Learn Algebra with Free Lessons & Tips

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.

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.

Understanding Linear Equations: Linear equations are fundamental in mathematics, representing straight lines on a coordinate plane. They're expressed in the form of ax+b=0ax+b=0, where aa and bb are constants.

Identifying Axis: In the context of linear equations, the term "axis" typically refers to either the x-axis or the y-axis on a Cartesian plane.

Analyzing the Equation: The linear equation provided is x−2=0x−2=0.

Finding the Axis: To determine which axis the given linear equation is parallel to, let's analyze the equation:

1. Equation Form:

   • x−2=0x−2=0

2. Solving for x:

   • x=2x=2

3. Interpretation:

   • This equation indicates that no matter what value y takes, x will always be 2. This implies that the line represented by this equation is parallel to the y-axis.

Conclusion: The linear equation x−2=0x−2=0 is parallel to the y-axis.

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.

Determining the Value of k

Introduction: To find the value of k when (x – 1) is a factor of the polynomial 4x^3 + 3x^2 – 4x + k, we'll utilize the Factor Theorem.

Factor Theorem: If (x – c) is a factor of a polynomial, then substituting c into the polynomial should result in zero.

Procedure:

1. Substitute x=1x=1 into the polynomial to make (x – 1) a factor.
2. Equate the result to zero.
3. Solve for k.

Step-by-Step Solution:

1. Substitute x=1x=1:

   • 4(1)3+3(1)2–4(1)+k=04(1)3+3(1)2–4(1)+k=0
   • 4+3–4+k=04+3–4+k=0

2. Solve for k:

   • 3+k=03+k=0
   • k=−3k=−3

Conclusion: The value of k when (x – 1) is a factor of the given polynomial is k=−3k=−3.

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.

Perimeter Calculation for Rectangle with Given Area

Given Information:

• Area of the rectangle: 25x2−35x+1225x2−35x+12

Step 1: Determine the Dimensions

To calculate the perimeter of a rectangle, we need to know its length and width. We can find these dimensions using the area provided.

Step 2: Factorize the Area

Factorize the given quadratic expression 25x2−35x+1225x2−35x+12 to find its factors, which represent the possible lengths and widths of the rectangle.

Step 3: Use Factorization to Find Dimensions

Once the quadratic expression is factorized, identify the pairs of factors that, when multiplied, give the area of the rectangle. These pairs represent possible lengths and widths.

Step 4: Calculate Perimeter

With the length and width of the rectangle known, calculate the perimeter using the formula:

Perimeter=2×(Length+Width)Perimeter=2×(Length+Width)

Step 5: Finalize

Plug in the values of length and width into the perimeter formula to obtain the final result.

Let's proceed with these steps to find the perimeter.