Learn Unit 2-Algebra with Free Lessons & Tips

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 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.

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.