Introduction: In this problem, we are tasked with verifying whether the values x=2x=2 and y=1y=1 satisfy the linear equation 2x+3y=72x+3y=7.

Verification: We'll substitute the given values of xx and yy into the equation and check if it holds true.

Given Equation: 2x+3y=72x+3y=7

Substituting Given Values:

• Substitute x=2x=2 and y=1y=1 into the equation. 2(2)+3(1)=72(2)+3(1)=7

Solving the Equation: 4+3=74+3=7 7=77=7

Conclusion:

• Since the equation simplifies to 7=77=7, it confirms that x=2x=2 and y=1y=1 satisfy the linear equation 2x+3y=72x+3y=7.

Therefore, the given values x=2x=2 and y=1y=1 indeed satisfy the linear equation 2x+3y=72x+3y=7.

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.

Graph of the Equation x - y = 4

Graphing the Equation:

To draw the graph of the equation x−y=4x−y=4, we'll 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.

Given equation: x−y=4x−y=4

Rewriting in slope-intercept form:

y=x−4y=x−4

Now, let's plot the graph using this equation.

Plotting the Graph:

1. Find y-intercept:
   Set x=0x=0 in the equation y=x−4y=x−4
   y=0−4y=0−4
   y=−4y=−4
   So, the y-intercept is at the point (0,−4)(0,−4).

2. Find x-intercept:
   To find the x-intercept, set y=0y=0 in the equation y=x−4y=x−4.
   0=x−40=x−4
   x=4x=4
   So, the x-intercept is at the point (4,0)(4,0).

Drawing the Graph:

Now, plot the points (0,−4)(0,−4) and (4,0)(4,0) on the Cartesian plane and draw a straight line passing through these points. This line represents the graph of the equation x−y=4x−y=4.

Intersecting with the x-axis:

To find where the graph line meets the x-axis, we need to find the point where y=0y=0.

Substitute y=0y=0 into the equation x−y=4x−y=4:

x−0=4x−0=4

x=4x=4

So, when the graph line meets the x-axis, the coordinates of the point are (4,0)(4,0).

Graphing the Equation x + 2y = 6

To graph the equation x+2y=6x+2y=6, we'll first rewrite it in slope-intercept form (y=mx+by=mx+b):

x+2y=6x+2y=6 2y=−x+62y=−x+6 y=−12x+3y=−21x+3

Plotting the Graph

To plot the graph, we'll identify two points and draw a line through them:

1. Intercept Method:

   • y-intercept (when x = 0): y=−12(0)+3=3y=−21(0)+3=3 Therefore, the y-intercept is (0, 3).
   • x-intercept (when y = 0): 0=−12x+30=−21x+3 −12x=3−21x=3 x=−6x=−6 Therefore, the x-intercept is (-6, 0).

2. Slope Method: From the slope-intercept form y=−12x+3y=−21x+3, the slope is -1/2, meaning the line decreases by 1 unit in the y-direction for every 2 units in the x-direction.

Plotting the Points and Drawing the Line

Using the intercepts and the slope, we plot the points (0, 3) and (-6, 0), then draw a line through them.

Finding the Value of x when y = -3

Given y=−3y=−3, we substitute this value into the equation y=−12x+3y=−21x+3 and solve for x:

−3=−12x+3−3=−21x+3 −12x=−3−3−21x=−3−3 −12x=−6−21x=−6 x=−6×(−2)x=−6×(−2) x=12x=12

Conclusion

• The graph of the equation x+2y=6x+2y=6 is a straight line passing through points (0, 3) and (-6, 0).
• The value of xx when y=−3y=−3 is x=12x=12.