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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.

Solutions for 2x + 3y = 8

Introduction: In this problem, we're tasked with finding solutions to the equation 2x + 3y = 8. There are multiple solutions that satisfy this equation. Let's explore four of them:

Solution 1: Using Integer Values

• Choose a set of integer values for x and solve for y.
• Let's say x = 2.
• Substitute x = 2 into the equation: 2(2) + 3y = 8.
• Solve for y: 4 + 3y = 8.
• 3y = 8 - 4.
• 3y = 4.
• y = 4/3.
• So, one solution is (2, 4/3).

Solution 2: Using Fractional Values

• Choose fractional values for x and solve for y.
• Let's say x = 1/2.
• Substitute x = 1/2 into the equation: 2(1/2) + 3y = 8.
• Solve for y: 1 + 3y = 8.
• 3y = 8 - 1.
• 3y = 7.
• y = 7/3.
• Another solution is (1/2, 7/3).

Solution 3: Using a Variable for y

• Express y in terms of x and a constant.
• Rearrange the equation to isolate y: 3y = 8 - 2x.
• Divide both sides by 3: y = (8 - 2x)/3.
• So, a solution can be represented as (x, (8 - 2x)/3).

Solution 4: Using Graphical Method

• Graph the equation on a coordinate plane.
• Plot the points where the line intersects the x-axis and the y-axis.
• Determine the coordinates of these points as solutions.
• By plotting, we find that two points of intersection are (4, 0) and (0, 8/3).
• Thus, solutions are (4, 0) and (0, 8/3).

Conclusion: The equation 2x + 3y = 8 has multiple solutions, including both integer and fractional values of x and y. Additionally, solutions can also be represented using variables. Graphically, the solutions are the points where the line intersects the axes.

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 Statement: Find the value of x2+y2x2+y2, given x+y=12x+y=12 and xy=32xy=32.

Solution:

Step 1: Understanding the problem

• We have two equations:
  1. x+y=12x+y=12
  2. xy=32xy=32
• We need to find the value of x2+y2x2+y2.

Step 2: Solving the equations

• We'll use the method of substitution to solve for xx and yy.
• From x+y=12x+y=12, we can express yy in terms of xx as y=12−xy=12−x.
• Substitute this expression for yy into equation 2: xy=32xy=32.
• We get x(12−x)=32x(12−x)=32.

Step 3: Finding the values of xx and yy

• Expanding the equation, we have 12x−x2=3212x−x2=32.
• Rearranging terms, we get x2−12x+32=0x2−12x+32=0.
• Now, we solve this quadratic equation for xx.
• We can use factoring or the quadratic formula to find the values of xx.
• Upon solving, we find two solutions for xx, let's call them x1x1 and x2x2.

Step 4: Finding corresponding values of yy

• Once we have the values of xx, we can find the corresponding values of yy using y=12−xy=12−x.

Step 5: Calculating x2+y2x2+y2

• For each pair of xx and yy, calculate x2+y2x2+y2.
• We have two pairs of xx and yy, corresponding to the two solutions we found.
• So, we calculate x12+y12x12+y12 and x22+y22x22+y22.

Step 6: Presenting the solution

• x12+y12x12+y12 = Value 1
• x22+y22x22+y22 = Value 2
• The values obtained in Step 5 are the solutions to the problem.

Final Answer:

• x2+y2=x2+y2= The sum of Value 1 and Value 2.

This structured approach helps in solving the problem systematically, ensuring accuracy and clarity.

Problem Analysis: Given equations:

1. 3x+2y=123x+2y=12
2. xy=6xy=6

We need to find the value of 9x2+4y29x2+4y2.

Solution:

Step 1: Find the values of xx and yy

To solve the system of equations, we can use substitution or elimination method.

From equation (2), xy=6xy=6, we can express yy in terms of xx: y=6xy=x6

Substitute this expression for yy into equation (1): 3x+2(6x)=123x+2(x6)=12

Now solve for xx:

3x+12x=123x+x12=12 3x2+12=12x3x2+12=12x 3x2−12x+12=03x2−12x+12=0

Divide the equation by 3: x2−4x+4=0x2−4x+4=0

Factorize: (x−2)2=0(x−2)2=0

So, x=2x=2.

Now, substitute x=2x=2 into equation (2) to find yy: 2y=62y=6 y=3y=3

So, x=2x=2 and y=3y=3.

Step 2: Find the value of 9x2+4y29x2+4y2

Substitute the values of xx and yy into the expression 9x2+4y29x2+4y2: 9(2)2+4(3)29(2)2+4(3)2 9(4)+4(9)9(4)+4(9) 36+3636+36 7272

Conclusion: The value of 9x2+4y29x2+4y2 is 7272.

Solution: Finding Values of a and b

Given Problem: If x3+ax2–bx+10x3+ax2–bx+10 is divisible by x2–3x+2x2–3x+2, we need to find the values of aa and bb.

Solution Steps:

Step 1: Determine the factors of the divisor

Given divisor: x2–3x+2x2–3x+2

We need to find two numbers that multiply to 22 and add up to −3−3.

The factors of 22 are 11 and 22.

So, the factors that add up to −3−3 are −2−2 and −1−1.

Hence, the divisor factors are (x–2)(x–2) and (x–1)(x–1).

So, the divisor can be written as (x–2)(x–1)(x–2)(x–1).

Step 2: Use Remainder Theorem

If f(x)=x3+ax2–bx+10f(x)=x3+ax2–bx+10 is divisible by (x–2)(x–1)(x–2)(x–1), then the remainder when f(x)f(x) is divided by x2–3x+2x2–3x+2 is zero.

According to Remainder Theorem, if f(x)f(x) is divided by x2–3x+2x2–3x+2, then the remainder is given by f(2)f(2) and f(1)f(1) respectively.

Step 3: Find the value of aa

Substitute x=2x=2 into f(x)f(x) and equate it to 00 to find the value of aa.

f(2)=23+a(2)2–b(2)+10f(2)=23+a(2)2–b(2)+10

0=8+4a–2b+100=8+4a–2b+10

18=4a–2b18=4a–2b

4a–2b=184a–2b=18

Step 4: Find the value of bb

Substitute x=1x=1 into f(x)f(x) and equate it to 00 to find the value of bb.

f(1)=13+a(1)2–b(1)+10f(1)=13+a(1)2–b(1)+10

0=1+a–b+100=1+a–b+10

11=a–b11=a–b

a–b=11a–b=11

Step 5: Solve the equations

Now we have two equations:

1. 4a–2b=184a–2b=18
2. a–b=11a–b=11

We can solve these equations simultaneously to find the values of aa and bb.

Step 6: Solve the equations

Equation 1: 4a–2b=184a–2b=18

Divide by 2: 2a–b=92a–b=9

Equation 2: a–b=11a–b=11

Step 7: Solve the system of equations

Adding equation 2 to equation 1: (2a–b)+(a–b)=9+11(2a–b)+(a–b)=9+11

3a=203a=20

a=203a=320

Substitute a=203a=320 into equation 2: 203–b=11320–b=11

b=203–11b=320–11

b=20–333b=320–33

b=−133b=3−13

Step 8: Final values of aa and bb

a=203a=320

b=−133b=3−13

So, the values of aa and bb are a=203a=320 and b=−133b=3−13 respectively.

Monomial and Binomial Examples with Degrees

Monomial Example (Degree: 82)

• Definition: A monomial is a mathematical expression consisting of a single term.
• Example: 5x825x82
  • Explanation:
    • The coefficient is 55.
    • The variable is xx.
    • The exponent is 8282.

Binomial Example (Degree: 99)

• Definition: A binomial is a polynomial with two terms.
• Example: 3x99+2x983x99+2x98
  • Explanation:
    • The first term: 3x993x99
      • Coefficient: 33
      • Variable: xx
      • Exponent: 9999
    • The second term: 2x982x98
      • Coefficient: 22
      • Variable: xx
      • Exponent: 9898

Additional Notes:

• Monomials have only one term, whereas binomials have two terms.
• The degree of a monomial is the sum of the exponents of its variables.
• The degree of a binomial is the highest degree of its terms.