it by 3 to get y. The graph forms a straight line passing through the origin with slope 3.

Exam Tip: Lines passing through the origin like y = mx always have a constant slope and represent proportional relationships.

Question 2 (Additional interpretation from diagram). When x = 4, draw a line parallel to the y-axis at a distance of 4 units from the y-axis to its right, cutting the line at Q and through Q draw a line parallel to the x-axis cutting the y-axis which is found to be at a distance of 1 unit above the x-axis. Thus, y = 1 when x = 4.
Answer: By reading from the graph, position a vertical line at x = 4. Where this line intersects the equation's graph is point Q. Draw a horizontal line through Q until it meets the y-axis. The meeting point shows that y = 1 at x = 4. This method confirms that substituting x = 4 into y = 3x gives y = 12, but the question appears to ask for graphical reading showing y = 1, suggesting we locate y = -6 for the reciprocal reading. Take a point P on the left of the y-axis such that the distance from P to the y-axis is 2 units. Draw PQ parallel to the y-axis, cutting the line y = 3x at Q. Draw QN parallel to the x-axis, meeting the y-axis at N. So y = ON = -6.
In simple words: To find a y-value on the graph for a given x-value, draw a vertical line up to the curve, then draw horizontally to the y-axis to read off the answer.

Exam Tip: Graphical methods let you estimate solutions visually; always verify with substitution into the original equation.

Question 3. The given equation is x + 2y - 3 = 0. Rearranging, x = 3 - 2y. Putting y = 1, x = 3 - (2 × 1) = 1. Putting y = 0, x = 3 - (2 × 0) = 3.

Answer: Plot the points (1,1) and (3,0) on graph paper and join them to get the needed graph. Take a point Q on the x-axis such that OQ = 5. Draw QP parallel to the y-axis, meeting the line (x = 3 - 2y) at P. Through P, draw PM parallel to the x-axis, cutting the y-axis at M. So y = OM = -1.
In simple words: Rearrange the equation to express one variable in terms of the other, find two points by substitution, plot them, and draw the line connecting them. Use the graph to find unknown values by drawing perpendiculars to the axes.

Exam Tip: Always find at least two points to sketch a linear graph accurately; the intercepts (where the line crosses the axes) are often the easiest points to locate.

Question 4. (i) The given equation is y = x. Let x = 1, then y = 1 and let x = 2, then y = 2.

Answer: Plot the points (1,1) and (2,2) on graph paper and join them to get the needed graph.
In simple words: In this equation, x and y are always equal. The graph is a diagonal line running from the lower left to the upper right, passing through the origin at a 45-degree angle.

Exam Tip: The line y = x is the line of symmetry for reflecting points across the coordinate axes; it always passes through the origin with slope 1.

Question 4. (ii) The given equation is y = -x. Now, if x = 1, y = -1 and if x = 2, y = -2.

Answer: Plot the points (1,-1) and (2,-2) on graph paper and join them to get the needed graph.
In simple words: Here, y is the opposite of x. As x increases, y decreases by the same amount. The graph is a diagonal line going from upper left to lower right, passing through the origin.

Exam Tip: When the coefficient of x is negative, the line slopes downward; a slope of -1 makes a 45-degree angle with both axes.

Question 4. (iii) The given equation is y + 3x = 0, which rearranges to y = -3x. Now, if x = -1, then y = -3(-1) = 3. And, if x = 1, then y = -3(1) = -3.

Answer: Plot the points (1,-3) and (-1,3) on graph paper and join them to get the needed graph.
In simple words: Rearrange to get y = -3x. This line passes through the origin and has a steep negative slope, meaning as x moves right, y drops quickly downward.

Exam Tip: A larger coefficient in front of x (like -3 instead of -1) makes the line steeper; always pass through the origin when there is no constant term.

Question 4. (iv) The given equation is 2x + 3y = 0. Rearranging, y = \( -\frac{2}{3} \) x. Now, if x = 3, then y = \( -\frac{2}{3} \) × 3 = -2. And, if x = -3, then y = \( -\frac{2}{3} \) × (-3) = 2.

Answer: Plot the points (3,-2) and (-3,2) on graph paper and join them to get the needed graph.
In simple words: After rearranging, the slope is \( -\frac{2}{3} \), which is a gentle downward slope. For every 3 units you move right, the line drops 2 units down.

Exam Tip: Fractional slopes still pass through the origin in equations of the form ax + by = 0; pick x-values that are multiples of the denominator to get whole number y-values.

Question 4. (v) The given equation is 3x - 2y = 0. Rearranging, y = \( \frac{3}{2} \) x. Now, if x = 2, then y = \( \frac{3}{2} \) × 2 = 3. And, if x = -2, then y = \( \frac{3}{2} \) × (-2) = -3.

Answer: Plot the points (2,3) and (-2,-3) on graph paper and join them to get the needed graph.
In simple words: The equation gives y = \( \frac{3}{2} \) x, a positive slope. The line rises steeply from lower left to upper right, passing through the origin.

Exam Tip: Positive slopes climb upward as you move from left to right; a slope greater than 1 means the rise is steeper than the run.

Question 4. (vi) The given equation is 2x + y = 0. Rearranging, y = -2x. Now, if x = 1, then y = -2(1) = -2. And, if x = -1, then y = -2(-1) = 2.

Answer: Plot the points (1,-2) and (-1,2) on graph paper and join them to get the needed graph.
In simple words: After rearranging, y = -2x shows a negative slope of -2. For every unit increase in x, y drops by 2 units.

Exam Tip: A slope of -2 is twice as steep as a slope of -1; pick simple x-values to make calculations quick and avoid fractions.

Question 5. The given equation is 2x - 3y = 5. Rearranging, y = \( \frac{2x-5}{3} \). Now, if x = 4, then y = \( \frac{2(4)-5}{3} \) = \( \frac{8-5}{3} \) = \( \frac{3}{3} \) = 1. And, if x = -2, then y = \( \frac{2(-2)-5}{3} \) = \( \frac{-4-5}{3} \) = \( \frac{-9}{3} \) = -3.

Answer: Plot the points (4,1) and (-2,-3) on graph paper and join them to get the needed graph.
In simple words: Isolate y by rearranging the equation. Substitute values for x to find matching y-values. The resulting line has a positive slope and does not pass through the origin since there is a constant term.

Exam Tip: When an equation includes both variables and a constant, the line will not pass through the origin; finding the intercepts helps you sketch it quickly.

Question 5 (Additional - from diagram interpretation). (i) When x = 4, draw a line parallel to the y-axis at a distance of 4 units from the y-axis to its right, cutting the line at Q and through Q draw a line parallel to the x-axis cutting the y-axis which is found to be at a distance of 1 unit above the x-axis. Thus, y = 1 when x = 4.
Answer: Using the graph, draw a vertical line at x = 4. This intersects the plotted line at point Q. From Q, draw a horizontal line to the y-axis, which meets it at 1 unit above the origin. Therefore, when x = 4, y = 1.
In simple words: To read values from a graph, find the x-value on the horizontal axis, go up to meet the line, then go left to read the y-value.

Exam Tip: Graphical reading is a quick way to estimate solutions; mark your perpendiculars clearly to avoid misreading the grid.

Question 5 (Additional - continuation). (ii) When y = 3, draw a line parallel to the x-axis at a distance of 3 units from the x-axis and above it, cutting the line at point P. Through P, draw a line parallel to the y-axis meeting the x-axis at a point which is found to be 7 units to the right of the y-axis. Thus, when y = 3, x = 7.
Answer: Using the graph, draw a horizontal line at y = 3. This intersects the plotted line at point P. From P, draw a vertical line downward to the x-axis, which meets it at 7 units to the right of the origin. Therefore, when y = 3, x = 7.
In simple words: To find x when you know y, work the reverse way: find y on the vertical axis, go horizontally to the line, then go down to read the x-value.

Exam Tip: Practice reading graphs in both directions - from x to find y and from y to find x - to build confidence in interpreting coordinate geometry.

Question 6. The given equation is 2x + y = 6. Rearranging, y = 6 - 2x. Now, if x = 1, then y = 6 - 2(1) = 4. And, if x = 2, then y = 6 - 2(2) = 2.

Answer: Plot the points (1,4) and (2,2) on graph paper and join them to get the needed graph. We observe that the line crosses the x-axis at a point P which is 3 units to the right of the y-axis. So, the coordinates of P are (3,0).
In simple words: Find two points by substituting x-values into the rearranged formula. The line will slant downward from left to right because the slope is negative. The x-intercept (where y = 0) can be found by setting y = 0 and solving for x.

Exam Tip: To find where a line crosses an axis, set the other variable to zero and solve; x-intercepts are found by setting y = 0 and y-intercepts by setting x = 0.

Question 7. The given equation is 3x + 2y = 6. Rearranging, 2y = 6 - 3x, so y = \( \frac{6-3x}{2} \). Now, if x = 2, then y = \( \frac{6-3(2)}{2} \) = \( \frac{6-6}{2} \) = 0. And, if x = 4, then y = \( \frac{6-3(4)}{2} \) = \( \frac{6-12}{2} \) = \( \frac{-6}{2} \) = -3.

Answer: Plot the points (2,0) and (4,-3) on graph paper and join them to get the needed graph. We observe that the line cuts the x-axis at a point P which is 3 units to the right of the y-axis. So, the coordinates of P are (3,0).
In simple words: Rearrange the equation to isolate y. Choose x-values that make the arithmetic easy - here, x = 2 gives a whole number y. The slope is negative, so the line goes downward. When y = 0, we find the x-intercept, which is an important point to locate on any linear graph.

Exam Tip: When rearranging equations, choose x-values that divide evenly to avoid fraction arithmetic; one of your points should ideally be an intercept.

Notes on Linear Equations in Two Variables

• Linear Equation in Two Variables: An equation of the form ax + by + c = 0, where a, b, and c are real numbers and at least one of a or b is non-zero. Example: 2x + 3y - 9 = 0 is a linear equation because 2, 3, and -9 are all real numbers and both a = 2 and b = 3 are non-zero.
• Infinite Solutions: A linear equation in two variables always has infinitely many solutions. Each solution is an ordered pair (x, y) that satisfies the equation.
• Graphical Representation: The graph of every linear equation in two variables forms a straight line on the coordinate plane.
• Solution Verification: To check if an ordered pair is a solution, substitute the x and y values into the equation. If both sides are equal, it is a solution. For example, to verify if (3, 4) solves 2x + 3y = 18, substitute: 2(3) + 3(4) = 6 + 12 = 18. Since the equation is satisfied, (3, 4) is a solution.