Solve The System Of Equations Below.${ \begin{array}{l} 5x + 2y = 9 \ 2x - 3y = 15 \end{array} }$A. { (3, -3)$}$ B. { (-3, 3)$}$ C. { (12, -3)$}$ D. { (-3, 12)$}$

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Introduction

In mathematics, a system of linear equations is a set of two or more linear equations that are solved simultaneously to find the values of the variables. In this article, we will focus on solving a system of two linear equations with two variables. We will use the method of substitution and elimination to find the solution.

The System of Equations

The system of equations we will be solving is:

5x+2y=92x−3y=15{ \begin{array}{l} 5x + 2y = 9 \\ 2x - 3y = 15 \end{array} }

Method of Substitution

One way to solve this system of equations is by using the method of substitution. This method involves solving one of the equations for one variable and then substituting that expression into the other equation.

Let's start by solving the first equation for x:

5x+2y=9{ 5x + 2y = 9 }

Subtracting 2y from both sides gives:

5x=9−2y{ 5x = 9 - 2y }

Dividing both sides by 5 gives:

x=9−2y5{ x = \frac{9 - 2y}{5} }

Now, substitute this expression for x into the second equation:

2x−3y=15{ 2x - 3y = 15 }

Substituting x = (9 - 2y)/5 gives:

2(9−2y5)−3y=15{ 2(\frac{9 - 2y}{5}) - 3y = 15 }

Multiplying both sides by 5 to eliminate the fraction gives:

2(9−2y)−15y=75{ 2(9 - 2y) - 15y = 75 }

Expanding the left-hand side gives:

18−4y−15y=75{ 18 - 4y - 15y = 75 }

Combine like terms:

18−19y=75{ 18 - 19y = 75 }

Subtracting 18 from both sides gives:

−19y=57{ -19y = 57 }

Dividing both sides by -19 gives:

y=−5719{ y = -\frac{57}{19} }

y=−3{ y = -3 }

Method of Elimination

Another way to solve this system of equations is by using the method of elimination. This method involves adding or subtracting the equations to eliminate one of the variables.

Let's start by multiplying the first equation by 3 and the second equation by 2 to make the coefficients of y opposites:

3(5x+2y)=3(9){ 3(5x + 2y) = 3(9) }

2(2x−3y)=2(15){ 2(2x - 3y) = 2(15) }

Expanding the left-hand side gives:

15x+6y=27{ 15x + 6y = 27 }

4x−6y=30{ 4x - 6y = 30 }

Now, add the two equations to eliminate y:

(15x+6y)+(4x−6y)=27+30{ (15x + 6y) + (4x - 6y) = 27 + 30 }

Combine like terms:

19x=57{ 19x = 57 }

Dividing both sides by 19 gives:

x=5719{ x = \frac{57}{19} }

x=3{ x = 3 }

Conclusion

We have solved the system of linear equations using both the method of substitution and the method of elimination. The solution to the system is x = 3 and y = -3.

Answer

The correct answer is:

(3,−3){ (3, -3) }

This solution satisfies both equations:

5(3)+2(−3)=15−6=9{ 5(3) + 2(-3) = 15 - 6 = 9 }

2(3)−3(−3)=6+9=15{ 2(3) - 3(-3) = 6 + 9 = 15 }

Therefore, the solution to the system of equations is x = 3 and y = -3.

Discussion

This system of equations can be solved using either the method of substitution or the method of elimination. The method of substitution involves solving one of the equations for one variable and then substituting that expression into the other equation. The method of elimination involves adding or subtracting the equations to eliminate one of the variables.

In this example, we used the method of substitution to solve for y and then substituted that expression into the other equation to solve for x. We also used the method of elimination to add the two equations to eliminate y and then solve for x.

Both methods are valid and can be used to solve this system of equations. However, the method of elimination is often faster and more efficient, especially for larger systems of equations.

Final Answer

The final answer is:

(3,−3){ (3, -3) }

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Introduction

In our previous article, we solved a system of linear equations using the method of substitution and the method of elimination. In this article, we will answer some common questions related to solving systems of linear equations.

Q: What is a system of linear equations?

A system of linear equations is a set of two or more linear equations that are solved simultaneously to find the values of the variables.

A: How do I know if a system of linear equations has a solution?

A system of linear equations has a solution if the two equations are consistent, meaning that they do not contradict each other. If the two equations are inconsistent, there is no solution.

Q: What is the difference between the method of substitution and the method of elimination?

The method of substitution involves solving one of the equations for one variable and then substituting that expression into the other equation. The method of elimination involves adding or subtracting the equations to eliminate one of the variables.

A: Which method is faster and more efficient?

The method of elimination is often faster and more efficient, especially for larger systems of equations.

Q: How do I know if I have the correct solution?

To check if you have the correct solution, substitute the values of x and y back into both original equations. If the equations are true, then you have the correct solution.

A: What if I have a system of linear equations with three or more variables?

To solve a system of linear equations with three or more variables, you can use the method of substitution or the method of elimination, or you can use a matrix method.

Q: Can I use a calculator to solve a system of linear equations?

Yes, you can use a calculator to solve a system of linear equations. Many calculators have a built-in function to solve systems of linear equations.

A: What if I have a system of linear equations with fractions or decimals?

To solve a system of linear equations with fractions or decimals, you can multiply both sides of the equation by a common denominator to eliminate the fractions or decimals.

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

Yes, you can use the method of substitution to solve a system of linear equations with fractions or decimals.

A: What if I have a system of linear equations with absolute values or square roots?

To solve a system of linear equations with absolute values or square roots, you can use the method of substitution or the method of elimination, or you can use a matrix method.

Q: Can I use a graphing calculator to solve a system of linear equations?

Yes, you can use a graphing calculator to solve a system of linear equations. Many graphing calculators have a built-in function to solve systems of linear equations.

A: What if I have a system of linear equations with non-linear equations?

To solve a system of linear equations with non-linear equations, you can use a matrix method or a numerical method.

Conclusion

Solving a system of linear equations can be a challenging task, but with the right methods and tools, you can find the solution. Remember to check your work and use a calculator or graphing calculator to help you solve the system.

Final Answer

The final answer is:

  • A system of linear equations is a set of two or more linear equations that are solved simultaneously to find the values of the variables.
  • The method of substitution involves solving one of the equations for one variable and then substituting that expression into the other equation.
  • The method of elimination involves adding or subtracting the equations to eliminate one of the variables.
  • The method of elimination is often faster and more efficient, especially for larger systems of equations.
  • To check if you have the correct solution, substitute the values of x and y back into both original equations.
  • You can use a calculator or graphing calculator to solve a system of linear equations.
  • To solve a system of linear equations with fractions or decimals, you can multiply both sides of the equation by a common denominator to eliminate the fractions or decimals.
  • To solve a system of linear equations with absolute values or square roots, you can use the method of substitution or the method of elimination, or you can use a matrix method.
  • To solve a system of linear equations with non-linear equations, you can use a matrix method or a numerical method.