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.

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.

Given:

• x2+y2+z2=83x2+y2+z2=83
• x+y+z=15x+y+z=15

To Find:

• x3+y3+z3−3xyzx3+y3+z3−3xyz

Approach:

1. Use the identity (x+y+z)3=x3+y3+z3+3(x+y)(y+z)(z+x)(x+y+z)3=x3+y3+z3+3(x+y)(y+z)(z+x).
2. Given x+y+z=15x+y+z=15, find x3+y3+z3x3+y3+z3.
3. Also, find (x+y)(y+z)(z+x)(x+y)(y+z)(z+x).
4. Substitute the values in the expression x3+y3+z3−3xyzx3+y3+z3−3xyz.

Step-by-Step Solution:

1. Find x3+y3+z3x3+y3+z3:

   • Using the identity (x+y+z)3=x3+y3+z3+3(x+y)(y+z)(z+x)(x+y+z)3=x3+y3+z3+3(x+y)(y+z)(z+x), where x+y+z=15x+y+z=15.
   • (15)3=x3+y3+z3+3(x+y)(y+z)(z+x)(15)3=x3+y3+z3+3(x+y)(y+z)(z+x)
   • 3375=x3+y3+z3+3(xy+yz+zx+3xyz)3375=x3+y3+z3+3(xy+yz+zx+3xyz) (Expanding (x+y+z)3(x+y+z)3)
   • x3+y3+z3=3375−3(xy+yz+zx)x3+y3+z3=3375−3(xy+yz+zx) (Subtracting 3xyz3xyz from both sides)

2. Find (x+y)(y+z)(z+x)(x+y)(y+z)(z+x):

   • Given x+y+z=15x+y+z=15, let's find xy+yz+zxxy+yz+zx.
   • Squaring x+y+z=15x+y+z=15:
     • (x+y+z)2=(15)2(x+y+z)2=(15)2
     • x2+y2+z2+2(xy+yz+zx)=225x2+y2+z2+2(xy+yz+zx)=225 (Expanding (x+y+z)2(x+y+z)2)
     • 83+2(xy+yz+zx)=22583+2(xy+yz+zx)=225 (Given x2+y2+z2=83x2+y2+z2=83)
     • xy+yz+zx=225−832=71xy+yz+zx=2225−83=71
   • Using (x+y)(y+z)(z+x)=(xy+yz+zx)+xyz(x+y)(y+z)(z+x)=(xy+yz+zx)+xyz:
     • (x+y)(y+z)(z+x)=71+xyz(x+y)(y+z)(z+x)=71+xyz

3. Substitute values into the expression:

   • x3+y3+z3−3xyz=3375−3(71)−3xyzx3+y3+z3−3xyz=3375−3(71)−3xyz
   • x3+y3+z3−3xyz=3375−213−3xyzx3+y3+z3−3xyz=3375−213−3xyz
   • x3+y3+z3−3xyz=3162−3xyzx3+y3+z3−3xyz=3162−3xyz

Final Answer:

• x3+y3+z3−3xyz=3162−3xyzx3+y3+z3−3xyz=3162−3xyz