Graph of the Equation 2x – 3y = 12

To draw the graph of the equation 2x−3y=122x−3y=12, let's first rewrite it in slope-intercept form, which is y=mx+by=mx+b, where mm is the slope and bb is the y-intercept.

Rewrite Equation in Slope-Intercept Form

2x−3y=122x−3y=12
−3y=−2x+12−3y=−2x+12
y=23x−4y=32x−4

Plotting the y-intercept and Slope

1. Y-intercept: When x=0x=0,
   y=23(0)−4y=32(0)−4
   y=−4y=−4
   So, the y-intercept is (0, -4).

2. Slope: The coefficient of xx is 2332, which represents the slope.
   For every increase of 1 in xx, yy increases by 2332.
   For every decrease of 1 in xx, yy decreases by 2332.

Plotting Points and Drawing the Graph

Now, let's plot some points to draw the graph:

• x = 3: y=23(3)−4=2−4=−2y=32(3)−4=2−4=−2
  Point: (3, -2)

• x = 6: y=23(6)−4=4−4=0y=32(6)−4=4−4=0
  Point: (6, 0)

• x = -3: y=23(−3)−4=−2−4=−6y=32(−3)−4=−2−4=−6
  Point: (-3, -6)

Plotting the Graph

With these points, we can draw a straight line passing through them.

Points where the Graph Intersects the Axes

X-axis

To find where the graph intersects the x-axis, we set y=0y=0 and solve for xx:

0=23x−40=32x−4
23x=432x=4
x=4×32x=24×3
x=6x=6

So, the graph intersects the x-axis at x=6x=6, which corresponds to the point (6, 0).

Y-axis

To find where the graph intersects the y-axis, we set x=0x=0 and solve for yy:

y=23(0)−4y=32(0)−4
y=−4y=−4

So, the graph intersects the y-axis at y=−4y=−4, which corresponds to the point (0, -4).

Summary

• X-axis intersection: (6, 0)
• Y-axis intersection: (0, -4)

This information helps us visualize and understand the behavior of the equation 2x−3y=122x−3y=12 on the coordinate plane.

Graph of 9x – 5y + 160 = 0

To graph the equation 9x – 5y + 160 = 0, we'll first rewrite it in slope-intercept form, y = mx + b, where m is the slope and b is the y-intercept.

Step 1: Rewrite the equation in slope-intercept form

9x – 5y + 160 = 0

Subtract 9x from both sides:

-5y = -9x - 160

Divide both sides by -5 to isolate y:

y = (9/5)x + 32

Now we have the equation in slope-intercept form.

Step 2: Identify the slope and y-intercept

The slope (m) is 9/5 and the y-intercept (b) is 32.

Step 3: Plot the y-intercept and use the slope to find additional points

Now, let's plot the y-intercept at (0, 32). From there, we'll use the slope to find another point. The slope of 9/5 means that for every 5 units we move to the right along the x-axis, we move 9 units upwards along the y-axis.

So, starting from (0, 32), if we move 5 units to the right, we move 9 units up to get the next point.

Step 4: Plot the points and draw the line

Plot the y-intercept at (0, 32) and the next point at (5, 41). Then, draw a line through these points to represent the graph of the equation.

Finding the value of y when x = 5

To find the value of y when x = 5, we'll substitute x = 5 into the equation and solve for y.

9x – 5y + 160 = 0

9(5) – 5y + 160 = 0

45 – 5y + 160 = 0

Combine like terms:

-5y + 205 = 0

Subtract 205 from both sides:

-5y = -205

Divide both sides by -5 to solve for y:

y = 41

So, when x = 5, y = 41.

Finding Solutions of Line AB Equation

Given Information:

• Line AB is represented by the equation.
• A graph depicting Line AB is provided.

Procedure:

1. Identify Points on Line AB: Locate four points on the graph that lie on Line AB.
2. Determine Coordinates: Extract the coordinates of these points.
3. Substitute Coordinates: Substitute the coordinates into the equation of Line AB.
4. Verify Solutions: Confirm that the substituted coordinates satisfy the equation of Line AB.

1. Identify Points on Line AB:

• Locate four distinct points where the line intersects the axes or stands out on the graph.

2. Determine Coordinates:

• Note down the coordinates (x, y) of each identified point.

3. Substitute Coordinates:

• Use the coordinates obtained to substitute into the equation of Line AB.
• The equation of a line is typically in the form y = mx + b, where m is the slope and b is the y-intercept.

4. Verify Solutions:

• Confirm that the substituted coordinates satisfy the equation of Line AB.
• The substituted values should make the equation true when solved.

Example:

• Suppose the equation representing Line AB is y = 2x + 3.
• Points on the graph are (0, 3), (1, 5), (2, 7), and (-1, 1).
• Substituting these coordinates into the equation:
  • For (0, 3): 3 = 2(0) + 3 (True)
  • For (1, 5): 5 = 2(1) + 3 (True)
  • For (2, 7): 7 = 2(2) + 3 (True)
  • For (-1, 1): 1 = 2(-1) + 3 (True)
• All points satisfy the equation, confirming they lie on Line AB.

Conclusion:

• By following these steps, you can find solutions of the equation representing Line AB from the provided graph.

Writing a Linear Equation for Taxi Fare

Given Information:

• Initial fare: Rs 10 for the first kilometre
• Subsequent fare: Rs 6 per km
• Distance: xx km
• Total fare: Rs yy

Formulating the Linear Equation

Let's denote:

• xx: Distance travelled in kilometres
• yy: Total fare in rupees

Equation for Total Fare:

The total fare can be calculated as the sum of the initial fare and the fare for the subsequent distance.

So, the equation can be expressed as:

y=10+6(x−1)y=10+6(x−1)

Where:

• x−1x−1: Represents the distance after the first kilometre

Calculating Total Fare for 15 km

Now, let's substitute x=15x=15 into the equation to find the total fare for a 15 km journey.

y=10+6(15−1)y=10+6(15−1) y=10+6(14)y=10+6(14) y=10+84y=10+84 y=94y=94

Answer:

The total fare for a 15 km journey would be Rs. 94.

Problem Analysis:

Given the equation 2x−y=p2x−y=p and a solution point (1,−2)(1,−2), we need to find the value of pp.

Solution:

Step 1: Substitute the Given Solution into the Equation

Substitute the coordinates of the given solution point (1,−2)(1,−2) into the equation:

2(1)−(−2)=p2(1)−(−2)=p

Step 2: Solve for pp

2+2=p2+2=p 4=p4=p

Step 3: Final Result

p=4p=4

Conclusion:

The value of pp for the equation 2x−y=p2x−y=p when the point (1,−2)(1,−2) is a solution is 44.