Learn Unit 2-Algebra with Free Lessons & Tips

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.

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.

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.

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.

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Heading 1: The Problem at Hand - Unraveling the Equation: Let's tackle the given mathematical problem: Compute the value of 9x² + 4y² if xy = 6 and 3x + 2y = 12. Our seasoned tutors at UrbanPro are well-equipped to guide students through such challenges.

Heading 2: Step-by-Step Solution - Navigating the Equation Maze: To solve the problem, we follow a systematic approach:

Sub-heading 1: Expressing y in terms of x:

• Utilize the given information, xy = 6.
• Substitute xy = 6 into the equation 3x + 2y = 12.
• Solve for y in terms of x.

Sub-heading 2: Substitution into the Expression:

• Once y is expressed in terms of x, substitute the value into the expression 9x² + 4y².
• This substitution simplifies the expression and facilitates further calculations.

Sub-heading 3: Simplification and Calculation:

• After substitution, simplify the expression and perform the necessary calculations.
• This step-by-step approach ensures accuracy and clarity in arriving at the final result.

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