Solve The Given System Using The Substitution Method.${ \begin{array}{l} y = 3x - 7 \ 6x - 2y = 12 \end{array} }$A. (14, 12)B. (12, 14)C. There Is No Solution.D. There Are An Infinite Number Of Solutions.

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Introduction

The substitution method is a technique used to solve systems of linear equations. It involves solving one equation for a variable and then substituting that expression into the other equation. This method is particularly useful when one of the equations is already solved for one of the variables.

Step 1: Solve the First Equation for y

The first equation is already solved for y:

y=3x−7y = 3x - 7

Step 2: Substitute the Expression for y into the Second Equation

Substitute the expression for y into the second equation:

6x−2(3x−7)=126x - 2(3x - 7) = 12

Step 3: Simplify the Equation

Expand and simplify the equation:

6x−6x+14=126x - 6x + 14 = 12

Step 4: Solve for x

Solve for x:

14=1214 = 12

This is a contradiction, which means that there is no solution to the system of equations.

Conclusion

Since there is no solution to the system of equations, the correct answer is:

  • C. There is no solution.

Discussion

The substitution method is a powerful tool for solving systems of linear equations. However, it is not always the most efficient method, especially when the equations are complex. In such cases, other methods such as the elimination method or matrix operations may be more suitable.

Example 2

Consider the following system of equations:

y=2x+1x+2y=6\begin{array}{l} y = 2x + 1 \\ x + 2y = 6 \end{array}

Step 1: Solve the First Equation for y

The first equation is already solved for y:

y=2x+1y = 2x + 1

Step 2: Substitute the Expression for y into the Second Equation

Substitute the expression for y into the second equation:

x+2(2x+1)=6x + 2(2x + 1) = 6

Step 3: Simplify the Equation

Expand and simplify the equation:

x+4x+2=6x + 4x + 2 = 6

Step 4: Solve for x

Solve for x:

5x+2=65x + 2 = 6

5x=45x = 4

x=45x = \frac{4}{5}

Step 5: Find the Value of y

Substitute the value of x into the first equation to find the value of y:

y=2(45)+1y = 2\left(\frac{4}{5}\right) + 1

y=85+1y = \frac{8}{5} + 1

y=85+55y = \frac{8}{5} + \frac{5}{5}

y=135y = \frac{13}{5}

Conclusion

The solution to the system of equations is:

  • x = 4/5
  • y = 13/5

Discussion

The substitution method is a useful tool for solving systems of linear equations. However, it is not always the most efficient method, especially when the equations are complex. In such cases, other methods such as the elimination method or matrix operations may be more suitable.

Example 3

Consider the following system of equations:

y=x2−2x+1x+y=3\begin{array}{l} y = x^2 - 2x + 1 \\ x + y = 3 \end{array}

Step 1: Substitute the Expression for y into the Second Equation

Substitute the expression for y into the second equation:

x+(x2−2x+1)=3x + (x^2 - 2x + 1) = 3

Step 2: Simplify the Equation

Expand and simplify the equation:

x+x2−2x+1=3x + x^2 - 2x + 1 = 3

Step 3: Rearrange the Equation

Rearrange the equation to form a quadratic equation:

x2−x+1=0x^2 - x + 1 = 0

Step 4: Solve the Quadratic Equation

Solve the quadratic equation using the quadratic formula:

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

x=1±(−1)2−4(1)(1)2(1)x = \frac{1 \pm \sqrt{(-1)^2 - 4(1)(1)}}{2(1)}

x=1±1−42x = \frac{1 \pm \sqrt{1 - 4}}{2}

x=1±−32x = \frac{1 \pm \sqrt{-3}}{2}

Conclusion

Since the quadratic equation has no real solutions, there is no solution to the system of equations.

Discussion

The substitution method can be used to solve systems of linear equations. However, it is not always the most efficient method, especially when the equations are complex. In such cases, other methods such as the elimination method or matrix operations may be more suitable.

Example 4

Consider the following system of equations:

y=2x−3x+2y=7\begin{array}{l} y = 2x - 3 \\ x + 2y = 7 \end{array}

Step 1: Solve the First Equation for y

The first equation is already solved for y:

y=2x−3y = 2x - 3

Step 2: Substitute the Expression for y into the Second Equation

Substitute the expression for y into the second equation:

x+2(2x−3)=7x + 2(2x - 3) = 7

Step 3: Simplify the Equation

Expand and simplify the equation:

x+4x−6=7x + 4x - 6 = 7

Step 4: Solve for x

Solve for x:

5x−6=75x - 6 = 7

5x=135x = 13

x=135x = \frac{13}{5}

Step 5: Find the Value of y

Substitute the value of x into the first equation to find the value of y:

y=2(135)−3y = 2\left(\frac{13}{5}\right) - 3

y=265−3y = \frac{26}{5} - 3

y=265−155y = \frac{26}{5} - \frac{15}{5}

y=115y = \frac{11}{5}

Conclusion

The solution to the system of equations is:

  • x = 13/5
  • y = 11/5

Discussion

The substitution method is a useful tool for solving systems of linear equations. However, it is not always the most efficient method, especially when the equations are complex. In such cases, other methods such as the elimination method or matrix operations may be more suitable.

Example 5

Consider the following system of equations:

y=x2+2x−3x+y=5\begin{array}{l} y = x^2 + 2x - 3 \\ x + y = 5 \end{array}

Step 1: Substitute the Expression for y into the Second Equation

Substitute the expression for y into the second equation:

x+(x2+2x−3)=5x + (x^2 + 2x - 3) = 5

Step 2: Simplify the Equation

Expand and simplify the equation:

x+x2+2x−3=5x + x^2 + 2x - 3 = 5

Step 3: Rearrange the Equation

Rearrange the equation to form a quadratic equation:

x2+3x−8=0x^2 + 3x - 8 = 0

Step 4: Solve the Quadratic Equation

Solve the quadratic equation using the quadratic formula:

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

x=−3±(3)2−4(1)(−8)2(1)x = \frac{-3 \pm \sqrt{(3)^2 - 4(1)(-8)}}{2(1)}

x=−3±9+322x = \frac{-3 \pm \sqrt{9 + 32}}{2}

x=−3±412x = \frac{-3 \pm \sqrt{41}}{2}

Conclusion

The solution to the system of equations is:

  • x = (-3 + sqrt(41))/2
  • y = ((-3 + sqrt(41))/2)^2 + 2((-3 + sqrt(41))/2) - 3

Discussion

The substitution method can be used to solve systems of linear equations. However, it is not always the most efficient method, especially when the equations are complex. In such cases, other methods such as the elimination method or matrix operations may be more suitable.

Conclusion

The substitution method is a useful tool for solving systems of linear equations. However, it is not always the most efficient method, especially when the equations are complex. In such cases, other methods such as the elimination method or matrix operations may be more suitable.

References

  • [1] "Algebra and Trigonometry" by Michael Sullivan
  • [2] "Linear Algebra and Its Applications" by Gilbert Strang
  • [3] "Calculus" by Michael Spivak

Keywords

  • Substitution method
  • Systems of linear equations
  • Algebra
  • Trigonometry
  • Linear algebra
  • Calculus

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Introduction

The substitution method is a technique used to solve systems of linear equations. It involves solving one equation for a variable and then substituting that expression into the other equation. This method is particularly useful when one of the equations is already solved for one of the variables.

Q: What is the substitution method?

A: The substitution method is a technique used to solve systems of linear equations by substituting one equation into the other equation.

Q: When should I use the substitution method?

A: You should use the substitution method when one of the equations is already solved for one of the variables.

Q: How do I solve a system of equations using the substitution method?

A: To solve a system of equations using the substitution method, follow these steps:

  1. Solve one equation for one of the variables.
  2. Substitute the expression for the variable into the other equation.
  3. Simplify the equation and solve for the other variable.
  4. Substitute the value of the other variable back into one of the original equations to find the value of the first variable.

Q: What are some common mistakes to avoid when using the substitution method?

A: Some common mistakes to avoid when using the substitution method include:

  • Not solving one equation for one of the variables before substituting it into the other equation.
  • Not simplifying the equation after substituting the expression for the variable.
  • Not checking the solution to make sure it satisfies both original equations.

Q: Can I use the substitution method to solve systems of nonlinear equations?

A: No, the substitution method is only used to solve systems of linear equations. If you have a system of nonlinear equations, you will need to use a different method, such as the elimination method or matrix operations.

Q: How do I know if the substitution method will work for a particular system of equations?

A: The substitution method will work for a particular system of equations if one of the equations is already solved for one of the variables. If neither equation is solved for one of the variables, you will need to use a different method, such as the elimination method or matrix operations.

Q: Can I use the substitution method to solve systems of equations with more than two variables?

A: Yes, you can use the substitution method to solve systems of equations with more than two variables. However, it may be more complicated and require more steps.

Q: What are some real-world applications of the substitution method?

A: The substitution method has many real-world applications, including:

  • Solving systems of equations in physics and engineering to model real-world problems.
  • Solving systems of equations in economics to model economic systems.
  • Solving systems of equations in computer science to model computer algorithms.

Q: How do I check my solution to make sure it satisfies both original equations?

A: To check your solution, substitute the values of the variables back into both original equations and make sure they are true.

Q: What are some common errors to watch out for when using the substitution method?

A: Some common errors to watch out for when using the substitution method include:

  • Not simplifying the equation after substituting the expression for the variable.
  • Not checking the solution to make sure it satisfies both original equations.
  • Not using the correct method for solving the system of equations.

Q: Can I use the substitution method to solve systems of equations with fractions or decimals?

A: Yes, you can use the substitution method to solve systems of equations with fractions or decimals. However, you will need to follow the same steps as before, but with the added complexity of fractions or decimals.

Q: How do I know if the substitution method is the best method for a particular system of equations?

A: The substitution method is the best method for a particular system of equations if one of the equations is already solved for one of the variables. If neither equation is solved for one of the variables, you will need to use a different method, such as the elimination method or matrix operations.

Q: Can I use the substitution method to solve systems of equations with absolute values?

A: No, the substitution method is not typically used to solve systems of equations with absolute values. If you have a system of equations with absolute values, you will need to use a different method, such as the elimination method or matrix operations.

Q: How do I know if the substitution method will work for a particular system of equations with absolute values?

A: The substitution method will not work for a particular system of equations with absolute values. If you have a system of equations with absolute values, you will need to use a different method, such as the elimination method or matrix operations.

Q: Can I use the substitution method to solve systems of equations with complex numbers?

A: No, the substitution method is not typically used to solve systems of equations with complex numbers. If you have a system of equations with complex numbers, you will need to use a different method, such as the elimination method or matrix operations.

Q: How do I know if the substitution method will work for a particular system of equations with complex numbers?

A: The substitution method will not work for a particular system of equations with complex numbers. If you have a system of equations with complex numbers, you will need to use a different method, such as the elimination method or matrix operations.

Conclusion

The substitution method is a useful tool for solving systems of linear equations. However, it is not always the most efficient method, especially when the equations are complex. In such cases, other methods such as the elimination method or matrix operations may be more suitable.

References

  • [1] "Algebra and Trigonometry" by Michael Sullivan
  • [2] "Linear Algebra and Its Applications" by Gilbert Strang
  • [3] "Calculus" by Michael Spivak

Keywords

  • Substitution method
  • Systems of linear equations
  • Algebra
  • Trigonometry
  • Linear algebra
  • Calculus