Solving Systems of Equations By Elimination & Substitution With 2 Variables

Summary

This video tutorial covers two methods for solving systems of linear equations: Elimination and Substitution. It demonstrates step-by-step how to solve examples using both techniques, including the process of adding or manipulating equations in Elimination to cancel out variables, and substituting one equation into another in Substitution to isolate and solve for unknowns. The video also highlights how to manage different types of equations and gives multiple examples, helping viewers understand how to find the solution as an ordered pair. The approach is clear, concise, and suitable for learners seeking to grasp solving systems of equations.

Takeaways

• 😀 Elimination method involves adding or subtracting equations to cancel out one variable, simplifying the system to solve for the other variable.
• 😀 In the first example of elimination, adding the equations 2x + 3y = 8 and 5x - 3y = -1 cancels out the y terms, making it easy to solve for x.
• 😀 After solving for x in the first elimination example, substitute the value of x back into the first equation to find y.
• 😀 The solution to the first system using elimination is the ordered pair (1, 2).
• 😀 The elimination method can also be used with negative coefficients by multiplying one equation to align the coefficients of one variable for cancellation.
• 😀 In the second elimination example, multiplying the second equation by -2 cancels out the x terms when added to the first equation, allowing for easy solving of y.
• 😀 The solution to the second system using elimination is the ordered pair (2, 3).
• 😀 Substitution method involves solving one equation for one variable and substituting this into the other equation to solve for the remaining variable.
• 😀 In the first substitution example, substitute y = 5 - 2x into the second equation 4x + 3y = 13, simplify, and solve for x.
• 😀 After solving for x in the substitution example, substitute the value of x into the original equation to solve for y. The solution is (1, 3).
• 😀 The substitution method is particularly useful when one equation is already solved for one variable, making it easy to substitute and solve the system.

Q & A

What is the elimination method in solving systems of equations?

The elimination method, also known as the addition method, involves adding or subtracting two equations to eliminate one variable, making it easier to solve for the remaining variable.

How do you use the elimination method to solve the system of equations: 2x + 3y = 8 and 5x - 3y = -1?

To solve this system using elimination, add the two equations. The terms involving y (3y and -3y) cancel out, leaving 7x = 7. Then divide both sides by 7 to find x = 1. Substitute x = 1 into one of the original equations to find y = 2.

In the elimination method, how do you decide which variable to eliminate?

You generally choose to eliminate the variable that has smaller coefficients, as it is easier to work with. For example, if one variable has coefficients like 2 and -2, you can multiply the equations to cancel out that variable.

What is the second example of the elimination method, and how is it solved?

The second example is the system: 2x + 5y = 19 and x - 2y = -4. By multiplying the second equation by -2, you get -2x + 4y = 8. Adding this to the first equation gives 9y = 27, so y = 3. Substitute y = 3 into the first equation to find x = 2.

What is the substitution method in solving systems of equations?

The substitution method involves solving one of the equations for one variable and then substituting that expression into the other equation to solve for the second variable.

How is the system y = 5 - 2x and 4x + 3y = 13 solved using substitution?

In this example, substitute y = 5 - 2x into the second equation. This gives 4x + 3(5 - 2x) = 13. Simplify to get -2x + 15 = 13. Solving this results in x = 1. Substituting x = 1 into the first equation gives y = 3.

What are the steps to solve the system y = 3x + 2 and y = 7x - 6 using substitution?

Substitute 3x + 2 for y in the second equation. This gives 3x + 2 = 7x - 6. Simplify to get 8 = 4x, and then divide by 4 to find x = 2. Substituting x = 2 into the first equation gives y = 8.

What happens if you use substitution for the system 4x + 2y = 14 and 3x - 5y = -22?

You solve for y in the first equation: 2y = -4x + 14, so y = -2x + 7. Substitute this into the second equation to get 3x - 5(-2x + 7) = -22. Simplify and solve to find x = 1. Substituting x = 1 into the equation for y gives y = 5.

When is it better to use the elimination method over the substitution method?

The elimination method is often more efficient when the coefficients of one variable are opposites or easily manipulated to cancel each other out, while substitution is helpful when one equation is already solved for one variable.

How do you handle cases where neither elimination nor substitution immediately simplifies the system of equations?

If neither method immediately simplifies the system, you can manipulate the equations by multiplying one or both of them to align the coefficients of the variables for easier elimination or substitution. For example, multiplying one equation to make the coefficients of x or y opposites.